R.D. Sharma Solutions Class 10th: Ch 5 Trigonometric Ratios Exercise 5.3

Chapter 5 Trigonometric Ratios R.D. Sharma Solutions for Class 10th Math Exercise 5.3

Exercise 5.3

1. Evaluate the following .
(i) sin 20° / cos70° 

Solution

(ii) cos19°/sin71° 

Solution

(iii) sin21° / cos 69° 

Solution

(iv) tan10° / cot 80°  

Solution

(v) sec 11° / cosec 79°    

Solution

2. Evaluate the following : 

(i) (sin 49° / cos 41°)²  + (cos 41° / sin 49°)² 

Solution

(ii) cos 48° - sin 42°

Solution

(iii) cot 40°/ tan 50° - 1/2 (cos 35°/sin 55°)

Solution

(iv) (sin 27°/cos 63°)² - (cos 63°/sin 27°)²
Solution

(v) (tan 35°/ cot 55°) + (cot 78° / tan 12°) - 1

Solution

 

(vi) sec 70°/cosec 20° + sin 59°/cos 3°

Solution

(vii) cosec 31° + sec 59°

Solution

 

(viii) (sin72° + cos 18°)(sin 72° - cos 18° )

Solution

We have to find : (sin 72° + cos 18°)( sin 72° - cos 18° )
Since sin(90°- θ) = cosθ . So
(sin72° + cos 18°)(sin 72° - cos 18°) = (sin 72°)² – (cos 18°)²
= [sin(90° - 18°)]² - (cos 18°)²
= (cos 18°)² - (cos 18°)²
= cos² 18° - cos² 18° 

So value of (sin 72° + cos 18°) (sin 72° + cos 18°) is 0 .

(ix) sin 35° sin 55°- cos 35° cos 55°

Solution

We find : sin 35° sin 55°- cos 35° cos 55°
Since sin (90° - θ) = cos θ and cos (90°- θ) = sin θ
sin 35° sin 55° - cos 35° cos55° = sin (90°- 55°)sin 55° - cos(90°- 55°) cos 55°
= cos 55° sin 55° - sin 55° cos 55°
= 1 – 1
= 0
So value of sin 35° sin 55° - cos 35° cos 55° is 0.

(x) tan 48° tan 23° tan 42° tan 67°

Solution

We have to find tan 48° tan 23° tan × 42° tan 67°
Since tan (90° - θ) = cot θ . So
tan 48° tan 23° tan 42° tan 67° = tan(90° - 67°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= (tan 67°cot 67°) (tan 42° cot 42°)
= 1×1
= 1
So value of tan 48° tan 23° tan 42° tan 67° is 1 .

(xi) sec 50° sin 40° + cos 40° cosec 50°  

Solution

We find to find sec 50° sin 40° + cos 40° cosec 50° 

Since cos (90°- θ) = sin θ, sec(90°- θ) = cosec θ and sin θ cosec θ = 1. So
sec 50° sin 40° + cos 40° cosec 50° = sec(90° - 40°) sin 40° + cos(90° - 50°) cosec 50°
= cosec 40° sin 40° + sin 50° cosec 50°
= 1 + 1
= 2
So value of sec 50° sin 40° + cos 40° cosec 50° is 2.

3. Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°.

(i) Sin 59° + cos 56°

Solution

sin 59° = sin(90°-31°) = cos 31°
cos 56° = cos (90° -34°) = sin 34°
⇒ cos 31° + sin 34°

(ii) Tan 65° + cot 49°  

Solution

tan 65° = tan (90°-25°) cot 25°
cot 49° = cot(90°-41°) tan 41°
⇒ cot 25° + tan 41°

(iii) Sec 76° + cosec 52°

Solution

sec 76° = sec(90°-14°) = cosec 14°
cosec 52° = cosec(90°-88°) =  sec 38°
⇒ cosec 14° + sec 38°

(iv) Cos 78° + sec 78°

Solution

cos 78° = cos (90°-12°) = sin 12°

sec 78° = sec(90°-12°) = cosec 12°

⇒ sin 12° + cosec 12°

(v)  Cosec 54° + sin 72°

Solution

cosec 54° = cosec (90°-36°) = sec 36°

sin 72° = sin (90°-18°) cos 18°

⇒ sec 36° + cos 18°

(vi) Cot 85° + cos 75° 

Solution

cot 85° = cot (90°-5°) = tan 5°

cos 75° = cos (90°-15°) = sin 15°

tan 5° + sin 15°

(vii) Sin 67° + cos 75°

Solution

sin 67° = sin (90°-23°) = cos 23°
cos 75° = cos(90° - 15°) = sin 15°
= cos 23° + sin 15°

