If \tan A+\sec A=\frac{3}{2} and A is an acute angle, then the value of \frac{10 \cot A+13 \cos A}{12 \tan A+5 cosec A} is

SSC CHSL 12 Aug 2021 Shift 3 Paper

Section: Quantitative Aptitude

Question : 65 of 99

Marks: +1, -0

\tan A+\sec A=\;{3}/{2}

and

A

is an acute angle, then the value of

\;{10 \cot A+13 \cos A}/{12 \tan A+5 cosec A}

Solution:

Given,

\tan A+\sec A=\;{3}/{2}

⇒ \;\; \sec A=\;{3}/{2}-\tan A

We know that,

1+\tan ^{2} A=\sec ^{2} A

∴ 1+\tan ^{2} A=(\;{3}/{2}-\tan A)^{2}

⇒ 1+\tan ^{2} A=\;{9}/{4}+\tan ^{2} A

-2 × \;{3}/{2} × \tan A

⇒ \;\; 3 \tan A=\;{9}/{4}-1=\;{9-4}/{4}=\;{5}/{4}

⇒ \;\; \tan A=\;{5}/{4 × 3}=\;{5}/{12}=\;{P}/{B}

∴ \;\; \cot A=\;{12}/{5}

△ A B C

, Pythagoras theorem,

A C^{2} \;\; =(A B)^{2}+(B C)^{2}

⇒ \;\; A C^{2} \;\; =(5)^{2}+(12)^{2}

\;\; =25+144=169

⇒ \;\; A C \;\; =√{169}=13 {cm}

\;\text " Now, "\; \;\; \cos A \;\; =\;{B}/{H}=\;{12}/{13}

\;\text " and "\; cosec A \;\; =\;{H}/{P}=\;{13}/{5}

∴ \;\;{10 \cot A+13 \cos A}/{12 \tan A+5} cosec A

=\;{10 × {12}/{5}+13 × {12}/{13}}/{12 × \;{5}/{12}+5 × \;{13}/{5}}

=\;{24+12}/{5+13}=\;{36}/{18}=2

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