WBBSE Solutions For Class 10 Maths Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles

What is complementary angles? 

Complementary angles

If the sum of any two angles of a triangle be 90° or one right angle, then the angles are called complementary angles to each other.

For example, Let ABC be a right-angled triangle of which ∠ACB = θ.

Then, ∠BAC = 90°- θ.

So that ∠ACB + ∠BAC = θ + 90° – θ = 90°.

Hence ∠ACB and ∠BAC are complementary angles to each other.

WBBSE Solutions for Class 10 Maths

Determination of trigonometrical ratios of complementary

Trigonometric ratios of (90° – θ)

Let straight lines XOX’ and YOY’ intersect each other at right angles at O and ∠XOP = 90°-0. So, ∠POY = θ.

Let us draw PN ⊥ OX and PM ⊥ OY.

∴ PN = OM and PM = ON,

Now, sin (90° – θ) = \(\frac{\mathrm{PN}}{\dot{\mathrm{OP}}}\) [by the definition of sin θ]

= \(\frac{\mathrm{OM}}{\mathrm{OP}}=\frac{\text { base } .}{\text { hypotenuse }}=\cos \theta\)

Similarly, cos (90°-θ) = \(\frac{\mathrm{ON}}{\mathrm{OP}}\) [by definition] = \(\frac{\mathrm{PM}}{\mathrm{OP}}\)

= \(\frac{\text { perpendicular }}{\text { hypotenuse }}=\sin \theta\)

tan (90°- 0) = \(\frac{\mathrm{PN}}{\mathrm{ON}}\) [ by definition ]

= \(\frac{\mathrm{OM}}{\mathrm{PM}}=\frac{\text { base }}{\text { perpendicular }}=\cot \theta\)

Also, \(\tan \left(90^{\circ}-\theta\right)=\frac{\sin \left(90^{\circ}-\theta\right)}{\cos \left(90^{\circ}-\theta\right)}=\frac{\cos \theta}{\sin \theta}=\cot \theta\)

\({cosec}\left(90^{\circ}-\theta\right)=\frac{\mathrm{OP}}{\mathrm{PN}}\) [by definition]= \(\frac{\mathrm{OP}}{\mathrm{OM}}\)

= \(\frac{\text { hypotenuse }}{\text { base }}=\sec \theta\)

Also, \({cosec}\left(90^{\circ}-\theta\right)=\frac{1}{\sin \left(90^{\circ}-\theta\right)}=\frac{1}{\cos \theta}=\sec \theta\)

\(\sec \left(90^{\circ}-\theta\right)=\frac{\mathrm{OP}}{\mathrm{ON}}\) [by definition]

= \(\frac{\mathrm{OP}}{\mathrm{PM}}=\frac{\text { hypotenuse }}{\text { perpendicular }}= {cosec} \theta\)

Also, \(\sec \left(90^{\circ}-\theta\right)=\frac{1}{\cos \left(90^{\circ}-\theta\right)}=\frac{1}{\sin \theta}={cosec} \theta\)

Again, \(\cot\left(90^{\circ}-\theta\right)=\frac{\mathrm{ON}}{\mathrm{PN}}\) [by definition]

= \(\frac{\mathrm{PM}}{\mathrm{OM}}=\frac{\text { perpendicular }}{\text { base }}=\tan \theta\)

Also, \(\cot \left(90^{\circ}-\theta\right)=\frac{\cos \left(90^{\circ}-\theta\right)}{\sin \left(90^{\circ}-\theta\right)}=\frac{\sin \theta}{\cos \theta}=\tan \theta\)

Hence We get,

1. sin (90° – θ) = cos θ
2. cos (90° – θ) = sinθ
3. tan (90° – θ) = cot θ
4. cosec (90° – θ) – sec θ
5. sec (90° – θ) = cosec θ
6. cot (90° – θ) = tan θ

From the above feature we see that, the trigonometric ratios of complementary angles change as per the following rules:

sin ⇔ cos
cosec ⇔ sec
tan ⇔ cot

Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles Multiple Choice Questions

Example 1. The value of (sin 43° cos 47° + cos 43° sin 47°) is

1. 1
2. sin 4°
3. cos 4°
4. none of these

Solution: sin 43° cos 47° + cos 43° sin 47°

= sin 43° cos (90° – 43°) + cos 43° sin (90° – 43°)

= sin 43° sin 43° + cos 43° cos 43°

= sin2 43° + cos2 43° =1.

