SSC CHSL 11 Aug 2021 Shift 2 Paper

Section: Quantitative Aptitude

Question : 68 of 100

Marks: +1, -0

\cot A+\operatorname{cosec} A=2

and

A

is an acute angle, then the value of

\frac{9 \tan A+16 \operatorname{cosec} A}{5 \sin A+3 \tan A}

Solution:

Given,

\cot A+cosec A=2

⇒ \;\; cosec A=2-\cot A

On squaring both sides;

cosec^{2} A=4+\cot ^{2} A-4 \cot A

⇒ 1+\cot ^{2} A=4+\cot ^{2} A-4 \cot A

⇒ \;\; [ ∵ cosec^{2} A=1+\cot ^{2} A]

\;\;\; 4 \cot A=4-1

\;\;\; \cot A=\;{3}/{4}

∴ \tan A=\;{4}/{3}, cosec A=\;{5}/{4}, \sin A=\;{4}/{5}

∴ \;\text " The value of "\; \;{9 \tan A+16 cosec A}/{5 \sin A+3 \tan A}

9 × \;{4}/{3}+16 × \;{5}/{4}

\;\; =\;{12+20}/{5 × \;{4}/{5}+3 × \;{4}/{3}}=4

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