WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities
Definition of trigonometric ratios:
In your earlier classes you have studied about the ratios between two similar real quantities.
So, it now arises a very vital quiry to all of you that what trigonometric ratios are.
In reply, we can say that trigonometric ratios are the ratios between the lengths of any two sides of a triangle with respect to some preassigned angle of that triangle.
The triangle and its angle in respect to which the trigonometric ratios are to be determined may be of any type.
But in this class 10 as per syllabus we shall discuss only the trigonometric ratios of a right angled triangle and with respect to such an angle which lies in between the domain 0° ≤ θ ≤ 90° where θ is the angle.
WBBSE Solutions for Class 10 Maths
Now, it is clear that a triangle consists of three angles. So, we can determine at most 18 trigonometric ratios with respect to these three angles, i.e., with respect to each angle we can determine at most 6 trigonometric ratios, so far our knowledge permit us. Let us determine the ratios.
Let ABC be a right angled triangle of which ∠B is a right angle. So, the other two angles ∠A and ∠C are acute angles.
Here, AB ⊥ BC and it is obvious that AC is its hypotenuse. So, with respect to the ∠C. AB is the perpendicular and BC is the base.
However, if our consideration of angle be ∠A, then the opposite side of ∠A will be considered as the perpendicular, i.e., BC is the perpendicular and the adjacent side of ∠A will be considered as the base, i.e.. AB is the base with respect to the ∠A.
Let ∠C = θ (theta). Then the six trigonometric ratios with respect to θ are defined as follows:
sine of ∠C = sin θ (writing in short) = \(\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{\mathrm{AB}}{\mathrm{AC}}\)
cosine of ∠C = cos θ (writing in short) = \(\frac{\text { base }}{\text { hypotenuse }}=\frac{\mathrm{BC}}{\mathrm{AC}}\)
tangent of ∠C = tan θ (writing in short) = \(\frac{\text { perpendicular }}{\text { base }}=\frac{\mathrm{AB}}{\mathrm{BC}}\)
cosecant of ∠C = cosec θ (writing in short) = \(\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{\mathrm{AC}}{\mathrm{AB}}\)
secant of ∠C = sec θ (writing in short) = \(\frac{\text { hypotenuse }}{\text { base }}=\frac{\mathrm{AC}}{\mathrm{BC}}\)
cotangent of ∠C = cot θ (writing in short) = \(\frac{\text { base }}{\text { perpendicular }}=\frac{\mathrm{BC}}{\mathrm{AB}}\)
By definition, these are the six trigonometric ratios with respect to the ∠C.
Observe minutely that the trigonometric ratios cosecθ, secθ, and cotθ are clearly the inverse ratios, i.e., multiplicative opposite ratios of sinθ cosθ, and tanθrespectively.
Similarly, if the angle of consideration be ∠A = β (Beta, let), then the six trigonometric ratios are:
sin β = \(\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{\mathrm{BC}}{\mathrm{AC}}\)
cos β = \(\frac{\text { base }}{\text { hypotenuse }}=\frac{\mathrm{AB}}{\mathrm{AC}}\)
tan β = \(\frac{\text { perpendicular }}{\text { base }}=\frac{\mathrm{BC}}{\mathrm{AB}}\)
cosec β = \(\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{\mathrm{AC}}{\mathrm{BC}}\)
sec β = \(\frac{\text { hypotenuse }}{\text { base }}=\frac{\mathrm{AC}}{\mathrm{AB}}\)
cot β = \(\frac{\text { base }}{\text { perpendicular }}=\frac{\mathrm{AB}}{\mathrm{BC}}\)
What do we mean by the trigonometric term sin θ?
Trigonometric term sin θ
From the definition given above, we see that the trigonometric term sinθ, it means that sinθ is purely a ratio between the two lengths of the sides of a right-angled triangle and the ratio is simply a real fraction either proper or improper.
Since sinθ is a fraction, so it holds all the properties of a fraction of real numbers.
A question now arises:
Is sinθ a product of sin and θ?
In reply we shall say that no, sinθ is not a product of sin and θ. It means sine of the angle θ, i.e., it means the ratio of perpendicular and hypotenuse of a right-angled triangle with respect to the angle θ.
Similar is the case with cosθ, tanθ, cosecθ, secθ, and tanθ.
Meaning of square of sin θ:
Square of sinθ =(sin θ)2 = sin2θ ≠ sin θ2, i.e., square of sinθ means the square of the ratio as a whole, but not the square of the angle related.
Similarly, (cosθ)2 = cos2θ, (tanθ)2 = tan2θ,
(cosecθ)2 = cosec2θ, (secθ)2 = sec2θ, (cotθ)2 = cot2θ.
\(\sqrt{\sin \theta}=\sin ^{\frac{1}{2}} \theta\) and so on.
\(\sqrt[3]{\sin \theta}=\sin ^{\frac{1}{3}} \theta\) and so on.
(sin θ)3 = sin3θ and so on.
Can the value of sinθ be greater than 1?
In a right-angled triangle, we know that the hypotenuse is always greater than the other two sides of the triangle of which one is the perpendicular and the other is the base. We also know that
\(\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}\)
Thus, in the fraction of sinθ, the denominator is always greater than the numerator unless and until θ takes the value 90°.
If θ = 90°, then the perpendicular and hypotenuse become the same, but the hypotenuse can never be greater than the perpendicular.
So, the value can never be greater than 1, but it may take the value 1 when the angle considered is 90°.
Can the value of cos0 be greater than 1?
We know that cosθ = \(\frac{\text { base }}{\text { hypotenuse }}\) and the hypotenuse is always greater than or equal to the base of a right-angled triangle.
So, the value of cosθ may be at most 1. but can never be greater than 1.
All other trigonometric ratios like tanθ, cosecθ, secθ, and cotθ may take any real value.
We can now conclude the properties of trigonometric ratios as follows:
Properties of trigonometric ratios:
- Trigonometric ratios are the ratios between the length of any two sides of a right-angled triangle.
- Trigonometric, ratios obey the properties of real-valued numerical fractions.
- The trigonometric ratio sinθ may take the value 1 at most, but can never be greater than 1.
- The trigonometric ratio cosθ may take the value 1 at most, but can never be greater than 1.
- The trigonometric ratios tanθ, cosecθ, secθ, and cotθ may take any real value.
- The trigonometric ratio sinθ is not a product of sin and θ. Similar are the cases with all other trigonometric ratios.
- Trigonometric ratios are measured with respect to some angle of a triangle.
- Only the term sin does not mean anything unless and untill an angle, say, θ is joined to it.
Relation between trigonometric ratios:
By the definition of trigonometric ratios we have seen that
Moreover, tanθ = \(\frac{\text { perpendicular }}{\text { base }}\)=\(\frac{\text { perpendicular }}{\text { hypotenuse }}\) [ Dividing by hypotenuse]
= \(\frac{\sin \theta}{\cos \theta}\) [by definition of sin θ and cos θ]
Similarly, cotθ = \(\frac{\text { base }}{\text { perpendicular }}=\frac{\frac{\text { base }}{\text { hypotenuse }}}{\frac{\text { perpendicular }}{\text { hypotenuse }}}\)
= \(\frac{\cos \theta}{\sin \theta}\) [by definition of cosθ and sinθ]
Trigonometric ratios of some standard angles
In trigonometry, the following angles are of immense use : θ°, 30°, 45°, 60°, 90°. So these angles are known as standard angles in trigonometry.
We shall now find the trigonometric ratios of these angles in the following table:
Ratios |
0° |
30° |
45° |
60° |
90° |
sin |
0 |
\(\frac{1}{2}\) |
\(\frac{1}{\sqrt{2}}\) |
\(\frac{\sqrt{3}}{2}\) |
1 |
cos |
1 |
\(\frac{\sqrt{3}}{2}\) |
\(\frac{1}{\sqrt{2}}\) |
\(\frac{1}{2}\) |
0 |
tan |
0 |
\(\frac{1}{\sqrt{3}}\) |
1 |
√3 |
undefined |
cosec |
undefined |
2 |
√2 |
\(\frac{2}{\sqrt{3}}\) |
1 |
sec |
1 |
\(\frac{2}{\sqrt{3}}\) |
√2 |
2 |
undefined |
cot |
undefined |
√3 |
1 |
\(\frac{1}{\sqrt{3}}\) |
0 |
The student can easily remember these values by the short-cut procedure given below:
write → 0, 1, 2, 3, 4
Divide by 4 → \(\frac{0}{4}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4}\)
= \(0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1\)
Take square root →
All other trigonometric ratios can easily be found by the formulas given in the relation between the trigonometric ratios.
Trigonometric ratios when the angles are negative:
- sin(- θ) = – sin θ
- cos(- θ) = cos θ
- tan (- θ) = – tan θ
- cosec(- θ) = – cosec θ
- sec(- θ) = sec θ
- cot(- θ) = – cot θ
From the above formulas we see that only the cos θ and sec θ are not affected by the negative angles.
So these two trigonometric ratios are said to be even trigonometric functions and all others are known as odd trigonometric functions.
Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Multiple Choice Questions
Example 1. \(\frac{\frac{1}{3} \cos 30^{\circ}}{\frac{1}{2} \sin 45^{\circ}}+\frac{\tan 60^{\circ}}{\cos 30^{\circ}}\)
- \(\frac{6+\sqrt{6}}{\sqrt{3}}\)
- \(\frac{6-\sqrt{6}}{\sqrt{3}}\)
- \(\frac{6+\sqrt{6}}{3}\)
- \(\frac{6-\sqrt{6}}{3}\)
Solution: \(\frac{\frac{1}{3} \cos 30^{\circ}}{\frac{1}{2} \sin 45^{\circ}}+\frac{\tan 60^{\circ}}{\cos 30^{\circ}}\)
= \(\frac{\frac{1}{3} \times \frac{\sqrt{3}}{2}}{\frac{1}{2} \times \frac{1}{\sqrt{2}}}+\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}=\frac{\frac{1}{2 \sqrt{3}}}{\frac{1}{2 \sqrt{2}}}+\sqrt{3} \times \frac{2}{\sqrt{3}}\)
= \(\frac{1}{2 \sqrt{3}} \times \frac{2 \sqrt{2}}{1}+2=\frac{\sqrt{2}}{\sqrt{3}}+2=\frac{\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}+2\)
= \(\frac{\sqrt{6}}{3}+2=\frac{\sqrt{6}+6}{3}=\frac{6+\sqrt{6}}{3}\)
∴ 3. \(\frac{6+\sqrt{6}}{3}\) is correct.
Example 2. sin2 45° + cos245° =
- 1
- -1
- 0
- 2
Solution: sin2 45°+cos2 45° = \(\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2=\frac{1}{2}+\frac{1}{2}=1\)
∴ 1. 1 is correct.
