Chapter 4 Inverse Trigonometric Functions
Invertible Functions A one-one onto function is called an invertible function.
Inverse Of A Function Let f: X → Y be a one-one onto function. Then, for each y ∈ Y, there exists a unique element x ∈ X such that f(x) = y.
So, we define a new function, denoted by f-1, called the inverse of f, as f-1: Y → X: f-1(y) = x ⇔ f(x) = y.
Clearly, domain(f-1) = range(f) and range (f-1) = domain (f).
Trigonometric functions are, in general, not one-on-one.
Therefore, their inverses do not exist.
However, if we restrict their domains, we can make them one-on-one, enabling us to have their inverses.
Example The sine function restricted to any of the intervals [-π/2, π/2], [3π/2, 5π/2], etc., is one-one onto, and in each case, the range is [-1,1].
Therefore, we can define the inverse of the sine function, denoted by sin-1x, in each of these intervals. Corresponding to each such interval, we get a branch of sin-1x.
The branch [-π/2. π/2], is called the principal-value branch and the value belonging to it is called the principal value.
We denote the inverse of the sine function as sin-1x, called sine inverse x.
∴ sin-1: [-1,1] → \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and sin-1x = θ ⇔ x = sin θ.
Thus, sin-1x is a function whose domain is [-1,1] and range is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Note that sin-1x does not mean (sin x)-1.
Similarly, we can define the principal-value branches of the remaining five trigonometrical functions.
The following table shows the inverse trigonometric functions and their principal-value branches.
Read and Learn More Class 12 Math Solutions
Solved Examples
Example 1 Find the principal value of each of the following:
(1) \(\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
(2) \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
(3) tan-1(√3)
(4) cosec-1(2)
Solution
(1) We know that the range of the principal-value branch of sin-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Let \(\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\) = θ. Then,
Sin θ = \(\frac{1}{\sqrt{2}}=\sin \frac{\pi}{4} \Rightarrow \theta=\frac{\pi}{4} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \text {. }\)
Hence, the principal value of \(\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\) is \(\frac{\pi}{4}\).
(2) We know that the range of the principal-value branch of cos-1 is [0, π].
Let \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\) = θ. Then,
\(\cos \theta=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{6} \in[0, \pi] .\)Hence, the principal value of \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\) is \(\frac{\pi}{6}\).
(3) We know that the range of the principal-value branch of tan-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Let tan-1(√3) = θ. Then,
\(\tan \theta=\sqrt{3}=\tan \frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{3} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right) .\)Hence, the principal value of tan-1(√3) is \(\frac{\pi}{3}\).
(4) We know that the range of the principal-value branch of cosec-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) – {0}.
Let cosec-1(2) = θ. Then,
cosec θ = 2 = \({cosec} \frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\} .\)
Hence, the principal value of cosec-1(2) is \(\frac{\pi}{6}\).
Example 2 Find the principal value of each of the following:
(1) \(\sin ^{-1}\left(\frac{-1}{2}\right)\)
(2) \(\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)\)
(3) \(\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)\)
(4) tan-1(-)
(5) cosec-1(-2)
(6) tan-1(-√3)
Solution
(1) We know that the principal-value branch of sin-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Let \(\sin ^{-1}\left(\frac{-1}{2}\right)\) = θ. Then,
sin θ = \(-\frac{1}{2}=\sin \left(\frac{-\pi}{6}\right), \text { where } \frac{-\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \text {. }\)
Hence, the principal value of \(\sin ^{-1}\left(\frac{-1}{2}\right)\) is \(\frac{-\pi}{6} .\)
(2) We know that the principal-value branch of sin-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Let \(\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)\) = θ. Then,
\(\sin \theta=\frac{-\sqrt{3}}{2}=\sin \left(\frac{-\pi}{3}\right) \text {, where } \frac{-\pi}{3} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \text {. }\)Hence, the principal value of \(\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)\) is \(\frac{-\pi}{3}\).
(3) We know that the principal-value branch of tan-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Let \(\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)\) = θ. Then,
\(\tan \theta=\frac{-1}{\sqrt{3}}=\tan \left(\frac{-\pi}{6}\right) \text {, where } \frac{-\pi}{6} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right) \text {. }\)Hence, the principal value of \(\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)\) is \(\frac{-\pi}{6}\).
(4) We know that the principal-value branch of tan-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) – {0}.
Let tan-1(-1) = θ. Then,
tan θ = -1 = \(\tan \left(\frac{-\pi}{4}\right) \text {, where } \frac{-\pi}{4} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right).\)
Hence, the principal value of tan-1(-1) is \(\frac{-\pi}{4}\).
(5) We know that the principal-value branch of cosec-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) – {0}.
Let cosec-1(-2) = θ. Then,
cosec θ = -2 = \({cosec}\left(\frac{-\pi}{6}\right), \text { where } \frac{-\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\} \text {. }\)
Hence, the principal value of cosec-1(2) is \(\frac{-\pi}{6}\).
(6) We know that the principal-value branch of tan-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Let tan-1(-√3) = θ. Then,
tan θ = -√3 = \(\tan \left(\frac{-\pi}{3}\right) \text {, where } \frac{-\pi}{3} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right) \text {. }\)
Hence, the principal value of tan-1(-√3) is \(\frac{-\pi}{3}\).
Example 3 Find the principal value of each of the following:
(1) \(\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)\)
(2) \(\cos ^{-1}\left(\frac{-1}{2}\right)\)
(3) \(\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)\)
Solution
(1) We know that the range of the principal value of cos-1 is [0, π].
Let \(\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)\) = θ. Then,
\(\cos \theta=\frac{-1}{\sqrt{2}}=-\cos \frac{\pi}{4}=\cos \left(\pi-\frac{\pi}{4}\right)=\cos \frac{3 \pi}{4}\)∴ \(\theta=\frac{3 \pi}{4} \in[0, \pi]\)
Hence, the principal value of \(\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)\) is \(\frac{3 \pi}{4}\).
(2) We know that the range of the principal value of cos-1 is [0, π].
Let \(\cos ^{-1}\left(\frac{-1}{2}\right)\) = θ. Then,
\(\cos \theta=\frac{-1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3} .\)∴ \(\theta=\frac{2 \pi}{3} \in[0, \pi] .\)
Hence, the principal value of \(\cos ^{-1}\left(\frac{-1}{2}\right)\) is \(\frac{2 \pi}{3}\).
(3) We know that the range of the principal value of cot-1 is (0, π).
Let \(\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)\) = θ. Then,
\(\cot \theta=\frac{-1}{\sqrt{3}}=-\cot \frac{\pi}{3}=\cot \left(\pi-\frac{\pi}{3}\right)=\cot \frac{2 \pi}{3} .\)∴ \(\theta=\frac{2 \pi}{3} \in(0, \pi) .\)
Hence, the principal value of \(\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)\) is \(\frac{2 \pi}{3}\).
Example 4 Find the principal value of each of the following:
(1) cot-1(-√3)
(2) sec-1(-√2)
(3) cosec-1(-1)
Solution
(1) We know that the range of the principal value of cot-1 is (0, π).
Let cot-1(-√3) = θ. Then,
\(\cot \theta=-\sqrt{3}=-\cot \frac{\pi}{6}=\cot \left(\pi-\frac{\pi}{6}\right)=\cot \frac{5 \pi}{6} .\)∴ \(\theta=\frac{5 \pi}{6} \in(0, \pi) .\)
Hence, the principal value of cot-1(-√3) is \(\frac{5 \pi}{6}\).
(2) We know that the range of principal value of sec-1 is [0, π] \(– \left\{\frac{\pi}{2}\right\}\).
Let sec-1(-√2) = θ. Then,
\(\sec \theta=-\sqrt{2}=-\sec \frac{\pi}{4}=\sec \left(\pi-\frac{\pi}{4}\right)=\sec \frac{3 \pi}{4} .\)∴ \(\theta=\frac{3 \pi}{4} \in[0, \pi]-\left\{\frac{\pi}{2}\right\} .\)
Hence, the principal value of sec-1(-√2) is \(\frac{3 \pi}{4}\).
(3) We know that the range of principal value of cosec-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) – {0}.
Let cosec-1(-1) = θ. Then, cosec θ = -1.
cosec θ = -1 = \(-{cosec} \frac{\pi}{2}={cosec}\left(\frac{-\pi}{2}\right)\)
∴ \(\theta=\frac{-\pi}{2} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\} .\)
Hence, the principal value of cosec-1(-1) is \(\frac{-\pi}{2}\).
Example 5 Find the value of \(\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)\)
Solution:
Given:
\(\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)\)We know that the ranges of principal values of cos-1 and sin-1 are [0, π] and \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) respectively.
