If \cot \theta=\frac{15}{8}, \theta is an acute angle, then find the value of \frac{(1-\cos \theta)(2+2 \cos \theta)}{(2-2 \sin \theta)(1+\sin \theta)}.

SSC CGL 20 Aug 2021 Shift 2 Paper

Question : 64 of 100

Marks: +1, -0

\cot θ=\;{15}/{8}, θ

is an acute angle, then find the value of

\;{(1-\cos θ)(2+2 \cos θ)}/{(2-2 \sin θ)(1+\sin θ)}

Solution:

Given,

\cot θ=\;{15}/{8}=\;{\;\text " Base "\;}/{\;\text " Perpendicular "\;}

By pythagoras theorem, (Hypotenuse)

^{2}

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

⇒(H)^{2}=(15)^{2}+(8)^{2}=225+64

⇒(H)^{2}=289

⇒ \;\; H=√{289}=17

\;\text " Now, "\; \sin θ=\;{\;\text " Perpendicular "\;}/{\;\text " Hypotenuse "\;}=\;{8}/{17}

\cos θ=\;{\;\text " Base "\;}/{\;\text " Hypotenuse "\;}=\;{15}/{17}

∴ \;{(1-\cos θ)(2+2 \cos θ)}/{(2-2 \sin θ)(1+\sin θ)}

=\;{(1-{15}/{17}) × 2(1+{15}/{17})}/{2(1-\;{8}/{17})(1+\;{8}/{17})}

=\;{\;{2}/{17} × {32}/{17}}/{{9}/{17} × \;{25}/{17}}=\;{64}/{225}

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