Example 3. \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)
- 1
- 0
- √2
- √3
Solution: \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)
= \(\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^2}=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{2}=\sqrt{3}\)
∴ 4. √3 is correct.
Example 4. \(\tan ^2 \frac{\pi}{2} \sin \frac{\pi}{3} \tan \frac{\pi}{6} \tan ^2 \frac{\pi}{3}\)
- 1
- 1\(\frac{1}{2}\)
- 1\(\frac{1}{3}\)
- 1\(\frac{1}{4}\)
Solution: \(\tan ^2 \frac{\pi}{2} \sin \frac{\pi}{3} \tan \frac{\pi}{6} \tan ^2 \frac{\pi}{3}\)
= tan245° sin 60° tan 30° tan260° [ π = 180°]
= \((1)^2 \times \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}} \times(\sqrt{3})^2=1 \times \frac{1}{2} \times 3=\frac{3}{2}=1 \frac{1}{2}\)
∴ 2. 1\(\frac{1}{2}\) is correct.
Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities True Or False
Example 1. sinθ is a product of sin and θ.
Solution: False
sinθ is a ratio between the two sides of a right-angled triangle with respect to the angle θ.
Example 2. The value of cos0 can never be greater than 1.
Solution: True
Since cos θ = \(\frac{\text { base }}{\text { hypotenuse }}\) and hypotenuse of a right-angled mangle can never be greater than its base with respect to any angle of it.
Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Fill In The Blanks
Example 1. \(\tan \theta=\frac{\text { perpendicular }}{}\)
Solution: Base
Example 2. sec 45° = \(\frac{1}{}\)
Solution: \(\frac{1}{\sqrt{2}} ; \quad \text { since } \quad \sec 45^{\circ}=\frac{1}{\cos 45^{\circ}}=\frac{1}{\frac{1}{\sqrt{2}}}\)
Example 3. cot 0° = _______
Solution: undefined
since cot0°= \(\frac{\cos 0^{\circ}}{\sin 0^{\circ}}=\frac{1}{0}\) = undefined.
Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Short Answer Type Questions
Example 1. If cosθ = 0.6, then show that (5 sinθ – 3 tanθ) = 0.
Solution:
Given
If cosθ = 0.6
\(5 \sin \theta-3 \tan \theta=5 \sin \theta-\frac{3 \sin \theta}{\cos \theta}\)
= \(\sin \theta\left(5-\frac{3}{\cos \theta}\right)\)
= \(\sin \theta\left(5-\frac{3}{0.6}\right)\)
= \(\sin \theta\left(5-\frac{30}{6}\right)=\sin \theta(5-5)\)
= \(\sin \theta \times 0=0\) [Proved]
Example 2. Find the value of sin² 45° – cosec² 60° + sec² 30°.
Solution: sin2 45°- cosec260° + sec230°
= \(\left(\frac{1}{\sqrt{2}}\right)^2-\left(\frac{2}{\sqrt{3}}\right)^2+\left(\frac{2}{\sqrt{3}}\right)^2=\frac{1}{2}-\frac{4}{3}+\frac{4}{3}=\frac{1}{2}\)
Hence the required value \(\frac{1}{2}\).
Example 3. Prove that cos 60° = cos2230° – sin20°.
Solution: LHS = cos 60°
= \(\frac{1}{2}\)
RHS = cos230° – sin230°
= \(\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{1}{2}\right)^2=\frac{3}{4}-\frac{1}{4}=\frac{3-1}{4}\)
=\(\frac{2}{4}=\frac{1}{2}\)
∴ LHS= RHS. (Proved)
Example 4. Prove that \(\sqrt{\frac{1+\cos 30^{\circ}}{1-\cos 30^{\circ}}}\) = sec 60° + tan 60°
Solution: LHS \(\sqrt{\frac{1+\cos 30^{\circ}}{1-\cos 30^{\circ}}}\)
=\(\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}}\)
= \(\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{\frac{2-\sqrt{3}}{2}}}=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\)
= \(\sqrt{\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}}\)
= \(\sqrt{\frac{(2+\sqrt{3})^2}{4-3}}=\sqrt{(2+\sqrt{3})^2}=2+\sqrt{3}\)
RHS = sec 60° + tan 60°
= 2 + √3
∴ LHS = RHS. [Proved]
Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Long Answer Type Questions
Example 1. In the window of a house, there is a ladder at an angle of 60° with the ground. If the ladder is 2√3 m long, then calculate the height of the window above the ground.
Solution:
Given
In the window of a house, there is a ladder at an angle of 60° with the ground. If the ladder is 2√3 m long
Let the height of the ground be AB. Then from the right-angled triangle ABC. we get.
sin 60° = \(\frac{AB}{AC}\) [ by the definition of sin θ]
or, \(\frac{\sqrt{3}}{2}\) = \(\frac{A B}{2 \sqrt{3}}\)
or, AB = 3
Hence the window is 3 m high above the ground.
Example 2. ABC is a right-angled triangle w ith its ∠B is 1 right angle. If AB = 8√3 cm and BC = 8 cm, then calculate the values of ∠ACB and ∠BAC.
Solution:
Given
ABC is a right-angled triangle w ith its ∠B is 1 right angle. If AB = 8√3 cm and BC = 8 cm
In the right-angled triangle ABC, ∠B = 90°.
∴∠ACB + ∠BAC = 90°…….(1)
Now, by the definition of tan θ,
tan ∠ACB = \(\frac{\mathrm{AB}}{\mathrm{BC}}\left[\tan \theta=\frac{\text { perpendicular }}{\text { base }}\right]\)
= \(\frac{8 \sqrt{3}}{8}\)
= √3 = tan60°
∴ ∠ACB = 60°
From (1) we get, ∠BAC = 90° – ∠ACB = 90° – 60° = 30°.
Hence the required value of ∠ACB = 60° and ∠BAC = 30°.
Example 3. In a right-angled triangle ABC, ∠B = 90°, ∠A = 30°, and AC = 20 cm. Determine the lengths of two sides BC and AB.
Solution:
Given
In a right-angled triangle ABC, ∠B = 90°, ∠A = 30°, and AC = 20 cm
From the right-angled triangle ABC, cos 30°= \(\frac{AB}{AC}\)
or, \(\frac{\sqrt{3}}{2}=\frac{A B}{20}\)
\(2 A B=20 \sqrt{3}\) or, \(\quad \mathrm{AB}=\frac{20 \sqrt{3}}{2}=10 \sqrt{3}\)
Also in the triangle ABC, sin 30°= \(\frac{BC}{AC}\) or, \(\frac{1}{2}\) = \(\frac{BC}{20}\)
or, 2BC = 20 or, BC = \(\frac{20}{2}\) = 10
Hence the lengths of both BC is 10 cm and AB is 10√3
Example 4. In a right-angled triangle PQR, ∠Q = 90°, ∠R = 45°. If PR = 3√2 ,then find the lengths of two sides PQ and QR.
Solution:
In the right-angled triangle PQR, ∠Q = 1 right-angle and ∠R = 45°
∴ sin 45°= \(\frac{PQ}{PR}\) [ by definition ]
or, \(\frac{1}{\sqrt{2}}=\frac{P Q}{3 \sqrt{2}}\) or, PQ = 3
Again, cos 45°= \(\frac{QR}{PR}\) [by definition]
or, \(\frac{1}{\sqrt{2}}=\frac{\mathrm{QR}}{3 \sqrt{2}}\) or, QR = 3
Hence the lengths of both PQ and QR is 3 m.
Example 5. Calculate: sin245° – cosec260° + sec230°.
Solution: sin245°- cosec20° + sec20°.
= \(\left(\frac{1}{\sqrt{2}}\right)^2-\left(\frac{2}{\sqrt{3}}\right)^2+\left(\frac{2}{\sqrt{3}}\right)^2\)
= \(\frac{1}{2}\)
Example 6. Calculate: sec245° – cot245° – sin230° – sin260°.
Solution: sec245°- cot245°- sin230°- sin260°
= \((\sqrt{2})^2-(1)^2-\left(\frac{1}{2}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2\)
= \(2-1-\frac{1}{4}-\frac{3}{4}=\frac{8-4-1-3}{4}=\frac{0}{4}= 0\)
Example 7. Calculate: 3 tan245° – sin260° – \(\frac{1}{3}\)cot230°- \(\frac{1}{8}\)sec245°.
Solution: 3 tan245° – sin260° – \(\frac{1}{3}\)cot230°- \(\frac{1}{8}\)sec245°.
= \(3 \times(1)^2-\left(\frac{\sqrt{3}}{2}\right)^2-\frac{1}{3} \times(\sqrt{3})^2-\frac{1}{8} \times(\sqrt{2})^2\)
= \(\frac{3}{4}-\frac{1}{3} \times 3-\frac{1}{8} \times 2=3-\frac{3}{4}-1-\frac{1}{4}=\frac{12-3-4-1}{4}=\frac{4}{4}=1\)
Example 8. Calculate: \(\frac{4}{3}\) cot230° + 3 sin2 60°- 2 cosec260°- \(\frac{3}{4}\) tan230°.
Solution: \(\frac{4}{3}\) cot230° + 3 sin260°- 2 cosec260°- \(\frac{3}{4}\) tan230°
= \(\frac{4}{3} \times(\sqrt{3})^2+3 \times\left(\frac{\sqrt{3}}{2}\right)^2-2 \times\left(\frac{2}{\sqrt{3}}\right)^2-\frac{3}{4} \times\left(\frac{1}{\sqrt{3}}\right)^2\)
= \(\frac{4}{3} \times 3+3 \times \frac{3}{4}-2 \times \frac{4}{3}-\frac{3}{4} \times \frac{1}{3}=4+\frac{9}{4}-\frac{8}{3}-\frac{1}{4}\)
= \(\frac{48+27-32-3}{12}=\frac{75-35}{12}=\frac{40}{12}=\frac{10}{3}=3 \frac{1}{3}\)
Example 9. Calculate : cot230° – 2 cos260°- \(\frac{3}{4}\)sec245° – 4 sin230°.