The value of (sin 43° cos 47° + cos 43° sin 47°) is 1.

∴ 1. 1 is correct.

Example 2. The value of \(\left(\frac{\tan 35^{\circ}}{\cot 55^{\circ}}+\frac{\cot 78^{\circ}}{\tan 12^{\circ}}\right)\) is

1. 0
2. 1
3. 2
4. 3

Solution: \(\left(\frac{\tan 35^{\circ}}{\cot 55^{\circ}}+\frac{\cot 78^{\circ}}{\tan 12^{\circ}}\right)\)

= \(\frac{\tan 35^{\circ}}{\cot \left(90^{\circ}-35^{\circ}\right)}+\frac{\cot 78^{\circ}}{\tan \left(90^{\circ}-78^{\circ}\right)}=\frac{\tan 35^{\circ}}{\tan 35^{\circ}}+\frac{\cot 78^{\circ}}{\cot 78^{\circ}}\)

= 1 + 1 = 2

The value of \(\left(\frac{\tan 35^{\circ}}{\cot 55^{\circ}}+\frac{\cot 78^{\circ}}{\tan 12^{\circ}}\right)\) is 2

∴ 3. 2 is correct.

Example 3. The value of { cos (40° + θ) – sin (50° – θ) } is

1. 2 cos θ
2. 7 sin θ
3. 0
4. none of these

Solution: cos (40° + θ) – sin (50° – θ)

= cos (40° + θ) – sin { 90° – (40° + θ)}

= cos (40° + θ) – cos (40° + θ) = 0

The value of { cos (40° + θ) – sin (50° – θ) } is 0

∴ 2. 0 is correct

Example 4. ABC is a triangle, sin\(\left(\frac{B+C}{2}\right)\)

1. sin\(\frac{\mathbf{A}}{2}\)
2. cos\(\frac{\mathbf{A}}{2}\)
3. sin A
4. cos A

Solution: Since ABC is a triangle,

∴ ∠A + ∠B + ∠C = 180° or, ∠B + ∠C = 180° – ∠A

or, \(\frac{\angle \mathrm{B}+\angle \mathrm{C}}{2}=90^{\circ}-\frac{\angle \mathrm{A}}{2}\)

∴ \(\sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=\sin \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)=\cos \frac{\mathrm{A}}{2}\)

∴ 2. cos\(\frac{\mathbf{A}}{2}\) is correct.

Example 5. If A + B = 90° and tan A = \(\frac{3}{4}\), then the value of cot B is

1. \(\frac{3}{4}\)
2. \(\frac{4}{3}\)
3. 3
4. 4

Solution: Given that A + B = 90° or, A = 90° – B

∴ tan A = \(\frac{3}{4}\) ⇒ tan (90°-B) = \(\frac{3}{4}\) ⇒ cot B = \(\frac{3}{4}\)

The value of cot B is \(\frac{3}{4}\)

∴ 1. \(\frac{3}{4}\) is correct.

Example 6. If sin 5θ = cos 4θ, then the value of θ is

1. 1°
2. 10°
3. 30°
4. 45°

Solution: sin 5θ = cos 4θ

or, sin5θ = sin (90° – 4θ)

∴ 5θ = 90° – 4θ

or, 5θ + 4θ = 90°

or, 9θ = 90°

or, θ = \(\frac{90^{\circ}}{9}\) =10°

The value of θ is 10°

∴ 2. 10° is correct.

Example 7. If tan 2θ = cot (θ + 15°), then the value of θ is

1. 5°
2. 10°
3. 20°
4. 25°

Solution: tan 2θ = cot (θ + 15°)

or, cot (90° – 2θ) = cot (θ + 15°)

∴ 90° – 2θ = θ + 15°

or, – 2θ – θ = 15° – 90°

or, -3θ = – 75°

or,θ =  \(\frac{-75^{\circ}}{-3}\) = 25°

The value of θ is 25°

∴ 4. 25° is correct

Example 8. If A + B = 90°, then the value of sin²A + sin²B =

1. -1
2. 0
3. 1
4. \(\frac{1}{\sqrt{2}}\)

Solution: A + B = 90° ⇒ B = 90°- A.

