If sin θ = 3/4, prove that Root [(cosec 2θ − cot 2 θ} / (sec 2 θ − 1)]

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Q) If sin θ = $\frac{3}{4}$ , prove that $\sqrt {\frac {\csc ^2 \theta\minus -\cot ^2 \theta} {\sec ^2 \theta-1}} = \frac{\sqrt7}{3}$ .

Ans: Let’s start LHS:

$\sqrt {\frac {\csc ^2 \theta -− \cot ^2 \theta} {\sec ^2 \theta -− 1}}$

= $\sqrt {\frac {(\frac{1}{\sin ^2 \theta} - \frac{\cos ^2 \theta}{\sin^2\theta})} {(\frac{1}{\cos ^2 \theta} - 1)}}$

= $\sqrt {\frac {(\frac{1-\cos ^2 \theta}{\sin^2\theta})} {(\frac{1-\cos ^2 \theta}{\cos^2\theta})}$

∵ 1- cos² θ= sin² θ

∴ LHS = $\sqrt {\frac {(\frac{\sin ^2 \theta}{\sin^2\theta})} {(\frac{\sin ^2 \theta}{\cos^2\theta})}$

= $\sqrt {\frac {(1)} {(\frac{\sin ^2 \theta}{\cos^2\theta})}$

= $\sqrt {(\frac{\cos ^2 \theta}{\sin^2\theta})}$

= $\frac{\cos \theta}{\sin \theta}$ = cot θ

∵ sin θ = $\frac{3}{4}$ ∴ cot θ = $\frac{\sqrt 7}{3}$

∴ LHS = $\frac{\sqrt 7}{3}$

= RHS

Hence Proved !

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