SELINA Solution Class 9 Chapter 25 Complementary Angles Exercise 25

Question 1.1

Evaluate : ${\displaystyle \frac{\cos 22°}{\sin 68°}}$

Sol:

${\displaystyle \frac{\cos 22°}{\sin 68°}}$

= ${\displaystyle \frac{\cos (90°-68°)}{\sin 68°}}$

= ${\displaystyle \frac{\sin 68°}{\sin 68°}}$

 =1

Question 1.2

Evaluate: ${\displaystyle \frac{\tan 47°}{\cot 43°}}$

Sol:

${\displaystyle \frac{\tan 47°}{\cot 43°}}$

= ${\displaystyle \frac{\tan (90°- 43°)}{\cot 43°}}$

= ${\displaystyle \frac{\cot 43°}{\cot 43°}}$

= 1

Question 1.3

Evaluate: ${\displaystyle \frac{\sec 75°}{\text{cosec}15°}}$

Sol:

${\displaystyle \frac{\sec 75°}{\text{cosec}15°}}$

= ${\displaystyle \frac{\sec (90°-15°)}{\text{cosec}15°}}$

= ${\displaystyle \frac{\text{cosec}15°}{\text{cosec}15°}}$

= 1

Question 1.4

Evaluate: ${\displaystyle \frac{\cos 55°}{\sin 35°}+\frac{\cot 35°}{\tan 55°}}$

' Sol:

${\displaystyle \frac{\cos 55°}{\sin 35°}+\frac{\cot 35°}{\tan 55°}}$

= ${\displaystyle \frac{\cos (90°-35°)}{\sin 35°}+\frac{\cot (90°-55°)}{\tan 55°}}$

= ${\displaystyle \frac{\sin 35°}{\sin 35°}+\frac{\tan 55°}{\tan 55°}}$

= 1 + 1
= 2

Question 1.5

Evaluate: sin² 40° - cos² 50°

Sol:

sin² 40° - cos² 50°
= sin² (90° - 50°) - cos² 50°
= cos² 50° - cos² 50°
= 0

Question 1.6

Evaluate: sec² 18° - cosec² 72°

Sol:

sec² 18° - cosec² 72°
= [sec (90° - 72°)]² - cosec² 72°
= cosec² 72° - cosec² 72°
= 0

Question 1.7

Evaluate: sin 15° cos 15° - cos 75° sin 75°

Sol:

sin 15° cos 15° - cos 75° sin 75°
= sin(90° - 75°) cos15° - cos 75° sin (90° -15°)
= cos 75° cos 15° - cos 75° cos 15°
= 0.

Question 1.8

Evaluate: sin 42° sin 48° - cos 42° cos 48°

Sol:

sin 42° sin 48° - cos 42° cos 48°
= sin(90° - 48°)sin 48° - cos(90° - 48°)cos 48°
= cos 48° sin 48° - sin 48°cos 48°
= cos 48° sin 48° - cos 48° sin 48°
= 0

Question 2.1

Evaluate: sin (90° - A) sin A - cos (90° - A) cos A

Sol:

sin (90° - A) sin A - cos (90° - A) cos A
= cos A sin A - sin A cos A
= 0

Question 2.2

Evaluate: sin² 35° - cos² 55°

Sol:

sin² 35° - cos² 55°
= sin² 35° - [cos(90° - 35°)]²
= sin² 35° - sin² 35°
= 0

Question 2.3

Evaluate: ${\displaystyle \frac{\cot 54°}{\tan 36°}+\frac{\tan 20°}{\cot 70°}-2}$

Sol:

${\displaystyle \frac{\cot 54°}{\tan 36°}+\frac{\tan 20°}{\cot 70°}-2}$

= ${\displaystyle \frac{\cot (90°-36°)}{\tan 36°}+\frac{\tan (90°-70°)}{\cot 70°}-2}$

= ${\displaystyle \frac{\tan 36°}{\tan 36°}+\frac{\cot 70°}{\cot 70°}-2}$
= 1 + 1 - 2
= 2 - 2
= 0

Quesiton 2.4

Evaluate: ${\displaystyle \frac{2\tan 53°}{\cot 37°}-\frac{\cot 80°}{\tan 10°}}$

Sol:

