NCERT Solutions for Class 10 Maths Exercise 8.4 Chapter 8 Introduction to Trigonometry – FREE PDF Download
NCERT Class 10 Maths Ch 8 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths
Chapter 8 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Introduction to Trigonometry solutions will help you understand the chapter thoroughly.
NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry
1. Express the trigonometric ratios 
and 
in terms of 
By using identity 
For 
By using identity 
= 
For 
2. Write the other trigonometric ratios of A in terms of 
By using identity, 
= 
For 
For
By using identity
For 
= 
For 
3. Evaluate:
(i) 
(ii) 
= 
= 
= 
(ii)
= 
= 
= 
= 1
4. Choose the correct option. Justify your choice:
(i) 
=
(A) 1
(B) 9
(C) 8
(D) 0
(ii) 
=
(A) 0
(B) 1
(C) 2
(D) none of these
(iii) 
=
(A) 
(B) 
(C) 
(D) 
(iv) 
=
(A) 
(B) 
(C) 
(D) none of these
= 
= 
[Since sec2θ−tan2θ=1sec2θ−tan2θ=1]
(ii) (C)
= 
= 
= 
= 
= 
= 
= 2
(iii)(D)
= 
= 
= 
[Since (a+b)(a−b)=a2−b2(a+b)(a−b)=a2−b2]
= 
= 
(iv)(D) = 
= 
= 
= 
= 
5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
, using the identity
(vi) 
(vii) 
(viii) 
(ix) 
(x) 
= 
[Since (a−b)2=a2+b2−2ab(a−b)2=a2+b2−2ab]
= 
= 
= 
= 
= 
= 
= 
= R.H.S.
(ii) L.H.S. 
= 
= 
= 
= 
= 
= 
= 
= R.H.S
(iii) L.H.S. 
= 
= sinθcosθsinθ−cosθsinθ+cosθsinθcosθ−sinθcosθsinθcosθsinθ−cosθsinθ+cosθsinθcosθ−sinθcosθ
= 
= 
= 
= 
= 
= 
= 1sinθcosθ+sinθcosθsinθcosθ1sinθcosθ+sinθcosθsinθcosθ
= 
= 
= 
(iv) L.H.S. 
= cosA+1cosA1cosAcosA+1cosA1cosA
= 
= 
= 
= 
[Since (a+b)(a−b)=a2−b2(a+b)(a−b)=a2−b2]
= 
= R.H.S.
(v) L.H.S. 
Dividing all terms by
= 
= 
= 
[Since cosec2θ−cot2θ=1cosec2θ−cot2θ=1]
= 
= (cotA+cosecA)+(cotA+cosecA)(cotA−cosecA)(1+cotA−cosecA)(cotA+cosecA)+(cotA+cosecA)(cotA−cosecA)(1+cotA−cosecA)
= 
= 
= R.H.S.
(vi) L.H.S. 
= 
= 
= 
= 
= 
= 
= R.H.S.
(vii) L.H.S. 
= 
= 
= 
= 
= 
= 
= R.H.S
(viii) L.H.S.
= 
= 
= 
= 
= 
= 
= 
= R.H.S.
(ix) L.H.S. 
= 
= 
= 
= 
= 
Dividing all the terms by ,
= 
= 
= 
= R.H.S.
(x) L.H.S. 
=
= 
= = R.H.S.
Now, Middle side = 
= 
= 
= (1−tanA−(1−tanA)tanA)2(1−tanA−(1−tanA)tanA)2 = 
= = R.H.S.