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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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