${\displaystyle \frac{2\tan 53°}{\cot 37°}-\frac{\cot 80°}{\tan 10°}}$

= ${\displaystyle \frac{2\tan (90°-37°)}{\cot 37°}-\frac{\cot (90°-10°)}{\tan 10°}}$

= ${\displaystyle \frac{2\cot 37°}{\cot 37°}-\frac{\tan 10°}{\tan 10°}}$
= 2 - 1
= 1

Question 2.5

Evaluate: cos² 25° - sin² 65° - tan² 45°

Sol:

cos² 25° - sin² 65° - tan² 45°
= [cos(90° - 65°)]² - sin² 65° - (tan 45°)²
= sin² 65° - sin² 65° - (1)²
= 0 - 1
= - 1.

Question 2.6

Evaluate: ${\displaystyle {\left(\frac{\sin 77°}{\cos 13°}\right)}^2+{\left(\frac{\cos 77°}{\sin 13°}\right)}^2-2{\cos }^{2}45°}$

Sol:

${\displaystyle {\left(\frac{\sin 77°}{\cos 13°}\right)}^2+{\left(\frac{\cos 77°}{\sin 13°}\right)}^2-2{\cos }^{2}45°}$

= ${\displaystyle {\left(\frac{\sin (90°-13°)}{\cos 13°}\right)}^2+{\left(\frac{\cos (90°-13°)}{\sin 13°}\right)}^2-2{(\cos 45°)}^2}$

= ${\displaystyle {\left(\frac{\cos 13°}{\cos 13°}\right)}^2+{\left(\frac{\sin 13°}{\sin 13°}\right)}^2-2{\left(\frac{1}{\sqrt{2}}\right)}^2}$

= ${\displaystyle {\left(1\right)}^2+{\left(1\right)}^2-2\times \frac{1}{2}}$

= 1 + 1 - 1
= 1

Question 3.1

Show that: tan 10° tan 15° tan 75° tan 80° = 1.

Sol:

L.H.S.
= tan 10° tan 15° tan 75° tan 80°
= tan (90° - 80°) tan (90° - 75°) tan 75° tan 80°
= cot 80° cot 75 ° tan 75° tan 80°
= (cot 80° tan 80°)(cot 75° tan 75°)
= (1)(1)
= 1
= R.H.S.

Question 3.2

Show that: sin 42° sec 48° + cos 42° cosec 48° = 2.

Sol:

L.H.S.
= sin 42° sec 48° + cos 42° cosec 48°

= ${\displaystyle \sin (90°-48°)\times \frac{1}{\cos 48°}+\cos (90°-48°)\times \frac{1}{\sin 48°}}$

= ${\displaystyle \cos 48°\times \frac{1}{\cos 48°}+\sin 48°\times \frac{1}{\sin 48°}}$

= 1 + 1
= 2
= R.H.S.

Question 4.1

Express the following in term of angles between 0°and 45°:
sin 59°+ tan 63°

Sol:

sin 59° + tan 63°

= sin(90 - 31)° + tan(90 - 27)°

= cos 31° + cot 27°.

Question 4.2

Express the following in term of angles between 0°and 45°:
cosec 68° + cot 72°

Sol:

cosec 68° + cot 72°

= cosec(90 - 22)° + cot(90 - 18)°

(∵ cosec(90 - θ) = secθ and cot(90 - θ) = tanθ)

= sec 22° + tan 18°

Question 4.3

Express the following in term of angles between 0°and 45°:
cos 74°+ sec 67°

Sol:

cos 74°+ sec 67°

= cos(90 - 16)° + sec(90 - 23)°

= sin 16° + cosec 23°

Question 5.1

For triangle ABC, show that:

${\displaystyle \sin \left(\frac{\text{A + B}}{2}\right)=\cos \frac{\text{C}}{2}}$

Sol:

We know that for a triangle ΔABC

∠A + ∠B + ∠C = 180°

${\displaystyle \frac{\angle \text{B}+\angle \text{A}}{2}=90°-\frac{\angle \text{C}}{2}}$

${\displaystyle \sin \left(\frac{\text{A + B}}{2}\right)=\sin (90°-\frac{\text{C}}{2})}$