Solution: cot230° – 2 cos260°- \(\frac{3}{4}\)sec245° – 4 sin230°
= \((\sqrt{3})^2-2 \times\left(\frac{1}{2}\right)^2-\frac{3}{4} \times(\sqrt{2})^2-4 \times\left(\frac{1}{2}\right)^2\)
= \(3-2 \times \frac{1}{4}-\frac{3}{4} \times 2-4 \times \frac{1}{4}\)
= \(3-\frac{1}{2}-\frac{3}{2}-1=\frac{6-1-3-2}{2}=\frac{6-6}{2}=\frac{0}{2}=0\)
Example 10. Calculate: sec260°- cot230°- \(\frac{2 \tan 30^{\circ}{cosec} 60^{\circ}}{1+\tan ^2 30^{\circ}}\)
Solution: sec260°- cot230°- \(\frac{2 \tan 30^{\circ}{cosec} 60^{\circ}}{1+\tan ^2 30^{\circ}}\)
= \((2)^2-(\sqrt{3})^2-\frac{2 \times \frac{1}{\sqrt{3}} \times \frac{2}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^2}\)
= \(4-3-\frac{\frac{4}{3}}{1+\frac{1}{3}}\)
= \(1-\frac{\frac{4}{3}}{\frac{3+1}{3}}=1-\frac{\frac{4}{3}}{\frac{4}{3}}=1-1=0\)
Example 11. Calculate: \(\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \tan 30^{\circ}}\) + cos 60° cos 30° + sin 60° sin 30°.
Solution: \(\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \tan 30^{\circ}}\) + cos 60° cos 30° + sin 60° sin 30°
= \(\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3} \times \frac{1}{\sqrt{3}}}+\frac{1}{2} \times \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \times \frac{1}{2}=\frac{\frac{3-1}{\sqrt{3}}}{1+1}+\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\)
= \(\frac{\frac{2}{\sqrt{3}}}{2}+\frac{2 \sqrt{3}}{4}=\frac{2}{\sqrt{3}} \times \frac{1}{2}+\frac{2 \sqrt{3}}{4}=\frac{1}{\sqrt{3}}+\frac{2 \sqrt{3}}{4}\)
= \(\frac{4+6}{4 \sqrt{3}}=\frac{10}{4 \sqrt{3}}=\frac{5}{2 \sqrt{3}}\)
\end{aligned}
Example 12. Calculate: \(\frac{1-\sin ^2 30^{\circ}}{1+\sin ^2 30^{\circ}} \times \frac{\cos ^2 60^{\circ}+\cos ^2 30^{\circ}}{{cosec}^2 90^{\circ}-\cot ^2 90^{\circ}}\) ÷ (sin 60° tan 30°)
Solution: \(\frac{1-\sin ^2 30^{\circ}}{1+\sin ^2 30^{\circ}} \times \frac{\cos ^2 60^{\circ}+\cos ^2 30^{\circ}}{{cosec}^2 90^{\circ}-\cot ^2 90^{\circ}}\) ÷ (sin 60° tan 30°)
= \(\frac{1-\left(\frac{1}{2}\right)^2}{1+\left(\frac{1}{2}\right)^2} \times \frac{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}{(1)^2-0} \div\left(\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}}\right)\)
= \(\frac{1-\frac{1}{4}}{1+\frac{1}{4}} \times \frac{\frac{1}{4}+\frac{3}{4}}{1} \div \frac{1}{2}=\frac{\frac{3}{4}}{\frac{5}{4}} \times \frac{1}{1} \div \frac{1}{2}\)
= \(\frac{3}{4} \times \frac{4}{5} \times 1 \times \frac{2}{1}=\frac{6}{5}=1 \frac{1}{5}\)
Example 13.Prove that: \(\frac{2 \tan ^2 30^{\circ}}{1-\tan ^2 30^{\circ}}\) + sec245°-cot245° = sec 60°.
Solution: LHS = \(\frac{2 \tan ^2 30^{\circ}}{1-\tan ^2 30^{\circ}}\) + sec245°-cot245° + sec245°-cot245°
= \(\frac{2 \times\left(\frac{1}{\sqrt{3}}\right)^2}{1-\left(\frac{1}{\sqrt{3}}\right)^2}+(\sqrt{2})^2-(1)^2\)
= \(\frac{2 \times \frac{1}{3}}{1-\frac{1}{3}}+2-1\)
= \(\frac{\frac{2}{3}}{\frac{2}{3}}+1=1+1=2\)
RHS = sec 60° = 2
∴ LHS = RHS. [Proved]
Example 14. Prove that: sin\(\frac{\pi}{3}\) tan\(\frac{\pi}{6}\) + sin\(\frac{\pi}{2}\) cos \(\frac{\pi}{3}\) = 2 sin 2 \(\frac{\pi}{4}\)
Solution: LHS = sin\(\frac{\pi}{3}\) tan\(\frac{\pi}{6}\) + sin\(\frac{\pi}{2}\) cos \(\frac{\pi}{3}\) = 2 sin 2 \(\frac{\pi}{4}\)
= sin 60° tan 30° + sin 90° cos 60°
= \(\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}}+1 \times \frac{1}{2}=\frac{1}{2}+\frac{1}{2}=1\)
RHS = 2 sin2 \(\frac{\pi}{4}\) = 2 x \(\left(\frac{1}{\sqrt{2}}\right)^2\) = 2 x \(\frac{1}{2}\) = 1.
∴ LHS = RHS. [Proved]
Example 15. If x sin 45° cos 45° tan 60°= tan245° – cos 60°, then determine the value of x.
Solution:
Given
x sin 45° cos 45° tan 60° = tan245° – cos 60°
or, \(x \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \times \sqrt{3}=1-\frac{1}{2}\)
or, \(x \times \frac{\sqrt{3}}{2}=1-\frac{1}{2}\)
or, \(\quad x \times \frac{\sqrt{3}}{2}=\frac{1}{2}\)
or, \(\quad x=\frac{1}{\sqrt{3}}\)
Hence the required value of x = \(\frac{1}{\sqrt{3}}\).
Example 16. If x sin 60° cos230° = \(\frac{\tan ^2 45^{\circ} \sec 60^{\circ}}{{cosec} 60^{\circ}}\) then find the value of x.
Solution: Given that x sin 60° cos230° = \(\frac{\tan ^2 45^{\circ} \sec 60^{\circ}}{{cosec} 60^{\circ}}\)
or, \(x \times \frac{\sqrt{3}}{2} \times\left(\frac{\sqrt{3}}{2}\right)^2=\frac{(1)^2 \times 2}{\frac{2}{\sqrt{3}}}\)
or, \(x \times \frac{\sqrt{3}}{2} \times \frac{3}{4}=\frac{2 \sqrt{3}}{2}\)
or, \(\quad x=\frac{8}{3}=2 \frac{2}{3}\)
Hence the value of x = 2 \(\frac{2}{3}\)
Example 17. If x2 = sin230° + 4 cot245° – sec260°, determine the value of x.
Solution:
Given
x2 = sin230° + 4 cot245° – sec260°
or, \(x^2=\left(\frac{1}{2}\right)^2+4 \times(1)^2-(2)^2\)
or, \(x^2=\frac{1}{4}+4-4\)
or, \(x^2=\frac{1}{4}\)
or, \(x= \pm \frac{1}{2}\)
Hence the value of x = \(\pm \frac{1}{2}\)
Example 18. If x tan 30° + y cot 60° = 0 and 2x-y tan 45° = 1, then calculate the values of x and y.
Solution: Given that x tan 30° + y cot 60° = 0
or, \(x \times \frac{1}{\sqrt{3}}+y \times \frac{1}{\sqrt{3}}=0\)
or, x + y = 0 [ multiplying by √3 ] ……..(1)
Again, 2x- y tan 45° = 1 or, 2x- y x 1 = 1 or, 2x – y = 1 ……..(2)
Adding (1) and (2) we get, 3x = 1 or, x = \(\frac{1}{3}\)
∴ from (1) we get, \(\frac{1}{3}\) + y = 0
or, y = –\(\frac{1}{3}\)
Hence the values of x = \(\frac{1}{3}\) and y = –\(\frac{1}{3}\)
Example 19. If A = B= 45°, then justify
- sin (A + B) = sin A cos B + cos A sin B.
- cos (A + B) – cos A cos B – sin A sin B.
Solution:
1. Given that A = B = 45°.
Now, sin (A + B) = sin (45° + 45°) = sin 90° = 1.
Also, sin A cos B + cos A sin B = sin 45° cos 45° + cos 45° sin 45°
= \(\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{1}{2}+\frac{1}{2}=1 .\)
Hence, sin (A + B) = sin A cos B + cos A sin B. (Proved]
2. cos (A + B) = cos (45° + 45°) = cos 90° = 0
Also, cos A cos B – sin A sin B
= cos 45° cos 45° – sin 45° sin 45°
= \(\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{1}{2}-\frac{1}{2}\) = 0.
Hence cos (A + B) = cos A cos B – sin A sin B. (Proved]
Example 20. In an equilateral triangle ABC, BD is a median. Prove that tan ∠ABD = cot ∠BAD.
Solution:
Here, ∠ADB = 90°, ∴ AB = hypotenuse.
∴ tan ∠ABD = [by definition]……(1)
Also, cot ∠BAD = [ by definition ]……(2)
∴ from (1) and (2) we get, tan ∠ABD = cot ∠BAD.
Hence tan ∠ABD = cot ∠BAD. [Proved]
Example 21. In an isosceles triangle ABC, AB = AC and ∠BAC = 90°, the bisector of ∠BAC intersects the side BC at the point D. Prove that \(\frac{\sec \angle A C D}{\sin \angle C A D}\) cosec2 ∠CAD.
Solution:
Given
In an isosceles triangle ABC, AB = AC and ∠BAC = 90°, the bisector of ∠BAC intersects the side BC at the point D.
In ΔABC, AB = AC,
∴ ∠ABC = ∠ACB = 45° [∠BAC = 90° ]
AD is the bisector of ∠BAC = 90°.
∴ ∠BAD = 45° and ∠CAD = 45°
Now, \(\frac{\sec \angle A C D}{\sin \angle C A D}=\frac{\sec 45^{\circ}}{\sin 45^{\circ}}\)
[because \(\angle \mathrm{CAD}=\frac{1}{2} \angle \mathrm{BAC}=\frac{1}{2} \times 90^{\circ}=45^{\circ}\)]
= \(\frac{\sqrt{2}}{\frac{1}{\sqrt{2}}}=\sqrt{2} \times \sqrt{2}=2\)
Also, cosec2 ∠CAD = cosec245° = (√2)2 = 2
Hence \(\frac{\sec \angle \mathrm{ACD}}{\sin \angle C A D}\) = cosec2 ∠CAD. [Proved]
Example 22. Determine the value/values of θ (0° ≤ θ ≤; 90°) for which 2 cos2θ-3 cosθ + 1 = 0 will be true.