Now, sin²A + sin²B

= sin²A + sin² (90° – A)

= sin²A + cos²A = 1

The value of sin²A + sin²B =1

∴ 3. 1 is correct.

Example 9. \(\frac{\sin 30^{\circ} 17^{\prime}}{\cos 59^{\circ} 43^{\prime}}\) =

1. -1
2. \(\frac{1}{2}\)
3. \(\frac{1}{\sqrt{2}}\)
4. 1

Solution: \(\frac{\sin 30^{\circ} 17^{\prime}}{\cos 59^{\circ} 43^{\prime}}=\frac{\sin 30^{\circ} 17^{\prime}}{\cos \left(90^{\circ}-30^{\circ} 17^{\prime}\right)}\)

= \(\frac{\sin 30^{\circ} 17^{\prime}}{\sin 30^{\circ} 17^{\prime}}=1\)

∴ 4. 1 is correct

Example 10. tan 1° tan 2° tan 3° ………. tan 89° =

1. 0
2. 1
3. √3
4. undefined

Solution: tan 1° tan 2° tan 3° ………. tan 89°

= tan (90° – 89°) tan (90° – 88°) tan (90° – 87°)………..tan 87° tan 88° tan 89°.

= (cot 89° tan 89°) (cot 88° tan 88°) (cot 87° tan 87°)……… (cot 44° tan 44°) tan 45°

= \(\begin{aligned}
&\left(\cot 89^{\circ} \cdot \frac{1}{\cot 89^{\circ}}\right)\left(\cot 88^{\circ} \cdot \frac{1}{\cot 88^{\circ}}\right)\left(\cot 87^{\circ} \cdot \frac{1}{\cot 87^{\circ}}\right) \\
& \cdots \cdots\left(\cot 44^{\circ} \cdot \frac{1}{\cot 44^{\circ}}\right) \tan 45^{\circ}
\end{aligned}\)

= 1. 1..1. 1. ………1 = 1

∴ 2. 1 is correct.

Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles True Or False

Example 1. If the sum of two angles is 90°, then they are called complementary angles to each other.

Solution: True

Example 2. The value of cos 54° and sin 36° are equal.

Solution: True

since cos 54° = cos (90° – 36°) = sin 36°.

Example 3. The simplified value of (sin 12° – cos 78°) is 1.

Solution: False

since (sin 12° – cos 78°) = Sin 12° – cos (90° – 12°) = sin 12° – sin 12° = 0.

Example 4. The values of sin 72° and cos 108° are equal.

Solution: False

since sin- 72° = sin (180° – 108°) = sin 108°.

Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles Fill In The Blanks

Example 1. The value of (tan 15° x tan 45° x tan 60° x tan 75°) is ______

Solution: √3

since, We have, tan 15° x tan 45° x tan 60° x tan 75°

= (tan 15° x tan 75°) x tan 45° x tan 60°.

= tan 15° x cot 15° x 1 x √3 [∵ tan 75° = tan (90° – 15°) = cot 15°]

= tan 15° x \(\frac{1}{\tan 15^{\circ}}\) x √3 = 1 x √3 = √3.

The value of (tan 15° x tan 45° x tan 60° x tan 75°)  =√3.

Example 2. The value of (sin 12° x cos 18° x sec 78° x cosec 72°) is_______

Solution: 1

since, We have, sin 12° x cos 18° x sec 78° x cosec 12°

= sin 12° x cos 18° x sec (90°  – 12°) x cosec (90° – 18°)

= sin 12° x cos 18° x cosec 12° x sec 18°

= sin 12° x cos 18° x \(\frac{1}{\sin 12^{\circ}} \times \frac{1}{\cos 18^{\circ}}\) = 1 x 1 = 1.

The value of (sin 12° x cos 18° x sec 78° x cosec 72°)= 1.

