SSC CHSL 15 Oct 2020 Shift 3 Paper

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Question : 53 of 100

Marks: +1, -0

If

\sec \θ-\tan \θ=3

, then

\cos \θ

is equal to

${3}/{7}$
${2}/{5}$
${3}/{5}$
${4}/{9}$

Solution:

Given,

\sec \θ-\tan \θ=3

...(i)

(\sec \θ-\tan \θ) ×{(\sec \θ+\tan \θ)}/{(\sec \θ+\tan \θ)}=3

⇒

{\sec ^{2} \θ-\tan ^{2} \θ}/{\sec \θ+\tan \θ}=3

⇒

\sec \θ+\tan \θ={1}/{3}

...(ii)
On adding Eqs. (i) and (ii), we get

2 \sec \θ=3+{1}/{3}={10}/{3}

\sec \θ={10}/{3 ×2}={5}/{3}

∴

\cos \θ={3}/{5}

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