Solution: 2 cos2 θ – 3 cos θ + 1 = 0
or, 2 cos2 θ-2 cos θ – cos θ + 1 = 0
or, 2 cos θ (cos θ – 1) – 1 (cos θ – 1) = 0
or, (cos θ – 1)(2 cos θ – 1) = 0
∴ either cos θ – 1 = 0
⇒ cos θ= 1
⇒ cos θ = cos θ°
⇒ θ = 0°
or, 2 cos θ – 1 = 0
⇒ 2 cos θ=1
cos θ = \(\frac{1}{2}\) = cos 60°
Hence the values of θ are 0° and 60°.
Example 23. Prove that cot \(\frac{\pi}{8}\)cot\(\frac{3\pi}{8}\)cot\(\frac{5\pi}{8}\)cot\(\frac{7\pi}{8}\) =1.
Solution: LHS = cot\(\frac{\pi}{8}\)cot\(\frac{3\pi}{8}\)cot\(\frac{5\pi}{8}\)cot\(\frac{7\pi}{8}\)
= cot\(\frac{\pi}{8}\)cot\(\frac{7\pi}{8}\)cot\(\frac{3\pi}{8}\)cot\(\frac{5\pi}{8}\)
= cot\(\frac{\pi}{8}\)\(\cot \left(\frac{\pi}{2}+\frac{3 \pi}{8}\right)\)cot\(\frac{3\pi}{8}\)cot\(\frac{5\pi}{8}\)
= cot\(\frac{\pi}{8}\)tan\(\frac{3\pi}{8}\)cot\(\frac{3\pi}{8}\)tan\(\frac{\pi}{8}\)
= cot\(\frac{\pi}{8}\) x tan\(\frac{\pi}{8}\) x tan\(\frac{3\pi}{8}\) x cot\(\frac{3\pi}{8}\)
= 1 x 1 cot\(\frac{\pi}{8}\) x tan\(\frac{\pi}{8}\) = cot \(\frac{\pi}{8}\) x \(\frac{1}{\cot \frac{\pi}{8}}\) = 1
= 1
= RHS [Proved]
Example 24. If tan 42° = 0.9, then find the value of cot 492°.
Solution: cot 492° = cot (90° x 5 + 42°)
= – tan 42° = – 0.9
Example 25.If sec (α- β) = 2 and sin (α + β) = \(\frac{1}{2}\), then find the least positive values of α and β.
Solution: Given that sec (α-β) = 2
⇒ sec (α- β) = sec 45°
⇒ α- β = 45°…….(1)
Again, sin(α + β) = \(\frac{1}{2}\) ⇒ sin (α +β) = sin 30°
= α + β = 30°…….(2)
But (2) is impossible, since α -β = 45°.
∴ sin (α + β) = sin 30° = sin (180° – 30°)
⇒ sin (α + β) = sin 150°
⇒ α + β = 150°……..(3)
Now. adding (1) and (3) we get, 2α = 195°
or, α = \(\frac{195^{\circ}}{2}=97 \frac{1}{2}^{\circ}\)
From (3) we get, α + 13 = 150°
or, \(97 \frac{1}{2}^{\circ}\) + 3 = 150° or, =150° -97\(\frac{1}{2}^{\circ}\) = 52\(\frac{1}{2}^{\circ}\)
Hence the least positive values of α = 97\(\frac{1}{2}^{\circ}\) and β = 52\(\frac{1}{2}^{\circ}\).
Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric Identities
An equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of angles involved in it.
The basic trigonometric ratios are:
- sin2 θ+ cos2 θ=1
- sec2 θ=1+ tan2 θ
- cosec2 θ =1+ cot2 θ
We shall now proof these identities.
1. sin2 θ+ cos2 θ =1
Let ΔABC be a right-angled triangle, in which ∠B = 90° and ∠ACB = 0. So, with respect to θ, hypotenuse = AC, perpendicular = AB, and base = BC.
Now, in ΔABC by Pythagoras theorem we get, AB2 + BC2 = AC2.
or, \(\frac{A B^2}{A C^2}+\frac{B C^2}{A C^2}=\frac{A C^2}{A C^2}\) [Dividing both the sides by AC2]
or, \(\left(\frac{\mathrm{AB}}{\mathrm{AC}}\right)^2+\left(\frac{\mathrm{BC}}{\mathrm{AC}}\right)^2=1\)
or, \(\left(\frac{\text { perpendicular }}{\text { hypotenuse }}\right)^2+\left(\frac{\text { base }}{\text { hypotenuse }}\right)^2=1\)
or, \((\sin \theta)^2+(\cos \theta)^2=1\) [by definition of sin θ and cos θ]
or, \(\sin ^2 \theta+\cos ^2 \theta=1\)
Hence \(\sin ^2 \theta+\cos ^2 \theta=1\) [Proved]
2. sec2 θ= 1 + tan2 θ
Let ΔABC be a right-angled triangle in which ∠B = right angle and ∠ACB = θ. So, with respect to θ, hypotenuse = AC, perpendicular = AB, and base = BC.
Now, in ΔABC by pythagoras theorem we get, AC2 = AB2 + BC2
or, \(\frac{\mathrm{AC}^2}{\mathrm{BC}^2}=\frac{\mathrm{AB}^2}{\mathrm{BC}^2}+\frac{\mathrm{BC}^2}{\mathrm{BC}^2}\) [Dividing both the sides by BC2 ]
or, \(\left(\frac{A C}{B C}\right)^2=\left(\frac{A B}{B C}\right)^2+1\)
or, \(\left(\frac{\text { hypotenuse }}{\text { base }}\right)=\left(\frac{\text { perpendicular }}{\text { base }}\right)^2+1\)
or, (sec θ)2 = (tan θ)2 + 1 [by definition of sec θ and tan θ]
or, sec2 θ = tan2 θ + 1.
Hence sec2 θ=1+ tan2θ. [Proved]
3. cosec2 θ = 1 + cot2 θ
Let ΔABC be a right-angled triangle in which ∠B = right angle and ∠ACB = 0.
So, with respect to θ, hypotenuse = AC, perpendicular = AB, and base = BC.
Now in ΔABC by pythagoras theorem we get, AB2 + BC2= AC2
or, \(\frac{\mathrm{AB}^2}{\mathrm{AB}^2}+\frac{\mathrm{BC}^2}{\mathrm{AB}^2}=\frac{\mathrm{AC}^2}{\mathrm{AB}^2}\) [Dividing both the sides by AB2]
or, \(1+\left(\frac{\mathrm{BC}}{\mathrm{AB}}\right)^2=\left(\frac{\mathrm{AC}}{\mathrm{AB}}\right)^2\)
[Dividing both the sides by AB2]
or, \(1+\left(\frac{\text { base }}{\text { perpendicular }}\right)^2=\left(\frac{\text { hypotenuse }}{\text { perpendicular }}\right)^2\)
or, 1 + (cot θ)2 = (cosec θ)2 [by definition of cot θ and cosec θ]
or, 1 + cot2 θ = cosec2 θ.
Hence cosec2 θ = 1 + cot2 θ [Proved]
We can deduce many all other trigonometric identities from these three basic identities. Such as:
sin2 θ + cos2 θ = 1 ⇒ sin2 θ=1- cos2 θ
⇒ \(\sin ^2 \theta+\cos ^2 \theta=1\)
⇒ \(\sin ^2 \theta=1-\cos ^2 \theta\)
⇒ \(\sin \theta= \pm \sqrt{1-\cos ^2 \theta}\)
Again, \(\cos ^2 \theta=1-\sin ^2 \theta \Rightarrow \cos \theta= \pm \sqrt{1-\sin ^2 \theta}\)
\(\sec ^2 \theta=1+\tan ^2 \theta\)
⇒ \(\sec ^2 \theta-\tan ^2 \theta=1\)
⇒ \((\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1\)
\(\sec ^2 \theta=1+\tan ^2 \theta\) ⇒ \(\sec \theta= \pm \sqrt{1+\tan ^2 \theta}\)
⇒ \(\tan ^2 \theta=\sec ^2 \theta-1 \Rightarrow \tan \theta= \pm \sqrt{\sec ^2 \theta-1}\)
Similarly cosec2θ = 1 + cot2 θ
⇒ cosec2 θ- cot2 θ = 1
⇒ (cosecθ + cot θ)(cosecθ- cotθ) = 1
Again, cosec2θ = 1 + cot2θ
⇒ cosec θ = ± \(\sqrt{1+\cot ^2 \theta}\)
Also cot2 θ = cosec22 θ-1
⇒ cot θ = ± \(=\sqrt{{cosec}^2 \theta-1}\)
Besides these trigonometric identities, the most important trigonometric formulas are:
- sin (A + B) = sin A cos B + cos A sin B.
- sin (A- B) = sin A cos B – cos A sin B.
- cos (A + B) = cos A cos B – sin A sin B.
- cos (A – B) = cos A cos B + sin A sin B.
- tan (A + B) = \(\frac{\tan A+\tan B}{1-\tan A \tan B}\)
- tan (A – B) = \(\frac{\tan A-\tan B}{1+\tan A \tan B}\)
- 2 sin A cos B = sin (A + B) + sin (A – B).
- 2 cos A sin B = sin (A + B) – sin (A – B).
- 2 sin A sin B = cos (A – B) – cos (A + B).
- 2 cos A cos B = cos (A + B) + cos (A – B).
- sin 2A = 2 sin A cos A = \(\frac{2 \tan A}{1+\tan ^2 A}\)
- cos 2A = cos2 A – sin2 A = 2 cos2 A – 1 = 1 – 2 sin2 A = \(\frac{1-\tan ^2 A}{1+\tan ^2 A}\)
- tan 2A = \(\frac{2 \tan A}{1-\tan ^2 A}\)
- sin 3A = 3 sin A – 4 sin3 A.
- cos 3A = 4 cos3 A – 3 cos A.
- tan 3A = \(\frac{3 \tan A-\tan ^3 A}{1-3 \tan ^2 A}\)
Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric Identities Multiple Choice Questions
Example 1. If 3x = cosec a and \(\frac{3}{x}\) = cot α , then the value of 3 \(\left(x^2-\frac{1}{x^2}\right)\) is
- \(\frac{1}{27}\)
- \(\frac{1}{81}\)
- \(\frac{1}{3}\)
- \(\frac{1}{9}\)
Solution:
Given that 3x = cosec α
or, (3x)2 = cosec2 α [Squaring]
or, 9x2 = cosec2 α …….(1)
Again, \(\frac{3}{x}\) = cot a or, \(\frac{3}{x}^2\) = \(\cot ^2 \alpha\) [squaring]
or, \(\frac{9}{x^2}\) = cot2 α…….(2)
We know that cosec2 α – cot2 α – 1
or, \(9 x^2-\frac{9}{x^2}=1\) [from (1) and (2)]
or, \(9\left(x^2-\frac{1}{x^2}\right)=1\)
or, \(x^2-\frac{1}{x^2}=\frac{1}{9}\)
or, \(3\left(x^2-\frac{1}{x^2}\right)=3 \times \frac{1}{9}\)
or, \(3\left(x^2-\frac{1}{x^2}\right)=\frac{1}{3}\)
∴ 3. \(\frac{1}{3}\) is correct.
