RD Sharma Class 10 Solutions Chapter 5 Trigonometric Ratios

RD Sharma Class 10 Solutions Trigonometric Ratios Exercise 5.1

Question 1.
In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.

Solution:
∴ Perpendicular BC – 2 units and
Hypotenuse AC = 3 units
By Phythagoras Theorem, in AABC,
(Hypotenuse)² = (Base)² + (Perpendicular)²
AC² = AB² + BC²
⇒ (3)² = (AB)² + (2)²
⇒ 9 = AB² + 4 ⇒ AB² = 9-4 = 5
AB = √5 units

Question 2.
In a ΔABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine
(i) sin A, cos A
(ii) sin C, cos C.
Solution:

Question 3.
In the figure, find tan P and cot R. Is tan P = cot R ?

Solution:

Question 4.
If sin A = $\frac { 9 }{ 41 }$ , compute cos A and tan A.
Solution:

Question 5.
Given 15 cot A = 8, find sin A and sec A.
Solution:

Question 6.
In ΔPQR, right angled at Q, PQ = 4 cm and RQ = 3 cm. Find the values of sin P, sin R, sec P and sec R.
Solution:

Question 7.
If cot 0 = $\frac { 7 }{ 8 }$ , evaluate :

Solution:

Question 8.
If 3 cot A = 4, check whether $\frac { 1-{ tan }^{ 2 }A }{ 1+{ tan }^{ 2 }A }$ = cos² A – sin² A or not.
Solution:

Question 9.
If tan θ = a/b , Find the Value of $\frac { cos\theta +sin\theta }{ cos\theta -sin\theta }$ .
Solution:

Question 10.
If 3 tan θ = 4, find the value of 4cos θ – sin θ $\frac { 4cos\theta -sin\theta }{ 2cos\theta +sin\theta }$ .
Solution:

Question 11.
If 3 cot 0 = 2, find the value of $\frac { 4sins\theta -3cos\theta }{ 2sin\theta +6cos\theta }$ .
Solution:

Question 12.
If tan θ = $\frac { a }{ b }$ , prove that

Solution:

Question 13.
If sec θ = $\frac { 13 }{ 5 }$ , show that $\frac { 2sins\theta -3cos\theta }{ 4sin\theta -9cos\theta }$ =3.
Solution:

Question 14.
If cos θ $\frac { 12 }{ 13 }$ , show that sin θ (1 – tan θ) $\frac { 35 }{ 156 }$ .
Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.
If sec θ = $\frac { 5 }{ 4 }$ , find the value of $\frac { sins\theta -2cos\theta }{ tan\theta -cot\theta }$ .
Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.
If tan θ = $\frac { 24 }{ 7 }$ , find that sin θ + cos θ.
Solution:

Question 22.
If sin θ = $\frac { a }{ b }$ , find sec θ + tan θ in terms of a and b.
Solution:

Question 23.
If 8 tan A = 15, find sin A – cos A.
Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
∠A and ∠B are acute angles and cos A = cos B
Draw a right angle AABC, in which ∠C – 90°

Question 27.
In a ∆ABC, right angled at A, if tan C =√3 , find the value of sin B cos C + cos B sin C. (C.B.S.E. 2008)
Solution:

Question 28.
28. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = $\frac { 12 }{ 5 }$ for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = $\frac { 4 }{ 3 }$ for some angle θ.
Solution:
(i) False, value of tan A 0 to infinity.
(ii) True.
(iii) False, cos A is the abbreviation of cosine A.
(iv) False, it is the cotengent of angle A.
(v) Flase, value of sin θ varies on 0 to 1.

Question 29.

Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.
If sin θ = $\frac { 3 }{ 4 }$ , prove that

Solution:

Question 33.

Solution:

Question 34.

Solution:

Question 35.
If 3 cos θ-4 sin θ = 2 cos θ + sin θ, find tan θ.
Solution:

Question 36.
If ∠A and ∠P are acute angles such that tan A = tan P, then show that ∠A = ∠P.
Solution:
∠A and ∠P are acute angles and tan A = tan P Draw a right angled AAPB in which ∠B = 90°

RD Sharma Class 10 Solutions Chapter 5 Trigonometric Ratios Ex 5.2

RD Sharma Class 10 Solutions Trigonometric Ratios Exercise 5.2

Evaluate each of the following (1-19) : 1.
Question 1.
sin45° sin30° + cos45° cos30°.
Solution:
sin 45° sin 30° + cos 45° cos 30°

Question 2.
sin60° cos30° + cos60° sin30°.
Solution:

Question 3.
cos60° cos45° – sin60° sin45°.
Solution:

Question 4.
sin²30° + sin²45° + sin²60° + sin²290°.
Solution:

Question 5.
cos²30° + cos²45° + cos²60° + cos²90°.
Solution:

Question 6.
tan²30° + tan²60° + tan²45°.
Solution:

Question 7.
2sin²30° – 3cos²45° + tan²60°.
Solution:

Question 8.
sin²30°cos²4S° + 4tan²30° + $\frac { 1 }{ 2 }$ sin²90° -2cos²90° + $\frac { 1 }{ 24 }$ cos20°.
Solution:

Question 9.
4 (sin⁴ 60° + cos⁴ 30°) – 3 (tan² 60° – tan² 45°) + 5cos² 45°
Solution:

Question 10.
(cosecc² 45° sec² 30°) (sin² 30° + 4cot² 45° – sec² 60°).
Solution:

