## 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°- Î¸) = cosÎ¸ . 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° - Î¸) = cos Î¸ and cos (90°- Î¸) = sin Î¸
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° - Î¸) = cot Î¸ . 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°- Î¸) = sin Î¸, sec(90°- Î¸) = cosec Î¸ and sin Î¸ cosec Î¸ = 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 Î¸ sin (90 - Î¸) - cos Î¸ cos (90 - Î¸) = 0

Solution

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

Solution

(iii) (tan(90-A)cot A /cosecA) - cosA = 0

Solution

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

Solution

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

Solution

9. Evaluate :
(i) 2/3 (cos430°-sin445°)-3(sin60°-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°  + Î¸) – sec (25°  – Î¸) – tan (55°  – Î¸) + cot (35° + Î¸)

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 Î¸ = cos (Î¸ – 45°), where Î¸ – 45° are acute angles, find the degree measure of Î¸.

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

Solution

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

Solution

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

Solution

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

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