Let \(\cos ^{-1}\left(\frac{1}{2}\right)\) = θ1. Then,
\(\cos \theta_1=\frac{1}{2}=\cos \frac{\pi}{3} \Rightarrow \theta_1=\frac{\pi}{3} \in[0, \pi] .\)Let \(\sin ^{-1}\left(\frac{1}{2}\right)\) = θ2. Then,
\(\sin \theta_2=\frac{1}{2}=\sin \frac{\pi}{6} \Rightarrow \theta_2=\frac{\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)∴ \(\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}+\left(2 \times \frac{\pi}{6}\right)=\left(\frac{\pi}{3}+\frac{\pi}{3}\right)=\frac{2 \pi}{3} \text {. }\)
Example 6 Find the value of \(\tan ^{-1}(1)+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right) \text {. }\)
Solution
Given
\(\tan ^{-1}(1)+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right) \text {. }\)We know that the ranges of principal values of tan-1, cos-1 and sin-1 are \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), [0, π] and \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) respectively.
Let tan-1(1) = θ1. Then,
\(\tan \theta_1=1=\tan \frac{\pi}{4} \Rightarrow \theta_1=\frac{\pi}{4} \in[0, \pi]\)Let \(\cos ^{-1}\left(\frac{-1}{2}\right)\) = θ2. Then,
\(\cos \theta_2=\frac{-1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3} .\)∴ \(\theta_2=\frac{2 \pi}{3} \in[0, \pi] .\)
Let \(\sin ^{-1}\left(\frac{-1}{2}\right)\) = θ3. Then,
\(\sin \theta_3=\frac{-1}{2}=-\sin \frac{\pi}{6}=\sin \left(\frac{-\pi}{6}\right) \Rightarrow \theta_3=\frac{-\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] .\)∴ \(\tan ^{-1}(1)+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)=\left(\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6}\right)=\frac{3 \pi}{4}\)
Example 7 Find the value of tan-1√3 – sec-1(-2).
Solution
Given
tan-1√3 – sec-1(-2)
We know that the ranges of principal values of tan-1 and sec-1 are \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and [0, π] – \(\left\{\frac{\pi}{2}\right\}\) respectively.
Let tan-1√3 = θ1. Then,
\(\tan \theta_1=\sqrt{3}=\tan \frac{\pi}{3} \Rightarrow \theta_1=\frac{\pi}{3} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)Let sec-1(-2) = θ2. Then,
\(\sec \theta_2=-2=-\sec \frac{\pi}{3}=\sec \left(\pi-\frac{\pi}{3}\right)=\sec \frac{2 \pi}{3}\)∴ \(\theta_2=\frac{2 \pi}{3} \in[0, \pi]-\left\{\frac{\pi}{2}\right\} .\)
Hence, tan-1√3 – sec-1(-2) = \(\left(\frac{\pi}{3}-\frac{2 \pi}{3}\right)=\frac{-\pi}{3} .\)
Properties Of Inverse Functions
Theorem 1 Prove that:
(1) sin-1(sin x) = x, x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
(2) cos-1(cos x) = x, x ∈ [0, π]
(3) tan-1(tan x) = x, x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
(4) cosec-1(cosec x) = x, x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) – {0}
(5) sec-1(sec x) = x, x ∈ [0, π] – \(\left\{\frac{\pi}{2}\right\}\)
(6) cot-1(cot x) = x, x ∈ (0, π)
Proof
(1) Let sin x = y, where x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).
Then, sin-1y = x ⇒ sin-1(sin x) = x [∵ y = sin x]
∴ sin-1(sin x) = x.
(2) Let cos x = y, where x x ∈ [0, π].
Then, cos-1y = x ⇒ cos-1(cos x) = x [∵ y = cos x].
∴ cos-1(cos x) = x.
Similarly, the other results may be proved.
Theorem 2 Prove that:
(1) sin(sin-1x) = x, x ∈ [-1,1]
(2) cos(cos-1x) = x, x ∈ [-1,1]
(3) tan(tan-1x) = x, x ∈ R
(4) cosec(cosec-1x) = x, x ∈ R – [-1,1]
(5) sec(sec-1x) = x, x ∈ R – [-1,1]
(6) cot(cot-1x) = x, x ∈ R
Proof
(1) Let sin-1x = θ, where x ∈ [-1,1].
Then, sin θ = x ⇒ sin(sin-1x) = x [∵ θ = sin-1 x].
∴ sin(sin-1x) = x.
(2) Let cos-1x = θ, where x ∈ [-1,1]
Then, cos θ = x ⇒ cos(cos-1x) = x [∵ θ = cos-1 x].
∴ cos(cos-1x) = x.
Similarly, the other results may be proved.
Theorem 3 Prove that:
(1) \(\sin ^{-1} \frac{1}{x}={cosec}^{-1} x\), (x ≥ 1 or x ≤ -1)
(2) \(\cos ^{-1} \frac{1}{x}=\sec ^{-1} x\), (x ≥ 1 or x ≤ -1)
(3) \(\tan ^{-1} \frac{1}{x}=\cot ^{-1} x\), (x > 0)
Proof
(1) \(\sin ^{-1} \frac{1}{x}\) = θ ⇒ sin θ = \(\frac{1}{2}\)
⇒ cosec θ = x
⇒ θ = cosec-1x
⇒ \(\sin ^{-1} \frac{1}{x}={cosec}^{-1} x\).
Hence, \(\sin ^{-1} \frac{1}{x}={cosec}^{-1} x\).
(2) \(\cos ^{-1} \frac{1}{x} = θ ⇒ cos θ = \frac{1}{x}\)
⇒ sec θ = x
⇒ θ = sec-1x
⇒ \(\cos ^{-1} \frac{1}{x}=\sec ^{-1} x\).
Hence, \(\cos ^{-1} \frac{1}{x}=\sec ^{-1} x\).
(3) \(\tan ^{-1} \frac{1}{x}\) = θ ⇒ tan θ = \(\frac{1}{x}\)
⇒ cot θ = x
⇒ θ = cot-1x
⇒ \(\tan ^{-1} \frac{1}{x}=\cot ^{-1} x\).
Hence, \(\tan ^{-1} \frac{1}{x}=\cot ^{-1} x\).
Theorem 4 Prove that:
(1) sin-1(-x) = -sin-1x, x ∈ [-1,1]
(2) tan-1(-x) = -tan-1x, x ∈ R
(3) cosec-1(-x) = -cosec-1x, |x| ≥ 1
Proof
(1) Let sin-1(-x) = θ. Then,
sin-1(-x) = θ ⇒ -x = sin θ
⇒ x = -sin θ = sin(-θ)
⇒ -θ = sin-1x
⇒ θ = -sin-1x
⇒ sin-1(-x) = -sin-1x.
∴ sin-1(-x) = -sin-1x.
Similarly, the other results may be proved.
Theorem 5 Prove that:
(1) cos-1(-x) = π – cos-1x, x ∈ [-1,1]
(2) sec-1(-x) = π – sec-1x, |x| ≥ 1
(3) cot-1(-x) = π – cot-1x, x ∈ R
Proof
(1) Let cos-1(-x) = θ. Then,
cos-1(-x) = θ ⇒ -x = cos θ
⇒ x = -cos θ = cos(π – θ)
⇒ cos-1x = (π – θ) = π – cos-1(-x)
⇒ cos-1(-x) = π – cos-1x.
∴ cos-1(-x) = π – cos-1x.
Similarly, the other results may be proved.
Theorem 6 Prove that:
(1) \(\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}, x \in[-1,1]\)
(2) \(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}, x \in R\)
(3) \({cosec}^{-1} x+\sec ^{-1} x=\frac{\pi}{2}, \quad|x| \geq 1\)
Proof
(1) Let sin-1x = θ. Then,
sin-1x = θ ⇒ x = sin θ = \(\cos \left(\frac{\pi}{2}-\theta\right)\)
⇒ \(\cos ^{-1} x=\frac{\pi}{2}-\theta\)
⇒ \(\cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x\)
⇒ \(\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\).
∴ \(\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\).
Similarly, the other results may be proved.
Theorem 7 Prove that:
(1) \(\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \text {, if } x y<1\)
(2) \(\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right) \text {, if } x y>-1\)
(3) \(2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right), \text { if }|x|<1\)
Proof
(1) Let xy < 1, tan-1x = θ and tan-1y = Φ. Then,
tan θ = x and tan Φ = y
⇒ \(\tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}=\frac{x+y}{1-x y}\)
⇒ \(\theta+\phi=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\)
⇒ \(\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\)
∴ \(\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\).
(2) Let xy > -1.
On replacing y by -y in (1), we get
\(\tan ^{-1} x+\tan ^{-1}(-y)=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\)⇒ \(\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\).
(3) Let |x| < 1.
Replacing y by x in (1), we get
\(2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\)Theorem 8 Prove that:
(1) \(2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right),|x|<1\)
(2) \(2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right),|x| \geq 0\)
(3) \(2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right),|x|<1\)
Proof
Let tan-1x = θ. Then x = tan θ.
(1) \(\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\) [∵ x = tan θ]
= sin-1(sin 2θ)
= 2θ = 2 tan-1x.
∴ \(2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)
(2) \(\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\) [∵ x = tan θ]
= cos-1(cos 2θ)
= 2θ = 2tan-1x.
∴ \(2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)
(3) \(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)=\tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)\) [∵ x = tan θ]
= tan-1(tan 2θ)
= 2θ = 2 tan-1x.
∴ \(2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\).
Theorem 9 Prove that \(2 \sin ^{-1} x=\sin ^{-1}\left[2 x \sqrt{1-x^2}\right], \frac{-1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} .\)
Proof
Given
\(2 \sin ^{-1} x=\sin ^{-1}\left[2 x \sqrt{1-x^2}\right], \frac{-1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} .\)Let sin-1x = θ. Then, x = sin θ.
∴ sin 2θ = 2 sinθ cosθ
= \(2 \sin \theta \cdot \sqrt{\left(1-\sin ^2 \theta\right)}\)
= \(2 x \sqrt{1-x^2}\)
⇒ \(2 \theta=\sin ^{-1}\left\{2 x \sqrt{1-x^2}\right\}\)
⇒ \(2 \sin ^{-1} x=\sin ^{-1}\left\{2 x \sqrt{1-x^2}\right\} \text {. }\)
Hence, \(2 \sin ^{-1} x=\sin ^{-1}\left\{2 x \sqrt{1-x^2}\right\} \text {. }\)
Theorem 10 Prove that \(2 \cos ^{-1} x=\cos ^{-1}\left(2 x^2-1\right), \frac{1}{\sqrt{2}} \leq x \leq 1 .\)
Proof
Let cos-1x = θ. Then, x = cos θ.
∴ cos 2θ = (2cos2θ – 1) = (2x2 – 1)
⇒ 2θ = cos-1(2x2 – 1)
⇒ 2cos-1x = cos-1(2x2 – 1) [∵ θ = cos-1x].
Hence, 2cos-1x = cos-1(2x2 – 1).
Theorem 11 Prove that 3sin-1x = sin-1(3x – 4x3), x ∈ \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
Proof
Given
3sin-1x = sin-1(3x – 4x3)
Put sin-1x = θ. Then, x = sinθ.
Now, sin 3θ = (3 sin θ – 4 sin3θ) = (3x – 4x3)
⇒ 3θ = sin-1(3x – 4x3)
⇒ 3sin-1x = sin-1(3x – 4x3) [∵ θ = sin-1x].
Hence, 3sin-1x = sin-1(3x – 4x3).
Theorem 12 Prove that 3cos-1x = cos-1(4x3 – 3x), x ∈ [\(\frac{1}{2}\), 1].
Proof
3cos-1x = cos-1(4x3 – 3x)
Put cos-1x = θ. Then, x = cosθ.
Now, cos 3θ = (4cos3θ – 3 cosθ) = (4x3 – 3x)
⇒ 3θ = cos-1(4x3 – 3x)
⇒ 3cos-1x = cos-1(4x3 – 3x) [∵ θ = cos-1x]
Hence, 3 cos-1x = cos-1(4x3 – 3x).
Theorem 13 Prove that:
(1) \(\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\}\)
(2) \(\sin ^{-1} x-\sin ^{-1} y=\sin ^{-1}\left\{x \sqrt{1-y^2}-y \sqrt{1-x^2}\right\}\)
(3) \(\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left\{x y-\sqrt{\left(1-x^2\right)\left(1-y^2\right)}\right\}\)
(4) \(\cos ^{-1} x-\cos ^{-1} y=\cos ^{-1}\left\{x y+\sqrt{\left(1-x^2\right)\left(1-y^2\right)}\right\}\)
Proof
(1) Let sin-1x = θ1, and sin-1y = θ2. Then,
sinθ1 = x and sin θ2 = y
⇒ sin(θ1 + θ2) = sinθ1cosθ2 + cosθ1sinθ2
= \(\left\{\sin \theta_1 \cdot \sqrt{1-\sin ^2 \theta_2}\right\}+\left\{\sqrt{1-\sin ^2 \theta_1} \cdot \sin \theta_2\right\}\)
= \(\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\}\)
⇒ \(\left(\theta_1+\theta_2\right)=\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\}\)
⇒ \(\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\}\).
Hence, \(\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\}\).
The other results can be proved similarly.
Theorem 14 Prove that:
(1) \(\sin ^{-1} x=\cos ^{-1} \sqrt{1-x^2}=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)={cosec}^{-1}\left(\frac{1}{x}\right)\)
= \(\sec ^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)=\cot ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\)
(2) \(\cos ^{-1} x=\sin ^{-1} \sqrt{1-x^2}=\tan ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\)
= \({cosec}^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)=\sec ^{-1}\left(\frac{1}{x}\right)=\cot ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\)
(3) \(\tan ^{-1} x=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)=\cos ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\)
= \({cosec}^{-1}\left(\frac{\sqrt{1+x^2}}{x}\right)=\sec ^{-1}\left(\sqrt{1+x^2}\right)=\cot ^{-1}\left(\frac{1}{x}\right)\)
Proof
(1) Let sin-1x = θ. Then, sin θ = x.
∴ \(\cos \theta=\sqrt{1-x^2}, \tan \theta=\frac{x}{\sqrt{1-x^2}} \cdot {cosec} \theta=\frac{1}{x}\)
\(\sec \theta=\frac{1}{\sqrt{1-x^2}} \text { and } \cot \theta=\frac{\sqrt{1-x^2}}{x} \text {. }\)∴ \(\theta=\cos ^{-1} \sqrt{1-x^2}, \theta=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right), \theta={cosec}^{-1} \frac{1}{x}\)
\(\theta=\sec ^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right) \text { and } \theta=\cot ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) \text {. }\)Hence, \(\sin ^{-1} x=\cos ^{-1} \sqrt{1-x^2}=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\)
= \({cosec}^{-1} \frac{1}{x}=\sec ^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)\)
= \(\cot ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) \text {. }\)
Similarly, the other results can be proved.
Solved Examples
Example 1 Find the value of:
(1) \(\sin ^{-1}\left(\sin \frac{\pi}{3}\right)\)
(2) \(\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)\)
(3) \(\tan ^{-1}\left(\tan \frac{\pi}{4}\right)\)
Solution
We have
(1) \(\sin ^{-1}\left(\sin \frac{\pi}{3}\right)=\frac{\pi}{3}, \text { since } \frac{\pi}{3} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \text {. }\)
(2) \(\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)=\frac{2 \pi}{3}, \text { since } \frac{2 \pi}{3} \in[0, \pi] \text {. }\)
(3) \(\tan ^{-1}\left(\tan \frac{\pi}{4}\right)=\frac{\pi}{4}, \text { since } \frac{\pi}{4} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
Example 2 Find the value of:
(1) \(\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)\)
(2) \(\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)\)
(3) \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)\)
Solution
(1) We know that the principal-value branch of sin-1 is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).
∴ \(\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right) \neq \frac{2 \pi}{3}\)
Now, \(\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)=\sin ^{-1}\left\{\sin \left(\pi-\frac{\pi}{3}\right)\right\}\)
= \(\sin ^{-1}\left(\sin \frac{\pi}{3}\right)\) [∵ sin(π – θ) = sinθ]
= \(\frac{\pi}{3}\) {∵ \(\frac{\pi}{3} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)}
Hence, \(\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)\) = \(\frac{\pi}{3}\).
(2) We know that the principal-value branch of cos-1 is [0, π].
∴ \(\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right) \neq \frac{7 \pi}{6}\)
Now, \(\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=\cos ^{-1}\left\{\cos \left(2 \pi-\frac{5 \pi}{6}\right)\right\}\)
= \(\cos ^{-1}\left\{\cos \frac{5 \pi}{6}\right\}\) {∵ cos(2π – θ) = cosθ}
= \(\frac{5 \pi}{6}\).
Hence, \(\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)\) = \(\frac{5 \pi}{6}\).
(3) We know that the principal-value branch of tan-1 is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).
∴ \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right) \neq \frac{3 \pi}{4}\)
Now, \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)=\tan ^{-1}\left\{\tan \left(\pi-\frac{\pi}{4}\right)\right\}\)
= \(\tan ^{-1}\left\{-\tan \frac{\pi}{4}\right\}\) [∵ \(\tan \left(\pi-\frac{\pi}{4}\right)=-\tan \frac{\pi}{4}\)]
= \(\tan _{-\pi}^{-1}\left\{\tan \left(\frac{-\pi}{4}\right)\right\}\) [∵ -tanθ = tan(-θ)]
= \(\frac{-\pi}{4}\).
Hence, \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)\) = \(\frac{-\pi}{4}\).
Example 3 Find the value of:
(1) \(\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)\)
(2) \(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)\)
(3) \(\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)\)
Solution
(1) We know that the principal-value branch of sin-1 is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).
∴ \(\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right) \neq \frac{3 \pi}{5}\)
Now, \(\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\sin ^{-1}\left\{\sin \left(\pi-\frac{2 \pi}{5}\right)\right\}\).
= \(\sin ^{-1}\left(\sin \frac{2 \pi}{5}\right)\) [∵ sin(π – θ) = sinθ]
= \(\frac{2 \pi}{5}\) {∵ \(\frac{2 \pi}{5} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)}.
Hence, \(\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\frac{2 \pi}{5}\)
(2) We know that the principal-value branch of cos-1 is [0, π].
