If sin θ = 3/4, prove that Root [(cosec 2θ − cot 2 θ} / (sec 2 θ − 1)] = √7 / 3

Question

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

Sapling Academy · Accepted Answer

Let's start LHS:

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

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

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

∵ 1- cos2  θ= sin2 θ

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

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

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

=  $\frac{\cos 	heta}{\sin 	heta}$ = 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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