Example 2. If 2x = sec A and \(\frac{2}{x}\) = tan A, then the value of \(2\left(x^2-\frac{1}{x^2}\right)\)
- \(\frac{1}{2}\)
- \(\frac{1}{4}\)
- \(\frac{1}{8}\)
- \(\frac{1}{16}\)
Solution:
Given:
If 2x = sec A and \(\frac{2}{x}\) = tan A
We know that sec2 A- tan2 A = 1.
or, \((2 x)^2-\left(\frac{2}{x}\right)^2=1\)
[because \( \sec \mathrm{A}=2 x\) and \(\tan \mathrm{A}=\frac{2}{x}\)]
or, \(4 x^2-\frac{4}{x^2}\) =1
or, \(\quad 4\left(x^2-\frac{1}{x^2}\right)=1\)
or, \(4\left(x^2-\frac{1}{x^2}\right)=1\)
or, \(x^2-\frac{1}{x^2}=\frac{1}{4}\)
or, \(2\left(x^2-\frac{1}{x^2}\right)=2 \times \frac{1}{4}\)
or, \(2\left(x^2-\frac{1}{x^2}\right)=\frac{1}{2}\)
∴ 1. \(\frac{1}{2}\) is correct
Example 3. If tan α + cot α = 2, then the value of tan13 α + cot13α is
- 1
- 0
- 2
- None of these
Solution:
Given that \({tan} \propto+\frac{1}{\tan \alpha}\)=2
or, \(\frac{\tan ^2 \alpha+1}{\tan \alpha}=2\)
or, \(\tan ^2 \alpha+1=2 \tan \alpha\)
or, \(\tan ^2 \alpha-2 \tan \alpha+1=0\)
or, \((\tan \alpha-1)^2=0\)
or, \(\tan \alpha-1=0\)
or, \(\tan \alpha=1\)
∴ \(\quad \cot \alpha=\frac{1}{\tan \alpha}=\frac{1}{1}=1\)
∴ tan13 α+cot13 α = (1)13+(1)13 =1+1=2
∴ 3. 2 is correct.
Example 4. If sin θ – cos2 θ= 0 (0° < θ < 90°) and sec θ + cosec θ= x, then the value of x is
- 1
- 2
- √2
- 2√2
Solution:
Given:
If sin θ – cos2 θ= 0 (0° < θ < 90°) and sec θ + cosec θ= x,
x = sec θ + cosec θ
= \(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}=\frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta}\)
[sinθ-cosθ = 0
or, sin θ + cos θ- 2 sin θ cos θ= 0
or, 1-2 sin θ cos θ = 0
or, sinθcosθ = \(\frac{1}{2}a\)……..(1)
Again, sin θ + cos θ
= \(\sqrt{(\sin \theta+\cos \theta)^2}\)
= \(\sqrt{(\sin \theta-\cos \theta)^2+4 \sin \theta \cos \theta}\)
= \(\sqrt{0^2+4 \times \frac{1}{2}}\)
= \(\sqrt{0+2}=\sqrt{2}\)……(2)]
= \(\frac{\sqrt{2}}{\frac{1}{2}}\) [from (2) and (1)]
= \(2 \sqrt{2}\)
∴ 4. 2 √2 is correct
Example 5. If 2 cos 3θ = 1, then the value of 0 is
- 10°
- 15°
- 20°
- 30°
Solution: Given that 2 cos 3θ = 1
or, cos 3θ = \(\frac{1}{2}\) or, cos 3θ = cos 60°
⇒ 3θ= 60°
or, θ = \(\frac{60^{\circ}}{3}\)
or, θ = 20°
∴ 3. 20° is correct.
Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric Identities True Or False
Example 1. If 0° ≤ α ≤ 90°, then the least value of (sec2 α + cos2 α) is 2.
Solution: True
Since sec2 α + cos2 α
= (sec α – cos α )2 + 2sec α cos α
= (sec α – cos α)2 2 . \(\frac{1}{\cos \alpha}\). cos α
= (sec α – cos α)2 + 2
Now, the least value of (sec α – cos α)2 is 0.
∴ The least value of sec2 α + cos2 α = 0 + 2 = 2
Hence the given statement is true.
Example 2. The value of cos 0° x cos 1° x cos 2° x cos 3° x ……x cos 90° is 1.
Solution: False
Since the value of cos 0° x cos 1° x cos 2° x cos 3° x …….. x Cos 90°
= 1 x cos 1° x cos 2° x cos 3° x ……… x 0 [cos 0° = 1 and cos 90° = 0]
Hence the given statement is false.
Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric Identities Fill In The Blanks
Example 1. The value of \(\left(\frac{4}{\sec ^2 \theta}+\frac{1}{1+\cot ^2 \theta}+3 \sin ^2 \theta\right)\) is _______
Solution: 4
Since \(\frac{4}{\sec ^2 \theta}+\frac{1}{1+\cot ^2 \theta}+3 \sin ^2 \theta\)
= 4 cos2 θ + \(\frac{1}{{cosec}^2 \theta}\) + 3 sin2 θ
= 4 cos2 θ + sin2 θ + 3 sin2 θ
= 4 cos2 θ + 4 sin2 θ
= 4 (cos2 θ + sin2 θ)
= 4×1=4
Example 2. If sin (θ- 30° ) = \(\frac{1}{2}\) then the value, of value of cos θ is _______
Solution: \(\frac{1}{2}\)
Since sin (θ-30°) = \(\frac{1}{2}\)
⇒ sin (θ- 30° ) = sin 30°
⇒ θ- 30° = 30°
⇒ θ = 30° +30°
⇒ θ = 60°
∴ cos θ = cos 60° = \(\frac{1}{2}\)
Example 3. If cos2 θ- sin2 0 = 1 , then the value of cos4 θ – sin4 θ is ________
Solution: \(\frac{1}{2}\)
Since cos4 θ sin4 θ = (cos4 θ)2 – (sin2 θ)2
= (cos2 θ + sin2 θ) (cos2 θ – sin2 θ)
= 1 x (cos2 θ – sin2 θ)
= cos2 θ – sin2 θ.
= \(\frac{1}{2}\)
Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric Identities Short Answer Type Questions
Example 1. If r cos θ = 2√3, r sinθ= 2 and 0° < θ < 90°, then determine the values of both r and θ.
Solution: r cosθ = 2√3
or, r2cos2 θ= 12 [squaring]…….(1)
rsinθ = 2
or, r2 sin2 θ = 4 [squaring]……..(2)
Adding (1) and (2) we get,
r2 cos2 θ+ r2 sin2 θ = 12 + 4
or, r2 (cos2 θ + sin2 θ) = 16
or, r2 = 16 [cos2 θ+ sin2 θ = 1]
or, r = 4
Again, \(\frac{r \cos \theta}{r \sin \theta}=\frac{2 \sqrt{3}}{2}\)
or, cot θ=√3
or, cot θ = cot 30° [0°<θ<90°]
⇒ θ = 30°.
Hence r = 4 and θ = 30°.
Example 2. If sin A + sin B = 2 where 0° ≤ A ≤ 90° and- 0° ≤ B ≤ 90°, then find the value of (cos A + cos B).
Solution: Given that 0° ≤ A < 90° and 0° ≤, B ≤ 90°
Also, sin. A + sin B = 2,
∴ Obviously, sin A = 1 and sin B = 1. [the greatest values of both sin A and sin B is 1 .]
Now, sin A = 1
⇒ sin A = sin 90°
⇒ A = 90°
Again, sin B = 1
⇒ sin B = sin 90°
⇒ B = 90°
∴ cos A + cos B = cos 90° + cos 90° = 0 + 0 = 0.
Hence the required value of cos A + cos B = 0.
Example 3. If 0° < θ < 90°, then calculate the least value of (9 tan2 θ + 4 cot2 θ).
Solution: 9 tan2 θ + 4 cot2 θ
= (3 tan θ)2 + (2 cot θ)2
= (3 tan θ – 2 cot θ)2 + 2.3 tan θ.2 cot θ
= (3 tan θ – 2 cot θ)2 + 12 tan θ \(\frac{1}{\tan \theta}\)
= (3 tan θ – 2 cot θ)2 + 12.
Now, the least value of (3 tan θ – 2 cot θ) is 0.
∴ The least of (9 tan2 θ + 4 cot2 θ) is (0 + 12) = 12.
Example 4. Calculate the value of (sin6 α + cos6 α + 3 sin2 α cos6 α ).
Solution: sin6 α + cos6 α + 3 sin2 α cos6 α
= (sin2 α )3 + (cos2 α )3 – 3 sin2 α cos2 α
= (sin2 α + cos2 α )3 – 3 sin2 α cos2 α (sin2 α + cos2 α ) + 3 sin2 α cos2 α
= (1)3 – 3 sin2 α cos2 α 1+3 sin2 α cos2 α
= 1-3 sin2 α cos2 α + 3 sin2 α cos2 α
= 1.
Hence the required value = 1.
Example 5. If cosec2 θ= 2 cot θ and 0° < θ < 90°, then determine the value of θ.
Solution: Given that cosec2 θ = 2 cot θ
or, 1 + cot2 θ – 2 cot θ = 0
or, (1)2 – 2.1.cot θ + (cot θ)2 = 0
or, (1 – cot θ)2 = 0
or, 1 – cot θ = 0
or, cot 0 = 1 = cot 45° [0° <θ < 90°]
⇒ θ= 45°.
Hence the value of θ is 45°.
Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric Identities Long Answer Type Questions
Example 1. If tan θ = \(\frac{3}{4}\), then show that \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\frac{1}{2}\).
Solution: Given that tan θ = \(\frac{3}{4}\)
or, cot θ = \(\frac{4}{3}\)
or, cot 2 θ = \(\frac{16}{9}\) [squaring]
or, cosec2 θ- 1 = \(\frac{16}{9}\)
or, cosec2 θ = \(\frac{16}{9}\) + 1
or, cosec2 θ = \(\frac{25}{9}\)
or, \(\frac{1}{\sin ^2 \theta}=\frac{25}{9}\)
or, sin2 θ= \(\frac{9}{25}\)
or, sin θ = \(\frac{3}{5}\)
Now, \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sqrt{\frac{1-\frac{3}{5}}{1+\frac{3}{5}}}=\sqrt{\frac{\frac{2}{5}}{\frac{8}{5}}}=\sqrt{\frac{2}{8}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)
Hence \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\) = \(\frac{1}{2}\)
Example 2. Express cosecθ and tanθ in terms of sinθ.