∴ \(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) \neq \frac{13 \pi}{6} .\)
Now, \(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left\{\cos \left(2 \pi+\frac{\pi}{6}\right)\right\}\)
= \(\cos ^{-1}\left\{\cos \frac{\pi}{6}\right\}\) [∵ cos(2π + θ) = cos θ]
= \(\frac{\pi}{6}\).
Hence, \(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\frac{\pi}{6}\)
(3) We know that the principal-value branch of tan-1 is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).
∴ \(\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right) \neq \frac{7 \pi}{6}\)
Now, \(\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=\tan ^{-1}\left\{\tan \left(\pi+\frac{\pi}{6}\right)\right\}\)
= \(\tan ^{-1}\left(\tan \frac{\pi}{6}\right)\) [∵ tan(π + θ) = tan θ]
= \(\frac{\pi}{6}\).
Hence, \(\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=\frac{\pi}{6}\)
Example 4 (1) Show that \(\sin ^{-1}\left\{\sin \frac{3 \pi}{4}\right\} \neq \frac{3 \pi}{4}\) and find its value.
(2) Show that \(\cos ^{-1}\left\{\cos \left(\frac{-\pi}{3}\right)\right\}=\frac{-\pi}{3}\) and find its value.
(3) Show that \(\tan ^{-1}\left\{\tan \frac{5 \pi}{6}\right\} \neq \frac{5 \pi}{6}\) and find its value.
Solution
(1) We know that the principal-value branch of sin-1 is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).
∴ \(\sin ^{-1}\left\{\sin \frac{3 \pi}{4}\right\} \neq \frac{3 \pi}{4}\)
Now, \(\sin ^{-1}\left\{\sin \frac{3 \pi}{4}\right\}=\sin ^{-1}\left\{\sin \left(\pi-\frac{\pi}{4}\right)\right\}\)
= \(\sin ^{-1}\left\{\sin \frac{\pi}{4}\right\} [∵ \sin \left(\pi-\frac{\pi}{4}\right)=\sin \frac{\pi}{4}]\)
= \(\frac{\pi}{4}\) {∵ \(\frac{\pi}{4} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)}.
∴ \(\sin ^{-1}\left\{\sin \frac{3 \pi}{4}\right\}=\frac{\pi}{4}\).
(2) We know that the principal-value branch of cos-1 is [0, π].
∴ \(\cos ^{-1}\left\{\cos \left(\frac{-\pi}{3}\right)\right\}=\frac{-\pi}{3}\)
Now, \(\cos ^{-1}\left\{\cos \left(\frac{-\pi}{3}\right)\right\}=\cos ^{-1}\left\{\cos \frac{\pi}{3}\right\}\). [∵ cos(-θ) = cos θ]
= \(\frac{\pi}{3}\) [∵ \(\frac{\pi}{3} \in[0, \pi]\)].
∴ \(\cos ^{-1}\left\{\cos \left(\frac{-\pi}{3}\right)\right\}=\frac{\pi}{3}\)
(3) We know that the principal-value branch of tan-1 is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).
∴ \(\tan ^{-1}\left\{\tan \frac{5 \pi}{6}\right\} \neq \frac{5 \pi}{6}\).
Now, \(\tan ^{-1}\left\{\tan \frac{5 \pi}{6}\right\}=\tan ^{-1}\left\{\tan \left(\pi-\frac{\pi}{6}\right)\right\}\)
= \(\tan ^{-1}\left\{\tan \left(\frac{-\pi}{6}\right)\right\}\) [∵ tan(π – θ) = tan(-θ)]
= \(\frac{-\pi}{6}\) [∵ \(\frac{-\pi}{6} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)].
∴ \(\tan ^{-1}\left\{\tan \frac{5 \pi}{6}\right\}=\frac{-\pi}{6}\).
Example 5 Evaluate:
(1) \(\sin \left(\cos ^{-1} \frac{3}{5}\right)\)
(2) \(\cos \left(\tan ^{-1} \frac{3}{4}\right)\)
Solution
(1) Let cos-1\(\frac{3}{5}\) = θ, where θ ∈ [0, π].
Then, cos θ = \(\frac{3}{5}\).
Since θ ∈ [0, π], we have sin θ > 0.
∴ \(\sin \theta=\sqrt{1-\cos ^2 \theta}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5} \text {. }\)
Hence, \(\sin \left(\cos ^{-1} \frac{3}{5}\right)=\sin \theta=\frac{4}{5} .\)
(2) Let tan-1\(\frac{3}{4}\) = θ, where θ ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).
Then, tan θ = \(\frac{3}{4}\).
Since θ ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\), so cosθ > 0.
∴ \(\cos \theta=\frac{1}{\sec \theta}=\frac{1}{\sqrt{1+\tan ^2 \theta}}=\frac{1}{\sqrt{1+9 / 16}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\).
Hence, \(\cos \left(\tan ^{-1} \frac{3}{4}\right)=\cos \theta=\frac{4}{5} \text {. }\)
Example 6 Evaluate:
(1) \(\sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right\}\)
(2) \(\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)\)
(3) sin(cot-1x)
(4) \(\tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)\)
Solution
(1) We know that sin-1(-θ) = -sin-1θ.
∴ \(\sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right\}=\sin \left\{\frac{\pi}{3}+\sin ^{-1} \frac{1}{2}\right\}\)
= \(\sin \left(\frac{\pi}{3}+\frac{\pi}{6}\right)\) [∵ \(\sin ^{-1} \frac{1}{2}=\frac{\pi}{6}\)]
= sin\(\frac{\pi}{2}\) = 1.
(2) Let cos-1 \(\frac{4}{5}\) = θ, where θ ∈ [0, π].
Then, cosθ = \(\frac{4}{5}\).
Since θ ∈ [0, π] ⇒ \(\frac{1}{2}\)θ ∈ \(\left[0, \frac{\pi}{2}\right]\) ⇒ sin\(\frac{1}{2}\)θ > 0.
∴ \(\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)=\sin \frac{1}{2} \theta=\sqrt{\frac{1-\cos \theta}{2}}=\sqrt{\frac{1-(4 / 5)}{2}}=\frac{1}{\sqrt{10}}\)
(3) Let cot-1x = θ. Then, θ ∈ [0, π].
∴ sin(cot-1x) = sinθ > 0.
Now, \(\sin \theta=\frac{1}{{cosec} \theta}=\frac{1}{\sqrt{1+\cot ^2 \theta}}=\frac{1}{\sqrt{1+x^2}}\).
∴ \(\sin \left(\cot ^{-1} x\right)=\sin \theta=\frac{1}{\sqrt{1+x^2}} .\)
(4) Let cos-1\(\frac{\sqrt{5}}{3}\) = θ. Then, cos θ = \(\frac{\sqrt{5}}{3}\), where θ ∈ [0, π].
∴ \(\tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)=\tan \frac{1}{2} \theta=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\sqrt{\frac{(1-\sqrt{5} / 3)}{(1+\sqrt{5} / 3)}}\)
= \(\sqrt{\frac{(3-\sqrt{5})}{(3+\sqrt{5})} \times \frac{(3-\sqrt{5})}{(3-\sqrt{5})}}=\frac{(3-\sqrt{5})}{2}\).
Example 7 Evaluate \(\sin \left[2 \cos ^{-1}\left(\frac{-3}{5}\right)\right]\)
Solution
Given:
\(\sin \left[2 \cos ^{-1}\left(\frac{-3}{5}\right)\right]\)Let cos-1(\(\frac{3}{5}\)) = θ, where θ ∈ [0, π].
Then, cos θ = –\(\frac{3}{5}\).
since θ ∈ [0, π], we have sin θ > 0.
∴ \(\sin \theta=\sqrt{1-\cos ^2 \theta}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5} .\)
∴ \(\sin \left[2 \cos ^{-1}\left(\frac{-3}{5}\right)\right]=\sin 2 \theta\)
= \(2 \sin \theta \cos \theta=\left\{2 \times \frac{4}{5} \times\left(\frac{-3}{5}\right)\right\}=\frac{-24}{25}\).
\(\sin \left[2 \cos ^{-1}\left(\frac{-3}{5}\right)\right]\) = \(2 \sin \theta \cos \theta=\left\{2 \times \frac{4}{5} \times\left(\frac{-3}{5}\right)\right\}=\frac{-24}{25}\).
Example 8 Evaluate \(\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)\)
Solution
Given
\(\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)\)Let sin-1\(\frac{3}{5}\) = A and sin-1\(\frac{5}{13}\) = B. Then,
A, B ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) ⇒ cos A > 0 and cos B > 0.
∴ sin A = \(\frac{3}{5}\) and sin B = \(\frac{5}{13}\)
⇒ \(\cos A=\sqrt{1-\sin ^2 A}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)
and \(\cos B=\sqrt{1-\sin ^2 B}=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13} \text {. }\)
∴ \(\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)=\cos (A+B)\)
= cos A cos B – sin A sin B
= \(\left(\frac{4}{5} \times \frac{12}{13}\right)-\left(\frac{3}{5} \times \frac{5}{13}\right)\)
= \(\left(\frac{48}{65}-\frac{15}{65}\right)=\frac{33}{65}\).
\(\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)\) = \(\left(\frac{48}{65}-\frac{15}{65}\right)=\frac{33}{65}\).
