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UPSC CDS 2 2022 Math Paper

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Question : 44 of 100

Marks: +1, -0

If

x=\; \frac{1+\sin \theta}{\cos \theta}

\; \frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}

$- x$
$x$ .
$2 x$
$\; \frac{x}{2}$

Solution: 👈: Video Solution

We Know

-

\sec^2 \theta-\tan^2 \theta=1

-

\sec \theta = \frac{1}{\cos \theta}, \tan \theta = \frac{\sin \theta}{\cos \theta}

-

(a-b)(a+b)=a^2-b^2

By using above formula

We have,

\Rightarrow \; \frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}

\Rightarrow \; \frac{\sec \theta+\tan \theta-(\sec^2 \theta-\tan^2 \theta)}{\tan \theta-\sec \theta+1}

\Rightarrow \; \frac{\sec \theta+\tan \theta-[ (\sec \theta-\tan \theta)(\sec \theta+\tan \theta) ]}{\tan \theta-\sec \theta+1}

\Rightarrow \; \frac{(\sec \theta+\tan \theta)[1-(\sec \theta+\tan \theta)]}{\tan \theta-\sec \theta+1}

\Rightarrow \sec \theta+\tan \theta

\Rightarrow \; \frac{1}{\cos \theta}+\; \frac{\sin \theta}{\cos \theta}

\Rightarrow \; \frac{1+\sin \theta}{\cos \theta}

\Rightarrow x

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