# RD Sharma Solutions Class 10 Chapter 5 Trigonometric Ratios Exercise 5.1

In this chapter, we provide RD Sharma Solutions for Class 10 Chapter 5 Trigonometric Ratios Exercise 5.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 10 Chapter 5 Trigonometric Ratios Exercise 5.1 pdf, free RD Sharma Solutions for Class 10 Chapter 5 Trigonometric Ratios Exercise 5.1 book pdf download. Now you will get step by step solution to each question.

## RD Sharma Class 10 Solutions Chapter 5 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 { costheta +sintheta }{ costheta -sintheta }).
Solution:

Question 10.
If 3 tan θ = 4, find the value of 4cos θ – sin θ (frac { 4costheta -sintheta }{ 2costheta +sintheta }).
Solution:

Question 11.
If 3 cot 0 = 2, find the value of (frac { 4sinstheta -3costheta }{ 2sintheta +6costheta }).
Solution:

Question 12.
If tan θ = (frac { a }{ b }), prove that

Solution:

Question 13.
If sec θ = (frac { 13 }{ 5 }), show that (frac { 2sinstheta -3costheta }{ 4sintheta -9costheta }) =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 { sinstheta -2costheta }{ tantheta -cottheta }).
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°

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