Example 3. If A and B are complementary to each other, then sin A = _______

Solution: cos B

since A and B are complementary, then A + B = 90° or, A = 90° – B

⇒ sin A = sin (90°-B) = cos B.

sin A = cos B.

Example 4. cos 72° – sin 18° = ______

Solution: 0

since cos 72° – sin 18° = cos 72° – sin (90° – 12°) = cos 12° – cos 72° = 0.

cos 72° – sin 18° = 0.

Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles Short Answer Type Questions

Example 1. If tan 4θ x tan 6θ = 1 and 6θ is a positive acute angle, then find the value of θ.

Solution: Given that tan 4θ x tan 6θ = 1.

⇒ tan 4θ = \(\frac{1}{\tan 6 \theta}\) ⇒ tan 4θ = cot 6θ

⇒ tan 4θ ⇒ tan (90°-6θ)

⇒ 4θ = 90°-6θ ⇒ 4θ + 6θ = 90°

⇒ 10θ = 90°

⇒ θ = \(\frac{90^{\circ}}{10}\) ⇒ θ = 9°

Hence the value of θ is 9°

Example 2. If sec 5A = cosec (A + 36°) and 5A is a positive acute angle, then find the value of A.

Solution: Given that sec 5A = cosec (A + 36°)

or, cosec (90° – 5A) = cosec (A + 36°)

⇒ 90°- 5A = A + 36°

⇒ – 5A – A = 36°- 90°

⇒ -6A = -54°

⇒ A = \(\frac{-54^{\circ}}{-6}\)

⇒ A = 9°.

Hence the required value of A is 9°.

Example 3. Find the value of \(\frac{2 \sin ^2 63^{\circ}+1+2 \sin ^2 27^{\circ}}{3 \cos ^2 17^{\circ}-2+3 \cos ^2 73^{\circ}}\)

Solution: \(\frac{2 \sin ^2 63^{\circ}+1+2 \sin ^2 27^{\circ}}{3 \cos ^2 17^{\circ}-2+3 \cos ^2 73^{\circ}}\)

= \(\frac{2 \sin ^2 63^{\circ}+2 \sin ^2\left(90^{\circ}-63^{\circ}\right)+1}{3 \cos ^2 17^{\circ}-2+3 \cos ^2\left(90^{\circ}-17^{\circ}\right)}=\frac{2 \sin ^2 63^{\circ}+2 \cos ^2 63^{\circ}+1}{3 \cos ^2 17^{\circ}-2+3 \sin ^2 17^{\circ}}\)

= \(\frac{2\left(\sin ^2 63^{\circ}+\cos ^2 63^{\circ}\right)+1}{3\left(\cos ^2 17^{\circ}+\sin ^2 17^{\circ}\right)-2}=\frac{2 \times 1+1}{3 \times 1-2}\)

= \(\frac{2+1}{3-2}=\frac{3}{1}=3\)

Hence the required value is 3.

Example 4. Find the value of tan 1° x tan 2° x tan 3° x……..tan 89°

Solution: tan 1° x tan 2° x tan 3° x………x tan 89°

= tan 1° x tan 2° x tan 3° x…………x tan 87° x tan 88° x tan 89°.

= tan 1° x tan 2° x tan 3° x…….x tan (90° – 3°) x tan (90° – 2°)x tan (90° – 1°)

= tan 1° x’ tan 2° x tan 3° x………..x cot 3° x cot 2° x cot 1°

= (tan 1° x cot 1°) x (tan 2° x cot 2°) x (tan 3° x cot 3°) x……….x (tan 44° x cot 44°) x tan 45°x (tan 44° x cot 44°) x tan 45°

= 1 X 1 X 1 X……….x 1 x tan 45° =  1 x 1 = 1.

Hence the required value is 1.

Example 5. Find the value of cot 17° \(\left(\cot 73^{\circ} \cos ^2 22^{\circ}+\frac{1}{\tan 73^{\circ} \sec ^2 68^{\circ}}\right)\)

Solution: cot 17° \(\left(\cot 73^{\circ} \cos ^2 22^{\circ}+\frac{1}{\tan 73^{\circ} \sec ^2 68^{\circ}}\right)\)

= cot 17° (cot 73° cos² 22° + cot 73° cos² 68°)

= cot (90° – 73°) { cot 73° cos² 22° + cot 73° cos² (90° – 22°)}

= tan 73°. cot 73° (cos² 22° + sin² 22°)

= tan 73°. \(\frac{1}{\tan 73^{\circ}}\) x 1 = 1×1 = 1.

