SSC CGL 24 Aug 2021 Shift 3 Paper

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Question : 54 of 100

Marks: +1, -0

If

\;{\sin θ+\cos θ}/{\sin θ-\cos θ}=5

, then the value of

\;{4 \sin ^{2} θ+3}/{2 \cos ^{2} θ+2}

is

$\;{75}/{17}$
$\;{75}/{34}$
$\;{1}/{2}$
$\;{3}/{2}$

Solution:

Given,

\;{\sin θ+\cos θ}/{\sin θ-\cos θ}=\;{5}/{1}

⇒ \;\; \sin θ+\cos θ \;\; =5 \sin θ-5 \cos θ

⇒ \;\; 5 \sin θ-\sin θ \;\; =\cos θ+5 \cos θ

⇒ \;\; 4 \sin θ \;\; =6 \cos θ

⇒ \;\; \;\; \;{\sin θ}/{\cos θ} \;\; =\;{6}/{4}=\;{3}/{2}

{[ ∵ \tan θ=\;{\sin θ}/{\cos θ}] }

⇒ \;\; \tan θ=\;{3}/{2}

. . . (i)
We know that,

\tan θ=\;{\;\text " Perpendicular "\;}/{\;\text " Base "\;}=\;{3}/{2}

By Pythagoras theorem,

\Rightarrow

(Hypotenuse)

^{2}

\;\; =(\;\text " Perpendicular "\;)^{2}+(\;\text " Base "\;)^{2}

⇒ \;\; (H)^{2} \;\; =(3)^{2}+(2)^{2}=9+4

⇒ \;\; (H)^{2} \;\; =13

⇒ \;\; H \;\; =√{13}

\sin θ \;\; =\;{\;\text " Perpendicular "\;}/{\;\text " Hypotenuse "\;}=\;{3}/{√{13}} \;\; . . . \;\text " (ii) "\;

\cos θ \;\; =\;{\;\text " Base "\;}/{\;\text " Hypotenuse "\;}=\;{2}/{√{13}} \;\; . . . \;\text " (iii) "\;

Then,

\;{4 \sin ^{2} θ+3}/{2 \cos ^{2} θ+2}

= \;\; \;{4 ×({3}/{√{13}})^{2}+3}/{2 ×(\;{2}/{√{13}})^{2}+2}=\;{4 × {9}/{13}+3}/{2 × \;{4}/{13}+2}

= \;\; \;{36+39}/{{13}/{8+26}}=\;{75}/{34}

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