Example 9 Find the value of \(\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}\).
Solution
We have
\(\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}=\tan ^{-1}\left\{2 \cos \left(2 \times \frac{\pi}{6}\right)\right\}\) [∵ sin-1\(\frac{1}{2}\) = \(\frac{\pi}{6}\)]
= \(\tan ^{-1}\left\{2 \cos \frac{\pi}{3}\right\}\)
= \(\tan ^{-1}\left(2 \times \frac{1}{2}\right)=\tan ^{-1} 1=\frac{\pi}{4}\).
The value of \(\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}\) = \(\tan ^{-1}\left(2 \times \frac{1}{2}\right)=\tan ^{-1} 1=\frac{\pi}{4}\).
Example 10 If tan-1\(\frac{4}{3}\) = θ, find the value of cos θ.
Solution
tan-1\(\frac{4}{3}\) = θ, where θ ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).
∴ tan θ = \(\frac{4}{3}\).
We know that cos θ > 0 where θ ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).
∴ \(\cos \theta=\frac{1}{\sec \theta}=\frac{1}{\sqrt{1+\tan ^2 \theta}}=\frac{1}{\sqrt{1+\frac{16}{9}}}=\frac{3}{5} .\)
The value of \(\cos \theta=\frac{1}{\sec \theta}=\frac{1}{\sqrt{1+\tan ^2 \theta}}=\frac{1}{\sqrt{1+\frac{16}{9}}}=\frac{3}{5} .[/latex
Example 11 If [latex]\cot ^{-1}\left(-\frac{1}{5}\right)=\theta\), find the value of sin θ.
Solution
Given: \(\cot ^{-1}\left(-\frac{1}{5}\right)=\theta\), where θ ∈ (0, π).
∴ cot θ = –\(\frac{1}{5}\).
Sin θ > 0 in (0, π).
\(\sin \theta=\frac{1}{{cosec} \theta}=\frac{1}{\sqrt{1+\cot ^2 \theta}}=\frac{1}{\sqrt{1+\frac{1}{25}}}=\frac{5}{\sqrt{26}} .\)The value of \(\sin \theta=\frac{1}{{cosec} \theta}=\frac{1}{\sqrt{1+\cot ^2 \theta}}=\frac{1}{\sqrt{1+\frac{1}{25}}}=\frac{5}{\sqrt{26}} .\)
Example 12 Prove that \(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}=\tan ^{-1} \frac{2}{9} \text {. }\)
Solution
We know that tan-1 x + tan-1 y = \(\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\), xy < 1.
Let x = \(\frac{1}{7}\) and y = \(\frac{1}{13}\). Then, xy = \(\frac{1}{91}\) < 1.
∴ \(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}=\tan ^{-1}\left\{\frac{\left(\frac{1}{7}+\frac{1}{13}\right)}{1-\frac{1}{7} \times \frac{1}{13}}\right\}=\tan ^{-1}\left(\frac{20}{90}\right)=\tan ^{-1} \frac{2}{9}\)
Example 13 Prove that tan-1\(\frac{3}{4}\) + tan-1\(\frac{3}{5}\) – tan-1\(\frac{8}{19}\) = \(\frac{\pi}{4}\).
Solution
We have
LHS = \(\left\{\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{3}{5}\right\}-\tan ^{-1} \frac{8}{19}\)
= \(\tan ^{-1}\left\{\frac{\left(\frac{3}{4}+\frac{3}{5}\right)}{\left(1-\frac{3}{4} \times \frac{3}{5}\right)}\right\}-\tan ^{-1} \frac{8}{19}\)
= \(\tan ^{-1}\left(\frac{27}{11}\right)-\tan ^{-1} \frac{8}{19}\)
= \(\tan ^{-1} \frac{\left(\frac{27}{11}-\frac{8}{19}\right)}{\left(1+\frac{27}{11} \times \frac{8}{19}\right)}=\tan ^{-1}\left(\frac{425}{425}\right)=\tan ^{-1} 1=\frac{\pi}{4}\) = RHS.
∴ LHS = RHS.
Example 14 Prove that tan-1\(\frac{1}{3}\) + tan-1\(\frac{1}{5}\) + tan-1\(\frac{1}{7}\) + tan-1\(\frac{1}{8}\) = \(\frac{\pi}{4}\).
Solution
We have
LHS = \(\left(\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}\right)+\left(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}\right)\)
= \(\tan ^{-1} \frac{\left(\frac{1}{3}+\frac{1}{5}\right)}{\left(1-\frac{1}{3} \times \frac{1}{5}\right)}+\tan ^{-1} \frac{\left(\frac{1}{7}+\frac{1}{8}\right)}{\left(1-\frac{1}{7} \times \frac{1}{8}\right)}=\tan ^{-1} \frac{(9 / 5)}{(14 / 5)}+\tan ^{-1} \frac{(15 / 56)}{(55 / 56)}\)
= \(\tan ^{-1} \frac{8}{14}+\tan ^{-1} \frac{15}{55}=\tan ^{-1} \frac{4}{7}+\tan ^{-1} \frac{3}{11}\)
= \(\tan ^{-1} \frac{\left(\frac{4}{7}+\frac{3}{11}\right)}{\left(1-\frac{4}{7} \times \frac{3}{11}\right)}=\tan ^{-1} \frac{(65 / 77)}{(65 / 77)}=\tan ^{-1} 1=\frac{\pi}{4}\) = RHS.
∴ LHS = RHS.
Example 15 Prove that 2tan-1\(\frac{1}{2}\) + tan-1\(\frac{1}{7}\) = tan-1\(\frac{31}{17}\).
Solution
We have
LHS = 2tan-1\(\frac{1}{2}\) + tan-1\(\frac{1}{7}\)
= \(\tan ^{-1} \frac{\left(2 \times \frac{1}{2}\right)}{\left\{1-\left(\frac{1}{2}\right)^2\right\}}+\tan ^{-1} \frac{1}{7}\) [∵ \(2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\)]
= \(\tan ^{-1} \frac{1}{(3 / 4)}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{7}\)
= \(\tan ^{-1} \frac{\left(\frac{4}{3}+\frac{1}{7}\right)}{\left(1-\frac{4}{3} \times \frac{1}{7}\right)}=\tan ^{-1} \frac{(31 / 21)}{(17 / 21)}=\tan ^{-1} \frac{31}{17}\) = RHS.
∴ LHS = RHS.
Example 16 Prove that \(2\left(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}\right)=\tan ^{-1} \frac{4}{3} \text {. }\)
Solution
We have
LHS = 2tan-1\(\frac{1}{4}\) + 2tan-1\(\frac{2}{9}\)
= \(\tan ^{-1} \frac{\left(2 \times \frac{1}{4}\right)}{\left\{1-\left(\frac{1}{4}\right)^2\right\}}+\tan ^{-1} \frac{\left(2 \times \frac{2}{9}\right)}{\left\{1-\left(\frac{2}{9}\right)^2\right\}}\)
= \(\tan ^{-1} \frac{(1 / 2)}{(15 / 16)}+\tan ^{-1} \frac{(4 / 9)}{(7 / 81)}=\tan ^{-1}\left(\frac{1}{2} \times \frac{16}{15}\right)+\tan ^{-1}\left(\frac{4}{9} \times \frac{81}{77}\right)\)
= \(\tan ^{-1} \frac{8}{15}+\tan ^{-1} \frac{36}{77}\)
= \(\tan ^{-1} \frac{\left(\frac{8}{15}+\frac{36}{77}\right)}{\left(1-\frac{8}{15} \times \frac{36}{77}\right)}=\tan ^{-1} \frac{(616+540)}{(1155-288)}\)
= \(\tan ^{-1}\left(\frac{1156}{867}\right)=\tan ^{-1} \frac{4}{3}\) = RHS.
∴ LHS = RHS.
Example 17 Prove that cot-17 + cot-18 + cot-118 = cot-13.
Solution
We have
LHS = cot-17 + cot-18 + cot-118
= \(\left(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}\right)+\tan ^{-1} \frac{1}{18}\)
= \(\tan ^{-1} \frac{\left(\frac{1}{7}+\frac{1}{8}\right)}{\left(1-\frac{1}{7} \times \frac{1}{8}\right)}+\tan ^{-1} \frac{1}{18}=\tan ^{-1} \frac{(15 / 56)}{(55 / 56)}+\tan ^{-1} \frac{1}{18}\)
= \(\left(\tan ^{-1} \frac{3}{11}+\tan ^{-1} \frac{1}{18}\right)=\tan ^{-1} \frac{\left(\frac{3}{11}+\frac{1}{18}\right)}{\left(1-\frac{3}{11} \times \frac{1}{18}\right)}\)
= \(\tan ^{-1} \frac{(65 / 198)}{(195 / 198)}=\tan ^{-1} \frac{1}{3}=\cot ^{-1} 3\) = RHS.
∴ LHS = RHS.
Example 18 Prove that sin-1\(\frac{3}{5}\) – sin-1\(\frac{8}{17}\) = cos-1\(\frac{84}{85}\).