Question 11.
cosec³ 30° cos 60° tan³ 45° sin2 90° sec² 45° cot 30°.
Solution:

Question 12.
cot² 30° – 2cocs² 60° – $\frac { 3 }{ 4 }$ sec2 45° – 4sec² 30°.
Solution:

Question 13.
(cos0° + sin45° + sin30°) (sin90° – cos45° + cos60°)
Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.
4 (sin⁴ 30° + cos² 60°) – 3 (cos² 45° – sin² 90°) – sin² 60°
Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Find the value of x in each of the following : (20-25)

Question 20.
2sin 3x = √3
Solution:

Question 21.
2sin $\frac { x }{ 2 }$ = 1
Solution:

Question 22.
√3 sin x=cos x
Solution:

Question 23.
tan x = sin 45° cos 45° + sin 30°
Solution:

Question 24.
√3 tan 2x = cos 60° +sin 45° cos 45°
Solution:

Question 25.
cos 2x = cos 60° cos 30° + sin 60° sin 30°
Solution:

Question 26.
If θ = 30°, verify that :

Solution:

Question 27.
If A = B = 60°, verify that:
(i) cos (A – B) = cos A cos B + sin A sin B
(ii) sin (A – B) = sin A cos B – cos A sin B tan A – tan B
(iii) tan (A – B) = $\frac { tanA-tanB }{ 1+tanA-tanB }$
Solution:

Question 28.
If A = 30° and B = 60°, verify that :
(i) sin (A + B) = sin A cos B + cos A sin B
(ii) cos (A + B) = cos A cos B – sin A sin B
Solution:

Question 29.
If sin (A + B) = 1 and cos (A,-B) = 1,0° < A + B < 90°, A > B find A and B.
Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.
In a ∆ABC right angle at B, ∠A = ∠C. Find the values of
(i) sin A cos C + cos A sin C
(ii) sin A sin B + cos A cos B
Solution:

Question 33.
Find acute angles A and B, if sin (A + 2B)= $\frac { \sqrt { 3 } }{ 2 }$ and cos (A + 4B) = 0, A > B.
Solution:

Question 34.
In ΔPQR, right-angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠P and ∠R.
Solution:

Question 35.
If sin (A – B) = sin A cos B – cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values of sin 15° and cos 15°.
Solution:

Question 36.
In a right triangle ABC, right angled at ∠C if ∠B = 60° and AB – 15 units. Find the remaining angles and sides.
Solution:

Question 37.
If ΔABC is a right triangle such that ∠C = 90°, ∠A = 45° and BC = 7 units. Find ∠B, AB and AC.
Solution:

Question 38.
In a rectangle ABCD, AB = 20 cm, ∠BAC = 60°, calculate side BC and diagonals AC and BD.
Solution:

Question 39.
If A and B are acute angles such that

Solution:

Question 40.
Prove that (√3 + 1) (3 – cot 30°) = tan³ 60° – 2sin 60°. [NCERT Exemplar]
Solution:

RD Sharma Class 10 Solutions Chapter 5 Trigonometric Ratios Ex 5.3

RD Sharma Class 10 Solutions Trigonometric Ratios Exercise 5.3

Question 1.
Evaluate the following :

Solution:
Question 2.
Evaluate the following :

Solution:

Question 3.
Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°
(i) sin 59° + cos 56°
(ii) tan 65° + cot 49“
(iii) sec 76° + cosec 52°
(iv) cos 78° + sec 78°
(v) cosec 54° + sin 72°
(vi) cot 85″ + cos 75°
(vii) sin 67° + cos 75°
Solution:
(i) sin 59° + cos 56°
= sin (90° – 31°) + cos (90° – 34°)
= cos 31° +sin 34°
(ii) tan 65° + cot 49°
= tan (90° – 25°) + cot (90° – 41°)
= cot 25° + tan 41°
(iii) sec 76° + cosec 52°
= sec (90° – 14°) + cosec (90 0 – 38°)
= cosec 14° + sec 38°
(iv) cos 78° + sec 78°
= cos (90° – 12°) + sec (90°- 12°)
= sin 12° + cosec 12°
(v) cosec 54° + sin 72°
= cosec (90° – 36°) + sin (90°-18°)
= sec 36° + cos 18°
(vi) cot 85° + cos 75°
= cot (90° – 5°) + cos (90° – 15°)
= tan 5° + sin 15°
(vii) sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°

Question 4.
Express cos 75° + cot 75° in terms of angles between 0° and 30°.
Solution:
cos 75° + cot 75° = cos (90° – 15°) + cot (90°-15°)
= sin 15° + tan 15°

Question 5.
If sin 3A = cos (A – 26°), where 3A is an acute angle, And the value of A.
Solution:
sin 3A = cos (A – 26°)
⇒ cos (90° – 3A) = cos (A – 26°)
Comparing,
90° – 3A = A – 26°
⇒ 90° + 26° = A + 3A ⇒ 4A = 116°

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

Solution:

Question 7.
Prove that :

Solution:

Question 8.
Prove the following :

Solution:

Question 9.
Evaluate :

Solution:

Question 10.
If sin θ= cos (θ – 45°), where θ and (θ – 45°) are acute angles, find the degree measure of θ.
Solution:

Question 11.
If A, B, C are the interior angles of a AABC, show that :
(i) $sin\frac { B+C }{ 2 } cos\frac { A }{ 2 }$
(ii) $cos\frac { B+C }{ 2 } sin\frac { A }{ 2 }$
Solution: