SSC CHSL 5 Aug 2021 Shift 3 Paper

Section: Quantitative Aptitude

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Question : 53 of 100

Marks: +1, -0

If

\tan x=\cot (48^{∘}+2 x)

and

0^{∘}< x< 90^{∘}

, then what is the value of

x

?

$12^{∘}$
$14^{∘}$
$16^{∘}$
$21^{∘}$

Solution:

Given,

\tan x=\cot (48^{∘}+2 x)

⇒ \;\; \tan x=\tan [90^{∘}-(48^{∘}+2 x)]

[ ∵ 0^{∘} < x < 90^{∘}]

On comparison,

x=90^{∘}-(48^{∘}+2 x)

x=42^{∘}-2 x

3 x=42^{∘}

∴ \;\; x=14^{∘}

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