Hence the required value = 1.

Example 6. If sin 10θ = cos 8θ and 10θ is a positive acute angle, then find the value of tan 9θ.

Solution: Given that sin 10θ = cos 8θ and 10θ is acute.

∴ sin 10θ = sin (90° – 8θ)

⇒ 10θ = 90°-8θ

⇒10θ + 8θ = 90°

⇒ 18 θ = 90°

⇒ θ = \(\frac{90^{\circ}}{18}\)

⇒ θ = 5°

⇒ 9θ = 9×5° = 45°

⇒ tan 9θ = tan 45°

⇒ tan 9θ = 1.

Hence the value of tan 9θ = 1.

Example 7. If α + β = \(\frac{\pi}{2}\) then prove that \(\cos \alpha=\sqrt{\frac{\sin \alpha}{\cos \beta}-\sin \alpha \cos \beta}\)

Solution: LHS = cos α

RHS = \(\cos \alpha=\sqrt{\frac{\sin \alpha}{\cos \beta}-\sin \alpha \cos \beta}\)

= \(\sqrt{\frac{\sin \alpha}{\cos \left(90^{\circ}-\alpha\right)}-\sin \alpha \cos \left(90^{\circ}-\alpha\right)}\)

[∵ α + β = 90° ⇒ β = 90°-α]

= \(\sqrt{\frac{\sin \alpha}{\sin \alpha}-\sin \alpha \cdot \sin \alpha}\)

= \(\sqrt{1-\sin ^2 \alpha}=\sqrt{\cos ^2 \alpha}=\cos \alpha\)

∴ LHS = RHS. [Proved]

Example 8. If sin 17° = \(\frac{x}{y}\), then prove that sec 17°- sin 73° = \(=\frac{x^2}{y \sqrt{y^2-x^2}}\)

Solution: LHS = sec 17° – sin 73°

= \(\frac{1}{\cos 17^{\circ}}\) -sin (90° -17°)

= \(\frac{1}{\cos 17^{\circ}}\) -cos 17° = \(\frac{1-\cos ^2 17^{\circ}}{\cos 17^{\circ}}\)

= \(\frac{\sin ^2 17^{\circ}}{\sqrt{1-\sin ^2 17^{\circ}}}=\frac{\left(\frac{x}{y}\right)^2}{\sqrt{1-\left(\frac{x}{y}\right)^2}}\)

= \(\frac{\frac{x^2}{y^2}}{\sqrt{1-\frac{x^2}{y^2}}}\)

= \(\frac{\frac{x^2}{y^2}}{\sqrt{\frac{y^2-x^2}{y^2}}}\)

= \(\frac{\frac{x^2}{y^2}}{\frac{\sqrt{y^2-x^2}}{y}}\)

= \(\frac{x^2}{y^2} \times \frac{y}{\sqrt{y^2-x^2}}=\frac{x^2}{y \sqrt{y^2-x^2}}\)

Hence sec 17° – sin 73° = \(\frac{x^2}{y \sqrt{y^2-x^2}}\) [proved]

Example 9. If sin 51° = \(\frac{a}{\sqrt{a^2+b^2}}\), then find the value of tan 39°.

Solution: sin 51° = \(\frac{a}{\sqrt{a^2+b^2}}\)

or, \(\sin ^2 51^{\circ}=\frac{a^2}{a^2+b^2}\) [squaring]

or, \(\sin ^2\left(90^{\circ}-39^{\circ}\right)=\frac{a^2}{a^2+b^2}\)

or, \(\cos ^2 39^{\circ}=\frac{a^2}{a^2+b^2}\)

or, \(\sec ^2 39^{\circ}=\frac{a^2+b^2}{a^2}\)

or, \(1+\tan ^2 39^{\circ}=\frac{a^2+b^2}{a^2}\)

or, \(\tan ^2 39^{\circ}=\frac{a^2+b^2}{a^2}-1\)

or, \(\tan ^2 39^{\circ}=\frac{a^2+b^2-a^2}{a^2}\)

or, \(\tan ^2 39^{\circ}=\frac{b^2}{a^2}\)

or, \(\tan 39^{\circ}=\frac{b}{a}\)

Hence \(\tan 39^{\circ}=\frac{b}{a}\)

Example 10. If √2 sin (α-β) =1 and α, β are complementary, then find the value of α and β.