Solution
Let sin-1\(\frac{3}{5}\) = x and sin-1\(\frac{8}{17}\) = y. Then,
sin x = \(\frac{3}{5}\) and sin y = \(\frac{8}{17}\).
∴ \(\cos x=\sqrt{1-\sin ^2 x}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)
and \(\cos y=\sqrt{1-\sin ^2 y}=\sqrt{1-\frac{64}{289}}=\sqrt{\frac{225}{289}}=\frac{15}{17} \text {. }\)
∴ cos(x-y) = cos x cos y + sin x sin y
= \(\left(\frac{4}{5} \times \frac{15}{17}\right)+\left(\frac{3}{5} \times \frac{8}{17}\right)=\left(\frac{12}{17}+\frac{24}{85}\right)=\frac{84}{85}\)
⇒ \(x-y=\cos ^{-1}\left(\frac{84}{85}\right) ⇒ \sin ^{-1} \frac{3}{5}-\sin ^{-1} \frac{8}{17}=\cos ^{-1} \frac{84}{85} \text {. }\)
Example 19 Prove that \(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}\)
Solution
LHS = \(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}, where xy = \left(\frac{1}{4} \times \frac{2}{9}\right)=\frac{1}{18}<1\)
= \(\tan ^{-1}\left\{\frac{\left(\frac{1}{4}+\frac{2}{9}\right)}{\left(1-\frac{1}{4} \times \frac{2}{9}\right)}\right\}=\tan ^{-1}\left(\frac{17}{34}\right)=\tan ^{-1} \frac{1}{2}\)
Now, let \(\frac{1}{2}cos-1\frac{3}{5}\) = θ. Then, cos 2θ = \(\frac{3}{5}\).
∴ \(\tan \theta=\sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}=\sqrt{\frac{\left(1-\frac{3}{5}\right)}{\left(1+\frac{3}{5}\right)}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)
⇒ \(\tan ^{-1} \frac{1}{2}=\theta\)
⇒ \(\tan ^{-1} \frac{1}{2}=\frac{1}{2} \cos ^{-1} \frac{3}{5}\).
Hence, \(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}\).
Example 20 Prove that \(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{3}{5}=\tan ^{-1} \frac{27}{11} \text {. }\)
Solution
Let cos-1\(\frac{4}{5}\) = θ. Then cos θ = \(\frac{4}{5}\).
∴ \(\tan \theta=\frac{\sqrt{1-\cos ^2 \theta}}{\cos \theta}=\frac{\sqrt{1-\frac{16}{25}}}{(4 / 5)}=\left(\frac{3}{5} \times \frac{5}{4}\right)=\frac{3}{4}\)
⇒ θ = tan-1\(\frac{3}{4}\).
∴ \(\cos ^{-1} \frac{4}{5}=\tan ^{-1} \frac{3}{4} \text {. }\)
∴ \(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{3}{5}\)
= \(\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{3}{5}}{1-\frac{3}{4} \times \frac{3}{5}}\right)=\tan ^{-1} \frac{27}{11} \text {. }\)
Example 21 Solve tan-12x + tan-13x = \(\frac{\pi}{4}\).
Solution
tan-12x + tan-13x = \(\frac{\pi}{4}\)
⇒ \(\tan ^{-1}\left(\frac{2 x+3 x}{1-6 x^2}\right)=\frac{\pi}{4} ⇒ \tan ^{-1}\left(\frac{5 x}{1-6 x^2}\right)=\frac{\pi}{4}\)
⇒ \(\frac{5 x}{\left(1-6 x^2\right)}=\tan \frac{\pi}{4}=1\) ⇒ 1 – 6x2 = 5x
⇒ 6x2 + 5x – 1 = 0 ⇒ 6x2 + 6x – x – 1 = 0
⇒ 6x(x+1) – (x+1) = 0 ⇒ (x+1)(6x-1) = 0
⇒ x = -1 or x = \(\frac{1}{6}\).
⇒ x = \(\frac{1}{6}\) [∵ x = -1 makes LHS negative].
Example 22 Solve \(\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}\)
Solution
We have
\(\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}\)⇒ \(\tan ^{-1} \frac{\left\{\frac{(x-1)}{(x-2)}+\frac{(x+1)}{(x+2)}\right\}}{\left\{1-\frac{(x-1)}{(x-2)} \cdot \frac{(x+1)}{(x+2)}\right\}}=\frac{\pi}{4}\)
⇒ \(\tan ^{-1}\left\{\frac{(x-1)(x+2)+(x+1)(x-2)}{\left(x^2-4\right)-\left(x^2-1\right)}\right\}=\frac{\pi}{4}\)
⇒ \(\tan ^{-1}\left\{\frac{\left(x^2+x-2\right)+\left(x^2-x-2\right)}{-3}\right\}=\frac{\pi}{4}\)
⇒ \(\frac{2 x^2-4}{-3}=\tan \frac{\pi}{4}=1\) ⇒ 2x2 – 4 = -3
⇒ 2x2 = 1 ⇒ x2 = \(\frac{1}{2}\) ⇒ x = \(\pm \frac{1}{\sqrt{2}}\)
Example 23 Solve tan-1(x+1) + tan-1(x-1) = tan-1\(\frac{8}{31}\).
Solution
We have
tan-1(x+1) + tan-1(x-1) = tan-1\(\frac{8}{31}\)
⇒ \(\tan ^{-1}\left\{\frac{(x+1)+(x-1)}{1-(x+1)(x-1)}\right\}=\tan ^{-1} \frac{8}{31}\)
⇒ \(\tan ^{-1}\left(\frac{2 x}{2-x^2}\right)=\tan ^{-1} \frac{8}{31}\)
⇒ \(\tan \left\{\tan ^{-1}\left(\frac{2 x}{2-x^2}\right)\right\}=\frac{8}{31}\) ⇒ \(\frac{2 x}{2-x^2}=\frac{8}{31}\)
⇒ 8x2 + 62x – 16 = 0 ⇒ 4x2 + 31x – 8 = 0
⇒ (4x – 1)(x + 8) = 0 ⇒ x = \(\frac{1}{4}\) or x = -8.
But, x = -8 gives LHS = tan-1(-7) + tan-1(-9), which is negative, while RHS is positive. So, x = -8 is not possible.
Hence, x = \(\frac{1}{4}\).
Example 24 Solve \(\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x,(x>0)\)
Solution
We have
\(\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x,(x>0)\)⇒ \(\tan ^{-1} 1-\tan ^{-1} x=\frac{1}{2} \tan ^{-1} x\)
⇒ \(\frac{3}{2} \tan ^{-1} x=\tan ^{-1} 1=\frac{\pi}{4}\)
⇒ \(\tan ^{-1} x=\left(\frac{\pi}{4} \times \frac{2}{3}\right)=\frac{\pi}{6}\) ⇒ \(x=\tan \left(\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}}.
Hence, x = [latex]\frac{1}{\sqrt{3}}\).
Example 25 Solve 2 tan-1(cos x) = tan-1(2 cosec x).
Solution
We have
2 tan-1(cos x) = tan-1(2 cosec x)
⇒ \(\tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^2 x}\right)=\tan ^{-1}(2 {cosec} x)\)
⇒ \(\tan \left[\tan ^{-1}\left(\frac{2 \cos ^x x}{\sin ^2 x}\right)\right]=2 {cosec} x\)
⇒ \(\frac{2 \cos x}{\sin ^2 x}=2 {cosec} x\) ⇒ cos x = sin x
⇒ tan x = 1 ⇒ x = \(\frac{\pi}{4}\).
Problems Based On Trigonometric Formulae
Example 26 Prove that \(\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)=\frac{x}{2}, x<\pi\)
Solution
We have
LHS = \(\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)=\tan ^{-1} \sqrt{\frac{2 \sin ^2(x / 2)}{2 \cos ^2(x / 2)}}\)
= \(\tan ^{-1}\left(\tan \frac{x}{2}\right)=\frac{x}{2}=\text { RHS. }\)
∴ \(\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)=\frac{x}{2} \text {. }\)
Example 27 Prove that \(\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\left(\frac{\pi}{4}-x\right), x<\pi \text {. }\)
Solution
We have
LHS = \(\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\)
= \(\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)\) [dividing num. and denom. by cos x]
= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-x\right)\right\}=\left(\frac{\pi}{4}-x\right)=\text { RHS }\).
∴ \(\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\left(\frac{\pi}{4}-x\right)\)
Example 28 Prove that \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)=\left(\frac{\pi}{4}-\frac{x}{2}\right) \text {. }\)
Solution
We have
LHS = \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)\)
= \(\tan ^{-1}\left\{\frac{\sin \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right)}\right\}\)
= \(\tan ^{-1}\left\{\frac{2 \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right\}\)
= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right\}=\left(\frac{\pi}{4}-\frac{x}{2}\right)=\text { RHS }\)
∴ \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)=\left(\frac{\pi}{4}-\frac{x}{2}\right) \text {. }\)
Example 29 Prove that \(\tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right)=\left(\frac{\pi}{4}+\frac{x}{2}\right), x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
Solution
We have
LHS = \(\tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right)\)
= \(\tan ^{-1}\left\{\frac{\sin \left(\frac{\pi}{2}-x\right)}{1-\cos \left(\frac{\pi}{2}-x\right)}\right\}\)
= \(\tan ^{-1}\left\{\frac{2 \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right\}\)
= \(\tan ^{-1}\left\{\cot \left(\frac{\pi}{4}-\frac{x}{2}\right)\right\}=\tan ^{-1}\left[\tan \left\{\frac{\pi}{2}-\left(\frac{\pi}{4}-\frac{x}{2}\right)\right]\right\}\)
= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}\)
= \(\left(\frac{\pi}{4}+\frac{x}{2}\right)\) = RHS.