Solution: cosecθ = \(\frac{1}{\sin \theta}\)
and tanθ = \(\frac{\sin \theta}{\cos \theta}=\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\)
Hence cosecθ = \(\frac{1}{\sin \theta}\) and tanθ = \(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\)
Example 3. If sec θ + tan θ= 2, determine the value of (sec θ – tan θ)
Solution: We know that sec2 θ – tan2 θ=1.
or, (sec θ + tan θ)(sec θ – tan θ) = 1
or, 2 (sec θ – tan θ) = 1 [sec θ + tan θ = 2 (given)]
or, sec θ – tanθ = \(\frac{1}{2}\)
Example 4. If cosec θ – cot θ= √2 – 1, calculate the value of (cosec θ + cot θ).
Solution: We know that cosec2 θ – cot2 θ = 1
or, (cosec θ + cot θ)(cosec θ – cot θ) = 1
or, (√2 – 1)(cosec θ- cot θ) = 1 [cosec θ + cot θ = √2 – 1]
= \(\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{\sqrt{2}+1}{(\sqrt{2})^2-(1)^2}\)
= \(\frac{\sqrt{2}+1}{2-1}=\sqrt{2}+1\)
Hence cosec θ – cot θ = √2 + 1.
Example 5. If Sin θ + cos θ = 1, then find the value of sin θ x cos θ.
Solution: Given that sin θ + cos θ = 1
or, (sin θ+ cos θ)2 = i (squaring]
or, sin2 θ + cos2 θ + 2 sin θ cos θ =1
or, 1 + 2 sin θ cos θ = 1
or, 2 sin θ cos θ = 0
or, sin θ cos θ = \(\frac{0}{2}\) = 0.
Hence the required value of sin θ x cos θ is 0.
Example 6. If tan θ + cot θ = 2, then determine the value of tan θ – cot θ.
Solution: Given that tan θ + cot θ = 2
or, tan θ + \(\frac{1}{\tan \theta}\) = 2 or, tan2 θ + 1 = 2 tan θ
or, tan2 θ – 2 tan θ + 1 = 0
or, (tan θ – 1)2 = 0
or, tan θ = 1 = tan 45°
⇒ θ = 45°
tan θ – cot θ = tan 45° – cot 45°
= 1 – 1 = 0.
Hence the required value of tan θ – cot θ = 0.
Example 7. If sin θ- cos θ = \(\frac{7}{13}\), then determine the value of (sin θ + cos θ).
Solution: Given that sin θ- cos θ = \(\frac{7}{13}\)
or, \((\sin \theta-\cos \theta)^2=\left(\frac{7}{13}\right)^2\) [squaring]
or, sin2 θ + cos2 θ- 2 sin θ cos θ = \(\frac{49}{169}\)
or, 1- 2 sin θ cos θ = \(\frac{49}{169}\) or, 2 sinθ cos θ = 1 – \(\frac{49}{169}\)
or, 2 sin θ cos θ = \(\frac{120}{169}\)……..(1)
Now, (sin θ + cos θ)2 = sin2 θ + cos2 θ + 2 sin θ cos θ
= 1 + 2 sin θ cos θ
= 1 + \(\frac{120}{169}\) [from (1)]
= \(\frac{289}{169}=\left(\frac{17}{13}\right)^2\)
sinθ + cosθ = \(\frac{17}{13}\)
Hence the required value of sin θ + cos θ = \(\frac{17}{13}\)
Example 8. If sin θ cos θ = \(\frac{1}{2}\), then calculate the value of (sin θ + cos θ).
Solution: (sin θ + cos θ)2 = sin2 θ + cos2 θ + 2 sin θ cos θ
= 1 + 2 sin θ cos θ
= 1 + 2 x \(\frac{1}{2}\) [sin θ cos θ = \(\frac{1}{2}\)]
= 1 + 1=2.
∴ sin θ + cos θ = √2
Aliter : sin θ cos θ = \(\frac{1}{2}\)
or, 2 sin θ cos θ = 2 x \(\frac{1}{2}\)
or, sin 2θ = 1 [2 sin θ cos θ = sin 2θ]
or, sin 2θ = sin 90° [sin 90°= 1]
⇒ 2θ = 90° ⇒ \(\frac{90^{\circ}}{2}\) 45°
∴ sin θ + cos θ = sin 45°+ cos 45°
= \(=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{1+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Hence, sin θ + cos θ = √2 .
Example 9. If sec θ- tan θ = \(\frac{1}{\sqrt{3}}\), then determine the values of both sec θ and tan θ.
Solution: Given that sec θ- tan θ = \(\frac{1}{\sqrt{3}}\)……(1)
Also, sec2 θ – tan2 θ = 1
or, (sec θ + tan θ)(sec θ- tan θ) = 1
or, (sec θ + tan θ) x \(\frac{1}{\sqrt{3}}\) = 1 [from (1)]
or, sec θ + tan θ = √3 ………(2)
Now, adding (1) and (2) we get, 2secθ = \(\frac{1}{\sqrt{3}}\) + √3
or, 2sec θ = \(\frac{1+3}{\sqrt{3}}\)
or, 2 sec θ = \(2 \sec \theta=\frac{4}{\sqrt{3}}\)
or, sec θ = \(2 \sec \theta=\frac{2}{\sqrt{3}}\) = sec 30°
⇒ θ = 30°
∴ tan θ = tan 30° = \(\frac{1}{\sqrt{3}}\)
Hence sec θ = \(\frac{2}{\sqrt{3}}\) and tan 0 = \(\frac{1}{\sqrt{3}}\).
Example 10. If cosec θ + cot θ = √3, then determine the value of both cosec θ and cot θ.
Solution: We know that cosec2 θ – cot2 θ = 1
or, (cosec θ + cot θ)(cosec θ – cot θ) = 1
or, √3 (cosec θ – cot θ) = 1
or, cosec θ – cot θ = \(\frac{1}{\sqrt{3}}\)……(1)
Also given that cosec θ + cot θ = √3 ………(2)
Now, adding (1) and (2) we get,
2 cosec θ = \(\frac{1}{\sqrt{3}}\) + √3
or, 2 cosec θ = \(\frac{1+3}{\sqrt{3}}\) or, 2 cosec θ = \(\frac{4}{\sqrt{3}}\)
or, cosec θ = \(\frac{2}{\sqrt{3}}\) = cosec 60°
⇒ θ = 60°
∴ cot θ= cot 60° = \(\frac{1}{\sqrt{3}}\)
Hence cosec θ = \(\frac{2}{\sqrt{3}}\) and cot θ = \(\frac{1}{\sqrt{3}}\)
Example 11. If \(\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}\) = 7 , then find the value of tan θ.
Solution: Given that \(\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}\) = 7
or, \(\frac{\sin \theta+\cos \theta+\sin \theta-\cos \theta}{\sin \theta+\cos \theta-\sin \theta+\cos \theta}=\frac{7+1}{7-1}\)
[By componerido-dividendo process]
or, \(\frac{2 \sin \theta}{2 \cos \theta}=\frac{8}{6}\)
or, tan θ = \(\frac{4}{3}\)
Hence the value of tan θ = \(\frac{4}{3}\).
Example 12. If \(\frac{{cosec} \theta+\sin \theta}{{cosec} \theta-\sin \theta}\) = \(\frac{5}{3}\), calculate the valur of sin θ.
Solution: Given that \(\frac{{cosec} \theta+\sin \theta}{{cosec} \theta-\sin \theta}= [latex]\frac{5}{3}\)
or, \(\frac{{cosec} \theta+\sin \theta-{cosec} \theta+\sin \theta}{{cosec} \theta+\sin \theta+{cosec} \theta-\sin \theta}=\frac{5-3}{5+3}\)
[By dividendo- componendo process]
or, \(\frac{2 \sin \theta}{2{cosec} \theta}=\frac{2}{8} \quad \text { or, } \quad \frac{\sin \theta}{\frac{1}{\sin \theta}}=\frac{1}{4}\)
or, \(\sin ^2 \theta=\frac{1}{4} \quad \text { or, } \sin \theta=\sqrt{\frac{1}{4}}=\frac{1}{2}\).
Hence the value of sin θ = \(\frac{1}{2}\)
Example 13. If tan2 θ + cot2 θ = \(\frac{10}{3}\), then determine the value of (tan θ + cot θ) and (tan θ – cot θ). Also find the value of tan θ.
Solution: Given that tan2 θ + cot2 θ = \(\frac{10}{3}\)
or, (tan θ)2 + (cot θ)2 = \(\frac{10}{3}\)
or, (tanθ + cotθ)2 – 2 tan θ cot θ = \(\frac{10}{3}\)
or, (tan θ + cot θ)2 – 2 \(\tan \theta \cdot \frac{1}{\tan \theta}\) = \(\frac{10}{3}\)
or, (tan θ + cot θ)2 – 2 = \(\frac{10}{3}\)
or, (tan θ + cot θ)2 = \(\frac{10}{3}\) + 2
or, \((\tan \theta+\cot \theta)^2=\frac{10+6}{3}\)
or, (tan θ+ cot θ)2 = \(\frac{16}{3}\)
or, tan θ+ cot θ = \(\frac{4}{\sqrt{3}}\)…….(1)
or, \(\tan \theta+\cot \theta=\frac{\sqrt{16}}{3}\)
Again, \(\tan ^2 \theta+\cot ^2 \theta=\frac{10}{3}\)
or, \((\tan \theta-\cot \theta)^2+2 \tan \theta \cdot \cot \theta=\frac{10}{3}\)
or, \((\tan \theta-\cot \theta)^2+2 \tan \theta \frac{1}{\tan \theta}=\frac{10}{3}\)
or, \((\tan \theta-\cot \theta)^2+2=\frac{10}{3}\)
or, \((\tan \theta-\cot \theta)^2=\frac{10}{3}-2\)
or, \((\tan \theta-\cot \theta)^2=\frac{10-6}{3}\)
or, \((\tan \theta-\cot \theta)^2=\frac{4}{3}\)
or, \(\tan \theta-\cot \theta=\sqrt{\frac{4}{3}}\)
or, \(\tan \theta-\cot \theta=\frac{2}{\sqrt{3}}\)…..(2)
Also, by adding (1) and (2) we get, 2tan θ= \(\frac{4}{\sqrt{3}}+\frac{2}{\sqrt{3}}\)
or, 2 tan θ = \(\frac{6}{\sqrt{3}}\)= or, tan θ = \(\frac{6}{\sqrt{3}}\) = √3
Hence tan θ + cot θ = \(\frac{4}{\sqrt{3}}\), tan 0- cot 0 = \(\frac{2}{\sqrt{3}}\)= and tan θ = √3.
