in NCERT Solution For Class 10, Maths, Chapter 8, Introduction To Trigonometry, Exercise 8.4, we will discuss questions related trigonometric identity. A trigonometric identity is defined as an equation involving trigonometric ratios of an angle, if it is true for all values of the angle(s) involved. In Exercise 8.4, Chapter 8, class 10, Maths we will we will prove one trigonometric identity, and use it further to prove other useful trigonometric identities.
Class 10, Maths, Chapter 8, Exercise 8.4, Solutions
Q.1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Ans:
Q.2. Write all the other trigonometric ratios of ∠A in terms of sec A.
Ans:
Q.3. Evaluate:
(i) $\frac{{{\sin }^{2}}\,\,{{63}^{0}}+{{\sin }^{2}}{{27}^{0}}}{{{\cos }^{2}}{{17}^{0}}+{{\cos }^{2}}{{73}^{0}}}$
(ii) sin 25° cos 65° + cos 25° sin 65°
Ans:
Q.4. Choose the correct option. Justify your choice.
(i) 9 sec2 A – 9 tan2 A =
(A) 1
(B) 9
(C) 8
(D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =
(A) 0
(B) 1
(C) 2
(D) –1
(iii) (sec A + tan A) (1 – sin A) =
(A) sec A
(B) sin A
(C) cosec A
(D) cos A
(iv) $\frac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=$
(A) sec2 A
(B) –1
(C) cot2 A
(D) tan2 A
Ans:
(i) (B), 9 sec2 A – 9 tan2 A = 9
(ii) (C), (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =2
(iii) (D), (sec A + tan A) (1 – sin A) = cos A
(iv) (D), $\frac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\tan }^{2}}A$
Q.5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosec θ – cot θ )2 = $\frac{1-\cos \,\,\theta }{1+\cos \,\,\theta }$
(ii) $\frac{\cos \,\,A}{1+\sin \,\,A}+\frac{1+\sin \,\,A}{\cos \,\,A}$= 2 sec A
(iii) $\frac{\tan \,\,\theta }{1-\cot \,\,\theta }+\frac{\cot \,\,\theta }{1-\tan \,\,\theta }$ = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]
(iv) $\frac{1+\sec \,\,A}{\sec A}=\frac{{{\sin }^{2}}\,A}{1-\cos \,\,A}$ [Hint : Simplify LHS and RHS separately]
(v) $\frac{\cos \,\,A-\sin \,\,A+1}{\cos \,\,A+\sin \,\,A-1}$=cosec A + cot A using the identity cosec2 A = 1 + cot2 A.
(vi) $\sqrt{\frac{1+\sin \,\,A}{1-\sin \,\,A}}$ =sec A + tan A
(vii) $\frac{\sin \,\,\theta -2\,\,{{\sin }^{3}}\theta }{2\,\,{{\cos }^{3}}\theta -\cos \,\,\theta }$= tan θ
(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
(ix) (cosec A – sin A)(sec A – cos A) = $\frac{1}{\tan \,\,A\,+\,\cot \,A}$
[Hint : Simplify LHS and RHS separately]
(x) $\left( \frac{1+{{\tan }^{2}}A}{1+co{{\operatorname{t}}^{2}}A} \right)={{\left( \frac{1-\tan A}{1-\cot A} \right)}^{2}}$= tan2 A
Ans:
(i) (cosec θ – cot θ )2 = $\frac{1-\cos \,\,\theta }{1+\cos \,\,\theta }$
(ii) $\frac{cos\,\,A}{1+sin\,\,A}+\frac{1+sin\,\,A}{cos\,\,A}$= 2 sec A
(iii) $\frac{tan\,\,\theta }{1-cot\,\,\theta }+\frac{cot\,\,\theta }{1-tan\,\,\theta }$ = 1 + sec θ cosec θ
(iv) $\frac{1+\sec \,A}{\sec \,\,A}=\,\frac{{{\sin }^{2}}\,A}{1-\,\,\cos \,\,A}$
(v) $\frac{\cos \,\,A-\,\sin \,A+1}{\cos \,\,A+\,\sin \,A-1}=\cos ec\,A+\,\cot \,A$
(vi) we have, $\sqrt{\frac{1+\sin \,A}{1-\sin \,A}}\,\,=\,\,\sec \,A+\tan \,A$
(vii) we have, $\frac{\sin \,\,\theta -2\,\sin {{\,}^{3}}\,\theta }{2\,{{\cos }^{3}}\,\,\theta -\cos \,\,\theta }=\tan \,\theta $
(viii) we have, (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
(ix) we have, (cosec A – sin A)(sec A – cos A) = $\frac{1}{\tan \,A+\cot \,A}$
(x) we have, $\left( \frac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \frac{1-\tan \,A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A$