Hence, \(\tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right)=\left(\frac{\pi}{4}+\frac{x}{2}\right)\).
Example 30 Prove that \(\tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right)=3 \tan ^{-1} \frac{x}{a}\).
Solution
Putting x = atan θ, we get
\(\tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right)=\tan ^{-1}\left(\frac{3 a^3 \tan \theta-a^3 \tan ^3 \theta}{a^3-3 a^3 \tan ^2 \theta}\right)\)= \(\tan ^{-1}\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)=\tan ^{-1}(\tan 3 \theta)\)
= \(3 \theta=3 \tan ^{-1} \frac{x}{a}\)
Hence, \(\tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right)=3 \tan ^{-1} \frac{x}{a}\)
Example 31 Prove that \(\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x-a \sin x}\right)=\tan ^{-1}\left(\frac{a}{b}\right)-x\).
Solution
We have
LHS = \(\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x-a \sin x}\right)=\tan ^{-1}\left\{\frac{\frac{a \cos x-b \sin x}{b \cos x}}{\frac{b \cos x-a \sin x}{b \cos x}}\right\}\)
[on dividing num. and denom. by cos x]
= \(\tan ^{-1}\left\{\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b} \tan x}\right\}=\tan ^{-1}\left(\frac{p-q}{1+p q}\right)\), where \(\frac{a}{b}\) = p and tan x = q
= tan-1p – tan-1q = tan-1\(\frac{a}{b}\) – tan-1(tan x)
= \(\left(\tan ^{-1} \frac{a}{b}-x\right)\) = RHS.
Hence, \(\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x-a \sin x}\right)=\left(\tan ^{-1} \frac{a}{b}-x\right) \text {. }\)
Example 32 Prove that \(\cot ^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}=\frac{x}{2}, x \in\left(0, \frac{\pi}{4}\right)\)
Solution
We have
LHS = \(\cot ^{-1}\left\{\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})} \times \frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}\right\}\)
= \(\cot ^{-1}\left\{\frac{(1+\sin x)+(1-\sin x)+2 \sqrt{1-\sin ^2 x}}{(1+\sin x)-(1-\sin x)}\right\}\)
= \(\cot ^{-1}\left\{\frac{2(1+\cos x)}{2 \sin x}\right\}=\cot ^{-1}\left(\frac{1+\cos x}{\sin x}\right)\)
= \(\cot ^{-1}\left\{\frac{2 \cos ^2(x / 2)}{2 \sin (x / 2) \cos (x / 2)}\right\}=\cot ^{-1}\left(\cot \frac{x}{2}\right)=\frac{x}{2}\) = RHS.
Hence, LHS = RHS.
Example 33 Prove that \(\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x\)
Solution
\(\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x\)Putting x = cos θ, we get
LHS = \(\tan ^{-1}\left\{\frac{\sqrt{1+\cos \theta}-\sqrt{1-\cos \theta}}{\sqrt{1+\cos \theta}+\sqrt{1-\cos \theta}}\right\}\)
= \(\tan ^{-1}\left\{\frac{\sqrt{2 \cos ^2(\theta / 2)}-\sqrt{2 \sin ^2(\theta / 2)}}{\sqrt{2 \cos ^2(\theta / 2)}+\sqrt{2 \sin ^2(\theta / 2)}}\right\}=\tan ^{-1}\left\{\frac{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}\right\}\)
= \(\tan ^{-1}\left\{\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}}\right\}\) [dividing num. and denom. by cos \frac{θ}{2}]
= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right\}=\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\)
= \(\left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\right)\) = RHS.
Hence, LHS = RHS.
Example 34 Prove that \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\frac{1}{2} \tan ^{-1} x\).
Solution
\(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\frac{1}{2} \tan ^{-1} x\)Putting x = tan θ, we get
\(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right)\)= \(\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)\)
= \(\tan ^{-1}\left\{\frac{2 \sin ^2\left(\frac{\theta}{2}\right)}{2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}\right\}=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)\)
= \(\frac{θ}{2}\) = \(\frac{1}{2}tan-1x\).
∴ \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\frac{1}{2} \tan ^{-1} x\)
Example 35 Prove that \(\tan ^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)=\sin ^{-1} \frac{x}{a}\)
Solution
\(\tan ^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)=\sin ^{-1} \frac{x}{a}\)Putting x = asin θ, we get
LHS = \(\tan ^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)\)
= \(\tan ^{-1}\left(\frac{a \sin \theta}{\sqrt{a^2-a^2 \sin ^2 \theta}}\right)\)
= \(\tan ^{-1}\left(\frac{{asin} \theta}{a \cos \theta}\right)=\tan ^{-1}(\tan \theta)\)
= θ = sin-1 \(\frac{x}{a}\) = RHS.
∴ \(\tan ^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)=\sin ^{-1} \frac{x}{a}\)
Example 36 Prove that \(\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right) .\)
Solution
\(\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right) .\)Putting x = tan2θ, we get
RHS = \(\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)\)
= \(\frac{1}{2} \cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\)
= \(\frac{1}{2} \cos ^{-1}(\cos 2 \theta)\)
= \(\left(\frac{1}{2} \times 2 \theta\right)=\theta=\tan ^{-1} \sqrt{x}\) = LHS
[∵ x = tan2θ ⇒ tan θ = √x ⇒ θ = tan-1√x].
Hence, \(\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right).\)
Example 37 Prove that \(\tan ^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)=\left(\frac{\pi}{2}-\sec ^{-1} x\right)\)
Solution
\(\tan ^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)=\left(\frac{\pi}{2}-\sec ^{-1} x\right)\)Putting x = sec θ, we get
\(\tan ^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)=\tan ^{-1}\left(\frac{1}{\sqrt{\sec ^2 \theta-1}}\right)\)= \(\tan ^{-1}\left(\frac{1}{\tan \theta}\right)=\tan ^{-1}(\cot \theta)\)
= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\theta\right)\right\}=\left(\frac{\pi}{2}-\theta\right)=\left(\frac{\pi}{2}-\sec ^{-1} x\right) .\)
∴ \(\tan ^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)=\left(\frac{\pi}{2}-\sec ^{-1} x\right)\)
Example 38 Prove that \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2\)
Solution
\(\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2\)Putting x2 = cos 2θ, we get
\(\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}\right)\)= \(\tan ^{-1}\left(\frac{\sqrt{2 \cos ^2 \theta}+\sqrt{2 \sin ^2 \theta}}{\sqrt{2 \cos ^2 \theta}-\sqrt{2 \sin ^2 \theta}}\right)\)
= \(\tan ^{-1}\left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right)=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right)\)
[dividing num. and denom. by cos θ]
= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}+\theta\right)\right\}=\left(\frac{\pi}{4}+\theta\right)\)
= \(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2\)
∴ \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2\)
Example 39 Prove that \(2 \tan ^{-1} \frac{1}{x}=\sin ^{-1}\left(\frac{2 x}{x^2+1}\right) \text {. }\)
Solution
\(2 \tan ^{-1} \frac{1}{x}=\sin ^{-1}\left(\frac{2 x}{x^2+1}\right) \text {. }\)Let tan-1\(\frac{1}{x}\) = θ. Then, \(\frac{1}{x}\) = tan θ ⇒ x = cot θ.
∴ LHS = \(2 \tan ^{-1} \frac{1}{x}\) = 2θ.
RHS = \(\sin ^{-1}\left(\frac{2 \cot \theta}{\cot ^2 \theta+1}\right)=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\)
= sin-1(sin 2θ) = 2θ.
∴ LHS = RHS.
Hence, \(2 \tan ^{-1} \frac{1}{x}=\sin ^{-1}\left(\frac{2 x}{x^2+1}\right) \text {. }\)
Example 40 Prove that \(\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1-x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}=\left(\frac{x+y}{1-x y}\right)\), where |x| < 1, y > 0 and xy < 1.
Solution
\(\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1-x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}=\left(\frac{x+y}{1-x y}\right)\)Putting x = tan θ and y = tan Φ, we get
LHS = \(\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1-x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}\)
= \(\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-\tan ^2 \phi}{1+\tan ^2 \phi}\right)\right\}\)
= \(\tan \left\{\frac{1}{2} \sin ^{-1}(\sin 2 \theta)+\frac{1}{2} \cos ^{-1}(\cos 2 \phi)\right\}\)
= \(\tan \left\{\left(\frac{1}{2} \times 2 \theta\right)+\left(\frac{1}{2} \times 2 \phi\right)\right\}\)
= \(\tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}=\frac{(x+y)}{(1-x y)}\) = RHS.