Example 14. If sec2 θ + tan2 θ = \(\frac{13}{12}\) , then calculate the value of (sec4 θ – tan4 θ)
Solution: sec4 θ – tan4 θ
= (sec2 θ)2 – (tan2 θ)2
= (sec2 θ + tan2 θ)(sec2 θ- tan2 θ)
= \(\frac{13}{12}\) x 1 [sec2 θ – tan2 θ=1]
= \(\frac{13}{12}\)
Hence the value of sec4 θ- tan4 θ = \(\frac{13}{12}\)
Example 15. In ΔPQR, ∠Q is right angle. If PR = √5 units and PQ – RQ = 1 unit, then determine the value of cos P – cos R.
Solution:
In ΔPQR, ∠Q = 90°
∴ PR = hypotenuse.
Now, cos P – cos R
= \(\frac{P Q}{P R}-\frac{R Q}{P R}\)
[by definition of cos P and cos R]
= \(\frac{P Q-R Q}{P R}\)
= \(\frac{1}{\sqrt{5}}\) [PQ-RQ = 1 and PR = √5]
Hence cos P- cos R = \(\frac{1}{\sqrt{5}}\)
Example 16. If sec θ + cos θ = \(\frac{5}{2}\), then calculate the value of (sec θ – cos θ).
Solution: Given that sec θ + cos θ = \(\frac{5}{2}\)
or, (sec θ + cos θ)2 = \(\left(\frac{5}{2}\right)^2\) [squaring]
or, (sec θ- cos θ) + 4 sec θ. cos θ = \(\frac{25}{4}\)
or, (sec θ- cosθ)2 +4. \(\frac{1}{\cos \theta}\) cos θ = \(\frac{25}{4}\)
or, (sec θ- cos θ)2 + 4 = \(\frac{25}{4}\)
or, (sec θ- cos θ)2 = \(\frac{25}{4}\) – 4
or, (sec θ- cos θ)2 = \(\frac{25-16}{4}\)
or, (sec θ- cos θ)2 = \(\frac{9}{4}\)
or, sec θ- cos θ = \(\sqrt{\frac{9}{4}}\) or, sec θ-cos θ = \(\frac{3}{2}\)
Hence sec θ – cos θ = \(\frac{3}{2}\)
Example 17. Determine the value of tan θ from the relation 5 sin2 θ + 4 cos2 θ = \(\frac{9}{2}\)
Solution: Given that 5 sin2 θ + 4 cos2 θ = \(\frac{9}{2}\)
or, 10 sin2 θ+ 8 cos2 θ = 9
or, 10 sin2 θ + 8(1 – sin2 θ) = 9
or, 10 sin2 θ + 8- sin2 θ = 9
or, 2 sin2 θ = 1
or, sin2 θ = \(\frac{1}{2}\)
or, sin θ = \(\frac{1}{\sqrt{2}}\) = sin 45°
∴ θ = 45
∴ tan 2 θ = tan 45° = 1
Example 18. In ΔXYZ, ∠Y is right angle. If XY – 2√3 units and XZ – YZ = 2 units, then determine the value of (sec X – tan X).
Solution:
In ΔXYZ, ∠Y = 90°.
∴ XZ = hypotenuse.
Now, sec X – tan X
= \(\frac{X Z}{X Y}-\frac{Y Z}{X Y}\)
[ by definition of sec X and tan X ]
= \(\frac{X Z-Y Z}{X Y}\)
= \(\frac{2}{2 \sqrt{3}}\) [XZ – YZ = 2 and XY = 2√3]
= \(\frac{1}{\sqrt{3}}\)
Hence sec X- tan X = \(\frac{1}{\sqrt{3}}\)
Example 19. Eliminate θ: x = 2 sin θ, y = 3 cos θ.
Solution: Given that x = 2 sin θ
or, \(\frac{x}{2}\) = sin θ
or, \(\frac{x^2}{4}=\sin ^2 \theta\)…….(1)
Also, y = 3 cos θ
or, \(\frac{y}{3}\) = cos θ
or, \(\frac{y^2}{9}=\cos ^2 \theta\)…….(2)
Now, adding (1) and (2) we get,
\(\frac{x^2}{4}+\frac{y^2}{9}=\sin ^2 \theta+\cos ^2 \theta\)
or, \(\frac{x^2}{4}+\frac{y^2}{9}=1\)
Hence eliminating 0 we get, \(\frac{x^2}{4}+\frac{y^2}{9}\) = 1.
Example 20. Eliminate θ: 5x = 3 sec θ, y = 3 tan θ.
Solution: Given that 5x = 3 sec 0
or, \(\frac{5x}{3}\) = sec θ
or, \(\frac{25 x^2}{9}\) = sec θ…….(1)
Again, y = 3 tan θ
or, \(\frac{y}{3}\) = tan θ
or, \(\frac{y^2}{9}=\tan ^2 \theta\)……..(2)
We know that sec2 θ – tan2 θ = 1
or, \(\frac{25 x^2}{9}-\frac{y^2}{9}=1\)
or, \(\frac{25 x^2-y^2}{9}=1\)
or, \(25 x^2-y^2=9\)
Hence eliminating θ, we get \(25 x^2-y^2=9\).
Example 21. If sin α = \(\frac{5}{13}\), then show that tan α + sec α = 1.5
Solution: sin α = \(\frac{5}{13}\)
∴ cos α = \(\sqrt{1-\sin ^2 \alpha}\)
= \(\sqrt{1-\left(\frac{5}{13}\right)^2}=\sqrt{1-\frac{25}{169}}\)
= \(\sqrt{\frac{169-25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}\)
Now, tan α + sec α = \(\frac{\sin \alpha}{\cos \alpha}+\frac{1}{\cos \alpha}\)
= \(\frac{\frac{5}{13}}{\frac{12}{13}}+\frac{1}{\frac{12}{13}}=\frac{5}{13} \times \frac{13}{12}+\frac{13}{12}\)
= \(\frac{5}{12}+\frac{13}{12}=\frac{5+13}{12}\)
= \(\frac{18}{12}=\frac{3}{2}=1 \cdot 5\)
Hence tan α + sec α = 1.5. [Proved]
Example 22. If tan A=\(\frac{m}{n}\), then determine the values of both sin A and sec A.
Solution: Given that tan A = \(\frac{m}{n}\)
or, \(\tan ^2 \mathrm{~A}=\frac{n^2}{m^2}\) [squaring]
or, \(1+\tan ^2 \mathrm{~A}=1+\frac{n^2}{m^2}\)
or, \(\sec ^2 \mathrm{~A}=\frac{m^2+n^2}{m^2}\)
or, \(\sec \mathrm{A}=\sqrt{\frac{m^2+n^2}{m^2}}\)
or, \(\sec \mathrm{A}=\frac{\sqrt{m^2+n^2}}{m}\)
tan A = \(\frac{m}{n}\)
or, cot A = \(\frac{m}{n}\) or, cot2 A = \(\frac{m^2}{n^2}\) [squaring]
or, \({cosec}^2 \mathrm{~A}-1=\frac{m^2}{n^2}\)
or, \({cosec}^2 \mathrm{~A}=\frac{m^2}{n^2}+1\)
or, \({cosec}^2 \mathrm{~A}=\frac{m^2+n^2}{n^2}\)
or, \({cosec} \mathrm{A}=\sqrt{\frac{m^2+n^2}{n^2}}\)
or, \(\frac{1}{\sin \mathrm{A}}=\sqrt{\frac{m^2+n^2}{n}}\)
or, \(\sin \mathrm{A}=\frac{n}{\sqrt{m^2+n^2}}\)
Hence \(\text { sin } \mathrm{A}=\frac{n}{\sqrt{m^2+n^2}} \text { and } \sec \mathrm{A}=\frac{\sqrt{m^2+n^2}}{m} .\)
Example 23. If cos θ = \(\frac{x}{\sqrt{x^2+y^2}}\), then show that x sin θ = y cos θ.
Solution: Given that cos θ = \(\frac{x}{\sqrt{x^2+y^2}}\)
or, cos θ = x2 [squaring] or, sec2θ = x2 +y
or, \(\cos ^2 \theta=\frac{x^2}{x^2+y^2}\) [squaring]
or, \(\sec ^2 \theta=\frac{x^2+y^2}{x^2}\)
or, \(1+\tan ^2 \theta=\frac{x^2+y^2}{x^2}\)
or, \(\tan ^2 \theta=\frac{x^2+y^2}{x^2}-1\)
or, \(\tan ^2 \theta=\frac{x^2+y^2-x^2}{x^2}\)
or, \(\tan ^2 \theta=\frac{y^2}{x^2}\)
or, \(\tan \theta=\frac{y}{x}\)
or, \(\frac{\sin \theta}{\cos \theta}=\frac{y}{x}\)
or, \(x \sin \theta=y \cos \theta\)
Hence \(x \sin \theta=y \cos \theta\) [Proved]
Example 24. If \(\sin \alpha=\frac{a^2-b^2}{a^2+b^2}\), then show that \(\cot \alpha=\frac{2 a b}{a^2-b^2}\)
Solution: Given that \(\sin \alpha=\frac{a^2-b^2}{a^2+b^2}\)
or, \(\sin ^2 \alpha=\frac{\left(a^2-b^2\right)^2}{\left(a^2+b^2\right)^2}\) (squaring)
or, \({cosec}^2 \alpha=\frac{\left(a^2+b^2\right)^2}{\left(a^2-b^2\right)^2}\)
or, \(1+\cot ^2 \alpha=\frac{\left(a^2+b^2\right)^2}{\left(a^2-b^2\right)^2}\)
or, \(\cot ^2 \alpha=\frac{\left(a^2+b^2\right)^2}{\left(a^2-b^2\right)^2}-1\)
or, \(\cot ^2 \alpha=\frac{\left(a^2+b^2\right)^2-\left(a^2-b^2\right)^2}{\left(a^2-b^2\right)^2}\)
or, \(\cot ^2 \alpha=\frac{4 a^2 b^2}{\left(a^2-b^2\right)^2}\)
or, \((\cot \alpha)^2=\left(\frac{2 a b}{a^2-b^2}\right)^2\)
⇒ \(\cot \alpha=\frac{2 a b}{a^2-b^2}\)
Example 25. If \(\frac{\sin \theta}{x}=\frac{\cos \theta}{y}\), then show that \(\sin \theta-\cos \theta=\frac{x-y}{\sqrt{x^2+y^2}}\).