Solution: Given that √2 sin (α- β) = 1

or, sin (α- β) = \(\frac{1}{\sqrt{2}}\) = sin 45°

⇒ α – β = 45° …………(1)

Also, α + β = 90° …………(2) [∵ α, β are complementary. ]

Now, adding (1) and (2) we get,

2α = 135° or, α = \(\frac{135^{\circ}}{2}\) = 67.5°.

From (2) we get, β = 90° – α = 90° – 67.5° = 22.5°

Hence α = 67.5° and β = 22.5°.

Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles Long Answer Type Questions

Example 1. Prove that if two angles α and β are complementary angles, then

1. sin²α + sin²β = 1
2. cotβ + cosβ = \(\frac{\cos \beta}{\cos \alpha}\) (1 +sinβ).
3. \(\frac{\sec \alpha}{\cos \alpha}\) – cot²β = 1

Solution:

1. LHS = sin²α + sin²β

= sin²α + sin²(90° – α) [∵ α +β = 90° ]

= sin²α + cos²α = 1.

Hence sin²α + cos²α = 1. [Proved]

2. LHS = cotβ + cosβ = \(\frac{\cos \beta}{\cos \beta}\) + cos β = \(\frac{\cos \beta}{\sin \left(90^{\circ}-\alpha\right)}\) + cos β

= \(\frac{\cos \beta}{\cos \alpha}+\cos \beta=\frac{\cos \beta}{\cos \alpha}(1+\cos \alpha)\)

= \(\frac{\cos \beta}{\cos \alpha}\left\{1+\cos \left(90^{\circ}-\beta\right)\right\}=\frac{\cos \beta}{\cos \alpha}(1+\sin \beta)\)

Hence cot β + cos β = \(\frac{\cos \beta}{\cos \alpha}(1+\sin \beta)\) [proved]

3. LHS = \(\frac{\sec \alpha}{\cos \alpha}\) – cot²β

= sec. \(\frac{1}{\cos \alpha}\) – cot²(90°-α) [∵ α + β = 90° ]

= sec α.sec α- tan²α = sec²α- tan²α = 1.

Example 2. Prove that sec²12°- \(\frac{1}{\tan ^2 78^{\circ}}\) = 1.

Solution: sec²12° – \(\frac{1}{\tan ^2 78^{\circ}}\)

= sec²12° – cot²78° = sec²12° – cot² (90° – 12°)

= sec²12° – tan²12° = 1.

Hence sec²12° – \(\frac{1}{\tan ^2 78^{\circ}}\) = 1. [Proved]

Example 3. If A + B = 90°, then prove that 1 + \(\frac{\tan A}{\tan B}\)= sec²A.

Solution: A + B = 90° ⇒ B = 90°- A

∴ \(1+\frac{\tan \mathrm{A}}{\tan \mathrm{B}}=1+\frac{\tan \mathrm{A}}{\tan \left(90^{\circ}-\mathrm{A}\right)}\)

= \(1+\frac{\tan \mathrm{A}}{\cot \mathrm{A}}=1+\tan \mathrm{A} \cdot \tan \mathrm{A}\)

= 1 + tan²A = sec²A

Hence \(1+\frac{\tan \mathrm{A}}{\tan \mathrm{B}}\) = sec²A. ( Proved )

Example 4. Prove that

1. cosec²48° – tan²42° = 1
2. sec 70° sin 20° + cos 20° cosec 70° = 2.

Solution:

1. cosec²48° – tan²42°

= cosec²(90° – 42°) – tan²42

= sec²42° – tan²42° =1.

