SSC CHSL 6 Dec 2015 Shift 1 Paper

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Question : 140 of 200

Marks: +1, -0

If

{x−x tan^2 30°}/{1+tan^2 30°}

=sin^2 30°+4cot^2 45°-sec^2 60°

Then value of x is:

$1/√3$
$1/5$
$1/4$
$1/2$

Solution:

Expression :

{x−x tan^2 30°}/{1+tan^2 30°}

=sin^2 30°+4cot^2 45°-sec^2 60°

⇒

{x−x(1/√ 3 ) ^2}/{1+(1/√3)^2}

=( 1/2)^2+4(1)^2-(2)^2

⇒

{{2x}/3}/{4/3}=1/4+4-4

⇒

x/2=1/4

⇒

x=2/4=1/2

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