Prove that (cosec 2 θ – sec 2 θ) / (cosec 2 θ + sec 2 θ) = 3/4, if tan θ = 1/√7.

Question

Q) Prove that $\frac {\cos ec ^2 	heta – \sec ^2 	heta}{\cos ec ^2 	heta + \sec ^2 	heta} = \frac{3}{4}$, if $	an 	heta = \frac{1}{\sqrt 7}$.

Sapling Academy · Accepted Answer

Let's start from LHS:

LHS = $\frac {\cos ec ^2 	heta – \sec ^2 	heta}{\cos ec ^2 	heta + \sec ^2 	heta} $

= $\frac {\frac{1}{\sin ^2 	heta} – \frac{1}{\cos ^2 	heta}}{\frac{1}{\sin ^2 	heta} + \frac{1}{\cos ^2 	heta}} $

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

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

= $\frac {\cos ^2 	heta - \sin ^2 	heta}{\cos ^2 	heta + \sin ^2 	heta}$

= $\cos ^2 	heta - \sin ^2 	heta $

∵ $ 	an 	heta = \frac{1}{\sqrt 7}$.

∴ $\sin 	heta = \frac{1}{\sqrt 8}$

and $\cos 	heta = \frac {\sqrt 7}{\sqrt 8}$

∴ LHS = $\cos ^2 	heta - \sin ^2 	heta$

= $ [\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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