Hence cosec²48° – tan²42° = 1 . [ Proved ]

2. sec 70° sin 20° + cos 20° cosec 70°

= sec 70° sin (90° – 70°) + cos 20° cosec (90° – 20°)

= sec 70° cos 70° + cos 20° sec 20°

= \(\frac{1}{\cos 70^{\circ}}\). cos 70° + cos 20°. \(\frac{1}{\cos 20^{\circ}}\)

= 1 + 1=2.

Hence sec 70° sin 20° + cos 20° cosec 70° = 2. [ Proved ]

Example 5. Prove that cosec²22° cot²68° = sin²22° + sin²68° + cot²68°.

Solution: cosec²22° cot²68° = cosec²(90° – 68°) cot²68°

= sec²68° cot²68° = (1 + tan²68°) cot²68°

= cot²68° + tan²68° cot²68°

= cot²68° + tan²68°. \(\frac{1}{\tan ^2 68^{\circ}}\)

= cot²68° + 1

= cot²68° + sin²22° + cos²22°

= cot²68° + sin²22° + cos²(90° – 68°)

= cot²68° + sin²22° + sin²68°

= sin²22° + sin²68° + cot²68°.

Hence cosec²22° cot²68° = sin²22° + sin²68° + cot²68°. [Proved]

Example 6. Prove that cot 12° cot 38° cot 52° cot 78° cot 60°= \(\frac{1}{\sqrt{3}}\)

Solution: cot 12° cot 38° cot 52° cot 78° cot 60°

= (cot 12° cot 78°) (cot 38° cot 52°) cot 60°

= {cot 12° cot (90° – 12°)} [cot 38° cot (90° – 38°)} cot 60°.

= (cot 12° tan 12°) (cot 38° tan 38°) cot 60°.

= 1 x 1 x \(\frac{1}{\sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\)

Hence cot 12 cot 38° cot 52° cot 78° cot 60° = \(\frac{1}{\sqrt{3}}\) [proved]

Example 7. ABCD Is a rectangular figure, joining A, C to prove that

1. tan ∠ACD = cot ∠ACB
2.  tan² ∠CAD + 1 = \(\frac{1}{\sin ^2 \angle B A C}\)

Solution:

 

1. tan ∠ACD

= tan (90° – ∠ACB) [∵ ∠ACD + ∠ACB = 90°]

= cot ∠ACB.

Hence tan ∠ACD = cot ∠ACB. [Proved].

2. tan² ∠CAD + 1

= sec² ∠CAD

= \(\frac{1}{\cos ^2 \angle \mathrm{CAD}}\)

= \(\frac{1}{\cos ^2\left(90^{\circ}-\angle \mathrm{BAC}\right)}\)

[∵ ∠CAD + ∠BAC = 90°]

= \(\frac{1}{\sin ^2 \angle \mathrm{CAD}}\)

Hence tan² ∠CAD + 1 = \(\frac{1}{\sin ^2 \angle \mathrm{CAD}}\) [proved]

Example 8. Find the value of sin² 5° + sin² 10° + sin² 15° +…….+ sin² 90°.

Solution: sin² 5° + sin² 10° + sin² 15° + ………+ sin² 85° + sin² 90°.

= (sin² 5° + sin² 85°) + (sin² 10° + sin² 80°) + (sin² 15° + sin² 75°) + (sin² 20° + sin² 70°) + (sin² 25° + sin² 65°) + (sin² 30° + sin² 60°) + (sin² 35° + sin² 55°) + (sin² 40° + sin² 50°) + sin² 45° + sin² 90°

= {sin² 5° + sin² (90° – 5°)} + {sin² 10° + sin² (90° – 10°)} + {sin² 15° + sin² (90°-15°)} + ……… + sin² 45° + sin² 90°.

= (sin² 5° + cos² 5°) + (sin² 10° + cos² 10°) + (sin² 15° + cos² 15°) ……………..+ sin² 45° + sin² 90°.

= 1 + 1 + 1 + (upto 8 terms) + \(\left(\frac{1}{\sqrt{2}}\right)^2\) + 1

= 8 + \(\frac{1}{2}\) + 1 =9 \(\frac{1}{2}\)

Hence the required value = 9 \(\frac{1}{2}\)

Example 9. AOB is a diameter of a circle with center O and C is any point on the circle, joining A, C; B, C; and O, C, prove that

1. tan ∠ABC = cot ∠ACO
2. sin² ∠BCO + sin² ∠ACO = 1.
3. cosec² ∠CAB – 1 = tan² ∠ABC.

