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SSC CGL 23 Aug 2021 Shift 2 Paper

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Question : 54 of 100

Marks: +1, -0

If

\frac{\sin^2 \theta}{\tan^2 \theta - \sin^2 \theta} = 5, \theta

is an acute angle, then the value of

\frac{24 \sin^2 \theta - 15 \sec^2 \theta}{6 \csc^2 \theta - 7 \cot^2 \theta}

is

$-2$
2
$-14$
14

Solution:

Given,

\frac{\sin^2 \theta}{\tan^2 \theta - \sin^2 \theta} = 5

\Rightarrow \sin^2 \theta = 5 \tan^2 \theta - 5 \sin^2 \theta

\Rightarrow 6 \sin^2 \theta = 5 \tan^2 \theta

\Rightarrow 6 \sin^2 \theta = 5 \times \frac{\sin^2 \theta}{\cos^2 \theta}

\Rightarrow \cos^2 \theta = \frac{5}{6}

We know that,

\sin^2 \theta + \cos^2 \theta = 1

\Rightarrow \sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{5}{6}

\Rightarrow \sin^2 \theta = \frac{1}{6}

Now,

\sec^2 \theta = \frac{1}{\cos^2 \theta} = \frac{1}{\frac{5}{6}}

\Rightarrow \sec^2 \theta = \frac{6}{5}

and

\csc^2 \theta = \frac{1}{\sin^2 \theta} = \frac{1}{\frac{1}{6}}

\Rightarrow \csc^2 \theta = 6

\Rightarrow \cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{\frac{5}{6}}{\frac{1}{6}}

\Rightarrow \cot^2 \theta = 5

\therefore \frac{24 \sin^2 \theta - 15 \sec^2 \theta}{6 \csc^2 \theta - 7 \cot^2 \theta}

\Rightarrow \frac{24 \times \left(\frac{1}{6}\right) - 15 \times \left(\frac{6}{5}\right)}{6 \times (6) - 7 \times 5}

= \frac{4-18}{36-35} = -14

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