SSC CPO SI and ASI Model Paper 8

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Question : 149 of 200

Marks: +1, -0

If

\sec θ+\tan θ=1+√{3}

, then

\cos θ

is equal to

3
$\;{1}/{√{3}}$
$\;{4}/{1}+3 √{3}$
$6/ 7$

Solution:

\sec θ+\tan θ=1+√{3} . . . . . .(1)

\sec θ-\tan θ=

\;{1}/{1+√{3}}=\;{1-√{3}}/{(1+√{3})}(1-√{3})=\;{√{3}-1}/{2}

On solving

(1)

and

(2)

, we get

\sec θ=\;{1+3 √{3}}/{4}

∴ \cos θ=\;{4}/{1+3 √{3}}

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