Solution:

since AOB is a diameter and C is any point on the circle,

∴ ∠ACB is a semicircular angle.

∴ ∠ACB = 90°

∴ AB is a hypotenuse of the right-angled triangle ABC.

Again, since ∠ACB = 90°, ∴ ∠BAC + ∠CBA = 90°.

Now, 1. tan ∠ABC = tan (90° – ∠CAB)

= cot ∠CAB……….(1)

Again, in ΔAOC, OA = OC [∵ radii of same circle]

⇒ ∠OAC = ∠ACO

⇒ ∠CAB = ∠ACO

∴ from (1) we get, tan ∠ABC = cot ∠ACO. [ Proved ]

2. sin² ∠BCO + sin² ∠ACO.

= sin² ∠BCO + sin² ∠OAC [OC = OA, ∴ ∠ACO = ∠OAC. ]

= sin² ∠BCO + sin² ∠CAB

= sin² ∠BCO + sin² (90° – ∠ABC)

= sin² ∠BCO + cos² ∠ABC

= sin²∠BCO + cos² ∠OBC

= sin² ∠BCO + cos² ∠BCO [OB = OC, ∠OBC =∠BCO ]

= 1.

Hence sin² ∠BCO + sin²∠ACO = 1 [Proved]

3. cosec² ∠CAB – 1

= cot² ∠CAB

= cot² (90° – ∠ABC)

= tan² ∠ABC.

Hence cosec² ∠CAB – 1 = tan² ∠ABC. [ Proved ]

Example 10. If sin α – cos α = 0 find the value of \(\sec \left(\frac{\pi}{2}-\alpha\right)+ {cosec}\left(\frac{\pi}{2}-\alpha\right)\)

Solution: \(\sec \left(\frac{\pi}{2}-\alpha\right)+ {cosec}\left(\frac{\pi}{2}-\alpha\right)\)

= cosec α + sec α = \(=\frac{1}{\sin \alpha}+\frac{1}{\cos \alpha}=\frac{\cos \alpha+\sin \alpha}{\sin \alpha \cos \alpha}\)

= \(\frac{\sin \alpha+\sin \alpha}{\sin \alpha \sin \alpha}\) [∵ sinα – cosα= 0 ⇒  sin α= cos α]

Again, sin α = cos α = sin (90° – α)

⇒ α = 90°- α or, 2α= 90° or, α = \(\frac{90^{\circ}}{2}\) = 45°.

∴ \(\frac{\sin \alpha+\sin \alpha}{\sin \alpha \cdot \sin \alpha}\)

= \(\frac{2 \sin \alpha}{\sin ^2 \alpha}=\frac{2}{\sin \alpha}\)

= \(\frac{2}{\sin 45^{\circ}}=\frac{2}{\frac{1}{\sqrt{2}}}=2 \sqrt{2}\)

Hence \(\sec \left(\frac{\pi}{2}-\alpha\right)+ {cosec}\left(\frac{\pi}{2}-\alpha\right)\)= 2√2.

Example 11. Prove that tan 20° + tan 70° = \(\frac{\sec ^2 20^{\circ}}{\sqrt{\sec ^2 20^{\circ}-1}}\)

Solution: tan 20° + tan 70°

= tan 20° + tan (90° – 20°)

= tan 20° + cot 20°

= tan 20° + \(\frac{1}{\tan 20^{\circ}}=\frac{\tan ^2 20^{\circ}+1}{\tan 20^{\circ}}\)

= \(\frac{\sec ^2 20^{\circ}}{\sqrt{\tan ^2 20^{\circ}}}=\frac{\sec ^2 20^{\circ}}{\sqrt{\sec ^2 20^{\circ}-1}}\)

Hence tan 20° + tan 70° = \(\frac{\sec ^2 20^{\circ}}{\sqrt{\sec ^2 20^{\circ}-1}}\) [Proved]