SSC CHSL 19 Mar 2020 Shift 1 Paper

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Question : 62 of 100

Marks: +1, -0

If

{\sin x+\cos x}/{\sin x-\cos x}={6}/{5}

, then the value of

{\tan ^{2} x+1}/{\tan ^{2} x-1}

is

${61}/{35}$
${61}/{60}$
${35}/{61}$
${60}/{61}$

Solution:

Given,

{\sin x+\cos x}/{\sin x-\cos x}={6}/{5}

By using Componendo and Dividendo, we get

{(\sin x+\cos x)+(\sin x-\cos x)}/{(\sin x+\cos x)-(\sin x-\cos x)}

={6+5}/{6-5}={11}/{1}

⇒

{2 \sin x}/{2 \cos x}={11}/{1}

⇒

\tan x=11

∴

{\tan ^{2} x+1}/{\tan ^{2} x-1}={(11)^{2}+1}/{(11)^{2}-1}

={121+1}/{121-1}={122}/{120}

={61}/{60}

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