Q) Prove that \frac {\cos ec ^2 \theta - \sec ^2 \theta}{\cos ec ^2 \theta + \sec ^2 \theta} = \frac{3}{4}, if \tan \theta = \frac{1}{\sqrt 7}.

Ans: Let’s start from LHS:

LHS = \frac {\cos ec ^2 \theta - \sec ^2 \theta}{\cos ec ^2 \theta + \sec ^2 \theta}

= \frac {\frac{1}{\sin ^2 \theta} - \frac{1}{\cos ^2 \theta}}{\frac{1}{\sin ^2 \theta} + \frac{1}{\cos ^2 \theta}}

= \frac {\frac{\cos ^2 \theta - \sin ^2 \theta}{(\sin ^2 \theta)(\cos ^2 \theta)}}{\frac{\cos ^2 \theta + \sin ^2 \theta}{(\sin ^2 \theta)(\cos ^2 \theta)}}

= \frac {\frac{\cos ^2 \theta - \sin ^2 \theta}{\cancel{(\sin ^2 \theta)(\cos ^2 \theta)}}}{\frac{\cos ^2 \theta + \sin ^2 \theta}{\cancel{(\sin ^2 \theta)(\cos ^2 \theta)}}}

= \frac {\cos ^2 \theta - \sin ^2 \theta}{\cos ^2 \theta + \sin ^2 \theta}

= \cos ^2 \theta - \sin ^2 \theta

\tan \theta = \frac{1}{\sqrt 7}.

\sin \theta = \frac{1}{\sqrt 8}

and \cos \theta = \frac {\sqrt 7}{\sqrt 8}

∴ LHS = \cos ^2 \theta - \sin ^2 \theta

= [\frac {\sqrt 7}{\sqrt 8} ] ^2 - [\frac{1}{\sqrt 8} ] ^2

= \frac{7}{8} - \frac {1}{8}

= \frac{6}{8}

= \frac{3}{4} = RHS

Hence Proved !

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