${\displaystyle \sin \left(\frac{\text{A + B}}{2}\right)=\cos \left(\frac{\text{C}}{2}\right)}$

Question 5.2

For triangle ABC, show that:

${\displaystyle \tan \left(\frac{\text{B}+\text{C}}{2}\right)=\cot \left(\frac{\text{A}}{2}\right)}$

Sol:

We know that for a triangle ΔABC

∠A + ∠B + ∠C = 180°

∠B + ∠C = 180° - ∠A

${\displaystyle \frac{\angle \text{B}+\angle \text{C}}{2}=90°-\frac{\angle \text{A}}{2}}$

${\displaystyle \tan \left(\frac{\text{B}+\text{C}}{2}\right)=\tan (90°-\frac{\text{A}}{2})}$

${\displaystyle \tan \left(\frac{\text{B}+\text{C}}{2}\right)=\cot \left(\frac{\text{A}}{2}\right)}$

Question 6.1

Evaluate: ${\displaystyle 3\frac{\sin 72°}{\cos 18°}-\frac{\sec 32°}{\text{cosec}58°}}$.

Sol:

${\displaystyle 3\frac{\sin 72°}{\cos 18°}-\frac{\sec 32°}{\text{cosec}58°}}$

= ${\displaystyle 3\frac{\sin (90°-18°)}{\cos 18°}-\frac{\sec (90°-58°)}{\text{cosec}58°}}$

= ${\displaystyle 3\frac{\cos 18°}{\cos 18°}-\frac{\text{cosec}58°}{\text{cosec}58°}}$

= 3 - 1

= 2

Question 6.2

Evaluate: 3 cos 80° cosec 10°+ 2 sin 59° sec 31°.

Sol:

3 cos 80° cosec 10°+ 2 sin 59° sec 31°
= 3 cos (90° - 10°) cosec 10° + 2sin (90° - 31) sec 31°
= 3 sin 10° cosec 10° + 2 cos 31° sec 31°
= 3 x 1 + 2 x 1
= 3 + 2
= 5

Question 6.3

Evaluate: ${\displaystyle \frac{\sin 80°}{\cos 10°}}$+ sin 59° sec 31°

Sol:

${\displaystyle \frac{\sin 80°}{\cos 10°}}$+ sin 59° sec 31°

= ${\displaystyle \frac{\sin (90°-10°)}{\cos 10°}}$ + sin (90° - 31°)sec 31°

= ${\displaystyle \frac{\cos 10°}{\cos 10°}+\frac{\cos 31°}{\cos 31°}}$
= 1 + 1
= 2

Question 6.4

Evaluate: tan (55° - A) - cot (35° + A)

Sol:

tan (55° - A) - cot (35° + A)

= tan [90° - (35° + A)] - cot (35° + A)

= cot (35° + A) - cot (35° + A)

= 0.

Question 6.5

Evaluate: cosec (65° + A) - sec (25° - A)

Sol:

cosec (65° + A) - sec (25° - A)

= cosec [90° - (25° - A)] - sec (25° - A)

= sec (25° - A) - sec (25° - A)

= 0.

Question 6.6

Evaluate: ${\displaystyle 2\frac{\tan 57°}{\cot 33°}-\frac{\cot 70°}{\tan 20°}-\sqrt{2}\cos 45°}$

Sol:

${\displaystyle 2\frac{\tan 57°}{\cot 33°}-\frac{\cot 70°}{\tan 20°}-\sqrt{2}\cos 45°}$

= ${\displaystyle 2\frac{\tan (90°-33°)}{\cot 33°}-\frac{\cot (90°-20°)}{\tan 20°}-\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)}$

= ${\displaystyle 2\frac{\cot 33°}{\cot 33°}-\frac{\tan 20°}{\tan 20°}-1}$

= 2 - 1 - 1

= 0.