∴ \(\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1-x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}=\left(\frac{x+y}{1-x y}\right)\)
Example 41 Prove that \(\cot ^{-1}\left(\frac{a b+1}{a-b}\right)+\cot ^{-1}\left(\frac{b c+1}{b-c}\right)+\cot ^{-1}\left(\frac{c a+1}{c-a}\right)=0 \text {. }\)
Solution
We have
LHS = \(\tan ^{-1}\left(\frac{a-b}{1+a b}\right)+\tan ^{-1}\left(\frac{b-c}{1+b c}\right)+\tan ^{-1}\left(\frac{c-a}{1+c a}\right)\)
= (tan-1a – tan-1b) + (tan-1b – tan-1c) + (tan-1c – tan-1a)
= 0 = RHS.
∴ LHS = RHS.
Graphs Of Inverse Trigonometric Functions
1. Graph of sin-1x
Let f:[-1,1] → \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) : f(x) = sin-1x.
Here \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) is called the principal-value branch of sin-1x.
The other branches of sin-1x are
\(\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right],\left[\frac{3 \pi}{2}, \frac{5 \pi}{2}\right], \ldots \text { and }\left[\frac{-3 \pi}{2}, \frac{-\pi}{2}\right],\left[\frac{-5 \pi}{2}, \frac{-3 \pi}{2}\right]\), etc
Table for sin-1x
Also, sin-1(-x) = -sin-1x.
∴ \(\left(x=\frac{-1}{2} \Rightarrow \sin ^{-1} x=\frac{-\pi}{6}\right)\),
\(\left(x=-0.7 \Rightarrow \sin ^{-1} x=\frac{-\pi}{4}\right)\),
\(\left(x=-0.86 \Rightarrow \sin ^{-1} x=\frac{-\pi}{3}\right)\),
\(\left(x=-1 \Rightarrow \sin ^{-1} x=\frac{-\pi}{2}\right) .\)
O(0,0), A(\(\frac{1}{2}\), \(\frac{\pi}{6}\)), B(0.7, \(\frac{\pi}{4}\)), c(0.86, \(\frac{\pi}{3}\)), D(1, \(\frac{\pi}{2}\)), E(\(-frac{1}{2}\), \(\frac{-\pi}{6}\)), F(-0.7, \(\frac{-\pi}{4}\)). G(-0.86, \(\frac{-\pi}{3}\)) and H(-1, \(\frac{-\pi}{2}\)).
Join the points OA, AB, BC, CD, and OE, EF, FG, GH successively with a freehand to get the required graph, as shown in the given figure.
Moreover, we have
\(\sin \frac{2 \pi}{3}=\sin \left(\pi-\frac{\pi}{3}\right)=\sin \frac{\pi}{3}=0.86\) \(\sin \frac{5 \pi}{6}=\sin \left(\pi-\frac{\pi}{6}\right)=\sin \frac{\pi}{6}=0.5, \sin \pi=0 \text {. }\) \(\sin \left(\frac{-2 \pi}{3}\right)=-\sin \frac{2 \pi}{3}=-0.86 .\) \(\sin \left(\frac{-5 \pi}{6}\right)=-\sin \frac{5 \pi}{6}=-0.5\)Now, we may extend the graph as shown in the figure.
2. Graph of cos-1x
Let f : [-1,1] → [0, π] : f(x) = cos-1x.
Here, [0, π] is called the principal value branch of cos-1x.
The other branches of cos-1x are [π, 2π], [2π, 3π], …, [-π,0], [-2π, -π], etc.
Table for cos-1x
On a graph paper, we plot the points
A(1,0), B(0.87, \(\frac{\pi}{6}\)), C(0.7, \(\frac{\pi}{4}\)), D(0.5, \(\frac{\pi}{3}\)), E(0, \(\frac{\pi}{2}\)), F(-0.5, \(\frac{2 \pi}{3}\)), G(-0.7, \(\frac{3 \pi}{4}\)), H(-0.86, \(\frac{5 \pi}{6}\)) and K(-1, π).
Join AB, BC, CD, DE, EF, FG, GH and HK successively with a freehand to obtain the graph of cos-1x, as shown in the given figure.
3. Graph of tan-1x
Let f : R → \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) : f(x) = tan-1x.
Here \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) is the principal value branch of tan-1x.
The other branches of tan-1x are (\(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)), (\(\frac{3 \pi}{2}\), \(\frac{5 \pi}{2}\)), … , (\(-\frac{3 \pi}{2}\), \(-\frac{\pi}{2}\)), etc.
We take the positive values and then use tan-1(-x) = -tan-1x.
Table for tan-1x
Also, tan-1(-x) = -tan-1x.
∴ tan-1(-0.58) = \(-\frac{\pi}{6}\), tan-1(-1) = \(-\frac{\pi}{4}\), tan-1(-1.73) = \(-\frac{\pi}{3}\).
On a graph paper, we plot the points O(0,0), A(0.58, \(\frac{\pi}{6}\)), B(1, \(\frac{\pi}{4}\)), C(1.73, \(\frac{\pi}{3}\)), and D = (0.58, \(-\frac{\pi}{6}\)), E(-1, \(-\frac{\pi}{4}\)), F(-1.73, \(-\frac{\pi}{3}\)).
Join OA, AB, BC, and OD, DE, EF successively to get the graph.
Now, when x → ∞, then tan-1x → \(\frac{\pi}{2}\).
Also, when x → ∞, then tan-1x → \(-\frac{\pi}{2}\).
4. Graph of cot-1x
Let f : R → (0, π) : f(x) = cot-1x.
Here, (0,π) is the principal value branch of cot-1x.
The other branches of cot-1x are (π, 2π), (2π, 3π), …, (-π,0), etc.
Table for cot-1x
Using cot(π – x) = -cot x, we have
\(\cot \frac{2 \pi}{3}=\cot \left(\pi-\frac{\pi}{3}\right)=-\cot \frac{\pi}{3}=-0.58, \cot \frac{3 \pi}{4}=\cot \left(\pi-\frac{\pi}{4}\right)=-\cot \frac{\pi}{4}=-1\) \(\cot \frac{5 \pi}{6}=\cot \left(\pi-\frac{\pi}{6}\right)=-\cot \frac{\pi}{6}=-1.73 \text {. }\)On a graph paper, we plot the points \(A\left(1.73, \frac{\pi}{6}\right), B\left(1, \frac{\pi}{4}\right), C\left(0.58, \frac{\pi}{3}\right)\),
\(D\left(0, \frac{\pi}{2}\right), E\left(-0.58, \frac{2 \pi}{3}\right), F\left(-1, \frac{3 \pi}{4}\right), \text { and } G\left(-1.73, \frac{5 \pi}{6}\right) \text {. }\)We join the points AB, BC, CD, DE, EF, FG successively to get the graph.
As x → ∞, then cot-1x → 0.
And, as x → ∞, then cot-1x → π.
5. Graph of sec-1x
Let f : R – (-1,1) → [0, π] – \(\frac{\pi}{2}\) : f(x) = sec-1x.
The other branches of sec-1x are [π, 2π] – \(\frac{3 \pi}{2}\), …, [-π, 0] – \(\frac{\pi}{2}\), etc.
Table for sec-1x
sec π = -1 ⇒ sec-1(-1) = π.
On a graph paper, we plot the points
A(1,0), B(1.15, \(\frac{\pi}{6}\)), C(1.41, \(\frac{\pi}{4}\)), D(2, \(\frac{\pi}{3}\)) and E(-2, \(\frac{2 \pi}{3}\)), F(-1.41, \(\frac{3 \pi}{4}\)), G(-1.15, \(\frac{5 \pi}{6}\)), H(-1, π).
We join the points AB, BC, CD, DE, and HG, GF, FE.
6. Graph of cosec-1x
Let f : R -(-1,1) → \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) – {0} : f(x) = cosec-1x.
The other branches of cosec-1x are \(\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]-\{\pi\}\), …, etc.
Table for cosec-1x
Since cosec-1(-x) = -cosec-1x, we have
\({cosec}^{-1}\left(\frac{-\pi}{6}\right)=-2, {cosec}^{-1}\left(\frac{-\pi}{4}\right)=-1.41, {cosec}^{-1}\left(\frac{-\pi}{3}\right)=-1.15 and {cosec}^{-1}\left(\frac{-\pi}{2}\right)=-1\)On a graph paper, we plot the points \(A\left(2, \frac{\pi}{6}\right), B\left(1.41, \frac{\pi}{4}\right), C\left(1.15, \frac{\pi}{3}\right)\)
\(D\left(1, \frac{\pi}{2}\right) \text { and } E\left(-2, \frac{-\pi}{6}\right), F\left(-1.41, \frac{-\pi}{4}\right), G\left(-1.15, \frac{-\pi}{3}\right), H\left(-1, \frac{-\pi}{2}\right) \text {. }\)Join the points as shown in the given figure, to get the graph.
Also as x → 0 from +ve values, then cosec-1x → ∞.
As x → 0 from -ve values, then cosec-1x → ∞.