Solution: Given that \(\frac{\sin \theta}{x}=\frac{\cos \theta}{y}\) = k(let)
∴ sin θ = kx………(1) and cos θ = ky……..(2)
We know that sin2θ + cos2θ = 1
⇒ (kx)2 + (ky)2 = 1 [from (1) and (2)]
or, k2x2 + k2y2 = 1 or, k2(x2+y2)= 1
or, \(k^2=\frac{1}{x^2+y^2}\)
or, \(k=\sqrt{\frac{1}{x^2+y^2}}\)
or, \(k=\frac{\cdot 1}{\sqrt{x^2+y^2}}\)
∴ \(\quad \sin \theta=k x=\frac{1}{\sqrt{x^2+y^2}} \times x=\frac{x}{\sqrt{x^2+y^2}}\)……..(3)
and \(\cos \theta=k y=\frac{1}{\sqrt{x^2+y^2}} \times y=\frac{y}{\sqrt{x^2+y^2}}\)……(4)
∴ \(\sin \theta-\cos \theta=\frac{x}{\sqrt{x^2+y^2}}-\frac{y}{\sqrt{x^2+y^2}}\) [from (3) and (4)]
= \(\frac{x-y}{\sqrt{x^2+y^2}}\)
Hence \(\sin \theta-\cos \theta=\frac{x-y}{\sqrt{x^2+y^2}}\) [Proved]
Aliter: Given that \(\frac{\sin \theta}{x}=\frac{\cos \theta}{y}\)
or, \(\frac{\sin \theta}{\cos \theta}=\frac{x}{y}\)
or, \(\tan \theta=\frac{x}{y}\)
or, \(\tan ^2 \theta=\frac{x^2}{y^2}\)
or, \(\sec ^2 \theta-1,=\frac{x^2}{y^2}\)
or, \(\sec ^2 \theta=\frac{x^2}{y^2}+1=\frac{x^2+y^2}{y^2}\)
or, \(\sec \theta=\sqrt{\frac{x^2+y^2}{y^2}}\)
or, \(\sec \theta=\frac{\sqrt{x^2+y^2}}{y}\)
or, \(\cos \theta=\frac{y}{\sqrt{x^2+y^2}}\)……..(1)
∴ \(\sin \theta =\sqrt{1-\cos ^2 \theta}=\sqrt{1-\left(\frac{y}{\sqrt{x^2+y^2}}\right)^2}\)
= \(\sqrt{1-\frac{y^2}{x^2+y^2}}=\sqrt{\frac{x^2+y^2-y^2}{x^2+y^2}}=\sqrt{\frac{x^2}{x^2+y^2}}\)
= \(\frac{x}{\sqrt{x^2+y^2}}\)……….(2)
Now, \(\sin \theta-\cos \theta =\frac{x}{\sqrt{x^2+y^2}}-\frac{y}{\sqrt{x^2+y^2}}\)[from (2) and (1)]
= \(\frac{x-y}{\sqrt{x^2+y^2}}\)
Hence \(\sin \theta-\cos \theta\) = \(\frac{x-y}{\sqrt{x^2+y^2}}\) [Proved]
Example 26. If (1 + 4x2) cos A = Ax, then show that cosec A + cot A = \(\frac{1+2 x}{1-2 x}\)
Solution: Given that (1 + 4x2) cos A = 4x
or, \(\cos A=\frac{4 x}{1+4 x^2}\)
∴ sin A = \(\sqrt{1-\cos ^2 A}=\sqrt{1-\left(\frac{4 x}{1+4 x^2}\right)^2}\)
= \(\sqrt{1-\frac{16 x^2}{\left(1+4 x^2\right)^2}}\)
= \(\sqrt{\frac{\left(1+4 x^2\right)^2-16 x^2}{\left(1+4 x^2\right)^2}}\)
= \(\sqrt{\frac{1+8 x^2+16 x^4-16 x^2}{\left(1+4 x^2\right)^2}}\)
= \(\sqrt{\frac{1-8 x^2+16 x^4}{\left(1+4 x^2\right)^2}}\)
= \(\sqrt{\frac{\left(1-4 x^2\right)^2}{\left(1+4 x^2\right)^2}}\)
= \(\frac{1-4 x^2}{1+4 x^2}\)
∴ cosec A + cot A = \(\frac{1}{\sin A}+\frac{\cos A}{\sin A}=\frac{1+\cos A}{\sin A}\)
= \(\frac{1+\frac{4 x}{1+4 x^2}}{\frac{1-4 x^2}{1+4 x^2}}\)
= \(\frac{\frac{1+4 x^2+4 x}{1+4 x^2}}{\frac{1-4 x^2}{1+4 x^2}}\)
= \(\frac{1+2 \cdot 1 \cdot 2 x+(2 x)^2}{1-4 x^2}\)
= \(\frac{(1+2 x)^2}{(1+2 x)(1-2 x)}=\frac{1+2 x}{1-2 x}\)
Hence cosec A + cot A = \(\frac{1+2 x}{1-2 x}\)
Example 27. If x = a sin θ and y = b tan θ, then prove that \(\frac{a^2}{x^2}-\frac{b^2}{y^2}=1\)
Solution: Given that x = a sin θ
⇒ sin θ = \(\frac{x}{a}\)
⇒ cosec θ = \(\frac{a}{x}\)
⇒ \({cosec}^2 \theta=\frac{a^2}{x^2}\)……..(1)
And y = b tan θ ⇒ tan θ = \(\frac{y}{b}\)
⇒ cot θ = \(\frac{b}{y}\)
⇒ \(\cot ^2 \theta=\frac{b^2}{y^2}\)……..(2)
We know that cosec2θ – cot2θ=1
or, \(\frac{a^2}{x^2}-\frac{b^2}{y^2}=1\) = 1 [from (1) and (2)]
Hence \(\frac{a^2}{x^2}-\frac{b^2}{y^2}=1\) = 1. [Proved]
Example 28. If sin θ + sin2 θ=1, then prove that cos2 θ + cos4 θ=1.
Solution: Given that sin θ + sin2 θ=1.
or, sin θ = 1 – sin2 θ
or, sin2 θ = cos2 θ
or, sin2 θ = cos4 θ [squaring]
or, 1 – cos2 θ = cos4 θ
or, 1 = cos2 θ + cos4 θ or, cos2 θ + cos4 θ= 1
Hence cos2 θ + cos4 θ=1. [Proved]
Example 29. If 3 cot θ = 4, then find the value of \(\frac{5 \sin \theta+3 \cos \theta}{5 \sin \theta-3 \cos \theta}\)
Solution: Given that 3 cot θ = 4
or, cot θ = \(\frac{4}{3}\)
or, tan θ = \(\frac{3}{4}\)
Now, \(\frac{5 \sin \theta+3 \cos \theta}{5 \sin \theta-3 \cos \theta}=\frac{5 \cdot \frac{\sin \theta}{\cos \theta}+3}{5 \cdot \frac{\sin \theta}{\cos \theta}-3}\) [Dividing both numerator and denominator by cos θ]
= \(=\frac{5 \tan \theta+3}{5 \tan \theta-3}=\frac{5 \times \frac{3}{4}+3}{5 \times \frac{3}{4}-3}=\frac{\frac{27}{4}}{\frac{3}{4}}=9\)
Hence the required value is 9.
Example 30. If 3 tan θ = 4, then find the value of \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\)
Solution:
Given that 3 tan θ = 4
or, tan θ = \(\frac{3}{4}\)
or, tan2 θ= \(\frac{16}{9}\)
or, sec2 θ-1 = \(\frac{16}{9}\)
or, sec2 θ = \(\frac{25}{9}\)
or sec θ = \(\frac{5}{3}\) or, cos θ = \(\frac{3}{5}\)
∴ sin θ = \(\sqrt{1-\cos ^2 \theta}\)
= \(\sqrt{1-\left(\frac{3}{5}\right)^2}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)
Now, \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sqrt{\frac{(1-\sin \theta)(1+\sin \theta)}{(1+\sin \theta)(1+\sin \theta)}}=\sqrt{\frac{1-\sin ^2 \theta}{(1+\sin \theta)^2}}\)
= \(\sqrt{\frac{\cos ^2 \theta}{(1+\sin \theta)^2}}=\frac{\cos \theta}{1+\sin \theta}=\frac{\frac{3}{5}}{1+\frac{4}{5}}\)
= \(\frac{\frac{3}{5}}{\frac{9}{5}}=\frac{3}{9}=\frac{1}{3}\)
[Aliter: Given that 3 tan θ = 4 or, tanθ = \(\frac{4}{3}\)
Now, \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}} =\sqrt{\frac{(1-\sin \theta)^2}{(1+\sin \theta)(1-\sin \theta)}}=\sqrt{\frac{(1-\sin \theta)^2}{1-\sin ^2 \theta}}\)
= \(\sqrt{\frac{(1-\sin \theta)^2}{\cos ^2 \theta}}=\frac{1-\sin \theta}{\cos \theta}=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}\)
= \(\sec \theta-\tan \theta=\sqrt{1+\tan ^2 \theta}-\tan \theta\)
= \(\sqrt{1+\left(\frac{4}{3}\right)^2-\frac{4}{3}=\sqrt{\frac{25}{9}}-\frac{4}{3}}\)
= \(\frac{5}{3}-\frac{4}{3}=\frac{1}{3}\)]
Example 31. If sec θ + tan θ = p, then find the value of cos θ.
Solution: Given that sec θ + tan θ = p ….(1)
We know that sec2 θ – tan2 θ = 1
or, (sec θ + tan θ)(sec θ – tan θ) = 1
or, p (sec θ – tan θ) = 1
or, sec θ – tan θ = \(\frac{1}{p}\)…….(2)
Adding (1) and (2) we get, 2 sec θ = p + \(\frac{1}{p}\)
or, 2 sec θ = \(\frac{p^2+1}{p}\) or, sec θ = \(\frac{p^2+1}{2p}\) or, cos θ= \(\frac{2p}{p^2+1}\)
Hence cos θ = \(\frac{2p}{p^2+1}\)
Example 32. If 7 cos2 θ + 3 sin2 θ = 4 and 0° < θ < 90°, then what is the value of tan θ?
Solution: Given that 7 cos2 θ + 3 sin2 θ = 4
or, 7(1 – sin2 θ) + 3 sin2 θ = 4
or, 7-7 sin2 θ + 3 sin2 θ = 4
or, 7 – 4 sin2 θ = 4 or, -4 sin2 θ = 4-7
or, – 4 sin2 θ =- 3
or, \(\sin ^2 \theta=\frac{3}{4}\)
or, sin θ = \(\frac{\sqrt{3}}{2}\) [0°<θ<90°]
or, sin θ = sin 60°
⇒ θ = 60°
∴ tan θ = tan 60° = √3.