Question 6.7

Evaluate: ${\displaystyle \frac{{\cot }^{2}41°}{{\tan }^{2}49°}-2\frac{{\sin }^{2}75°}{{\cos }^{2}15°}}$

Sol 1:

${\displaystyle \frac{{\cot }^{2}41°}{{\tan }^{2}49°}-2\frac{{\sin }^{2}75°}{{\cos }^{2}15°}}$

= ${\displaystyle \frac{{\left[\cot (90°-49°)\right]}^2}{{\tan }^{2}49°}-2\frac{{\left[\sin (90°-15°)\right]}^2}{{\cos }^{2}15°}}$

= ${\displaystyle \frac{{\tan }^{2}49°}{{\tan }^{2}59°}-2\frac{{\cos }^{2}15°}{{\cos }^{2}15°}}$

= 1 - 2
= -1

Sol 2:

${\displaystyle \frac{{\cot }^{2}41°}{{\tan }^{2}49°}-2\frac{{\sin }^{2}75°}{{\cos }^{2}15°}}$

= ${\displaystyle \frac{{\left[\cot (90°-49°)\right]}^2}{{\tan }^{2}49°}-2\frac{{\left[\sin (90°-15°)\right]}^2}{{\cos }^{2}15°}}$

= ${\displaystyle \frac{{\tan }^{2}49°}{{\tan }^{2}49°}-2\frac{{\cos }^{2}15°}{{\cos }^{2}15°}}$

= 1 - 2
= -1

Question 6.8

Evaluate: ${\displaystyle \frac{\cos 70°}{\sin 20°}+\frac{\cos 59°}{\sin 31°}-8{\sin }^{2}30°}$

Sol:

${\displaystyle \frac{\cos 70°}{\sin 20°}+\frac{\cos 59°}{\sin 31°}-8{\sin }^{2}30°}$

= ${\displaystyle \frac{\cos (90°-20°)}{\sin 20°}+\frac{\cos (90°-31°)}{\sin 31°}-8{\left(\frac{1}{2}\right)}^2}$

= ${\displaystyle \frac{\sin 20°}{\sin 20°}+\frac{\sin 31°}{\sin 31°}-2}$

= 1 + 1 - 2
 = 0

Question 6.9

Evaluate: 14 sin 30°+ 6 cos 60°- 5 tan 45°.

Sol:

14 sin 30°+ 6 cos 60°- 5 tan 45°

= ${\displaystyle 14\left(\frac{1}{2}\right)+6\left(\frac{1}{2}\right)-5\left(1\right)}$

= 7 + 3 - 5

= 5.

Question 7

A triangle ABC is right-angled at B; find the value of ${\displaystyle \frac{\sec \text{A}.\sin \text{C}-\tan \text{A}.\tan \text{C}}{\sin \text{B}}}$.

Sol:

Since Δ ABC is a right angled triangle, right angled at B,
A + C = 90°

∴ ${\displaystyle \frac{\sec \text{A}.\sin \text{C}-\tan \text{A}.\tan \text{C}}{\sin \text{B}}}$

= ${\displaystyle \frac{\sec \text{A}(90°-\text{C})\sin \text{C}-\tan (90°-\text{C})\tan \text{C}}{\sin 90°}}$

= ${\displaystyle \frac{\text{cosec} \text{C}\sin \text{C}-\cot \text{C}\tan \text{C}}{1}}$

= ${\displaystyle \frac{1}{\sin \text{C}}\times \sin \text{C}-\frac{1}{\tan \text{C}}\times \tan \text{C}}$

= 1 - 1
= 0

Question 8.1

In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
sin (90° - 3A).cosec 42° = 1.

Sol:

Sin (90° - 3A). cosec 42° = 1

⇒ sin (90° - 3A) = ${\displaystyle \frac{1}{\text{cosec}42°}}$

⇒ cos 3A = ${\displaystyle \frac{1}{\text{cosec}(90°-48°)}}$

⇒ cos 3A = ${\displaystyle \frac{1}{\sec 48°}}$

⇒ cos 3A = cos 48°

⇒ 3A = 48°

⇒ A = 16°.

Question 8.2

In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
cos(90° - A).sec 77° = 1

Sol:

cos (90° - 3A). sec 77° = 1

⇒ cos(90° - 3A) = ${\displaystyle \frac{1}{\sec 77°}}$

⇒ sin 3A = ${\displaystyle \frac{1}{\sec (90-12°)}}$

⇒ sin 3A = ${\displaystyle \frac{1}{\text{cosec}12°}}$

⇒ sin 3A - sin 12°
⇒ 3A = 12°
⇒ A = 3°