4. Express Cos 75° + cot 75° in terms of angles between 0° and 30°.

Solution

cot 75° = cos (90°-15°) = sin 15°
cot 75° = cot (90° 15°) = tan 15°
= sin 15° + tan 15°

5. If Sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A = ?

Solution

6. If A, B, C are interior angles of a triangle ABC, prove that

(i) tan [c+A/2] = cot B/2

Solution

(ii) sin [B+C/2] = cos A/2

Solution

7. Prove that

(i) tan 20° tan 35° tan 45° tan 55° Tan 70° = 1

Solution

(ii) sin 48° sec 42° + cosec 42° = 2

Solution

(iii) sin 70°/ cos 20° +  cosec 20°/ sec 70° - 2 cos 70° cosec 20° = 0 

Solution

(iv) cos 80°/sin 10° + cos 59° cosec 31° = 2 

Solution

8. Prove the following .

(i) sin θ sin (90 - θ) - cos θ cos (90 - θ) = 0  

Solution

(ii) cos (90°-θ) sec (90°-θ) tan θ/cosec (90°-θ)sin(90°-θ)cot(90°-θ) + tan(90°-θ)/cot θ = 2  

Solution

(iii) (tan(90-A)cot A /cosec² A) - cos² A = 0 

Solution

(iv) cos(90°-A) sin(90°-A)/tan(90°-A) - sin² A = 0 

Solution  

(v) sin(50°+ θ) - cos(40° - θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° = 1Sol.

Solution

9. Evaluate : 

(i) 2/3 (cos⁴30°-sin⁴45°)-3(sin² 60°-sec²45°)+1/4cot²30° 

Solution

(ii) 4 (sin⁴30° + cos⁴60°) - 2/3(sin²60° - cos²45°) + 1/2 tan² 60° 

Solution

(iii) (sin 50°/cos40° + cosec 40°/sec 50°) - 4 cos 50° cosec 40°   

Solution

(iv) ) Tan 35° tan 40° tan 50° tan 55°

Solution

(v) Cosec (65°  + θ) – sec (25°  – θ) – tan (55°  – θ) + cot (35° + θ)    

Solution

(vi) tan 7° tan 23° tan 60° tan 67° tan 83° .

Solution

(vii) (2 sin 68°/ cos 22° ) - (2 cot 15°/5 tan 75°) - 3 tan 45° tan 20° tan 40° tan 50° tan 70°/ 5 .

Solution

(viii) 3 cos 55° / 7 sin 35° - 4(cos 70° cosec 20°) /7 (tan 5° tan 25° tan 45° tan 65° tan 85°) 

Solution

(ix) sin 18°/cos 72° + √3 { tan 10° tan 30°  tan 40° tan 50°  tan 80°}

Solution

(x) (cos 58°/sin 32°) + (sin 22°/cos 68°)- cos38°cosec52°/tan18°tan35°tan60°tan72°tan65°

Solution

10. If Sin θ = cos (θ – 45°), where θ – 45° are acute angles, find the degree measure of θ.

Solution

11. If A, B, C are the interior angles of a ∆ABC,  show that: 

(i) sin(B+C/2) = cos A/2

(ii) cos [B+C/2] = sin A/2

Solution

12. If 2θ + 45° and 30° − θ are acute angles, find the degree measure of θ satisfying Sin (20 + 45°) = cos (30 - θ°) 

Solution

13. If θ is a positive acute angle such that sec θ = cosec 60°, find 2 cos² θ – 1 . 

Solution

14. If cos 2θ = sin 4 θ where 2θ , 4θ  are acute angles, find the value of θ . 

Solution

15. If Sin 3θ  = cos (θ  – 6°) where 3 θ  and θ − 6° are acute angles, find the value of θ . 

Solution

16. If Sec 4A = cosec (A – 20°) where 4A is acute angle, find the value of A. 

Solution

17. If Sec 2A = cosec (A – 42°) where 2A is acute angle. Find the value of A .

Solution