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UPSC CDS 2 2022 Math Paper

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Question : 47 of 100

Marks: +1, -0

If

\frac{1}{\csc\theta-\cot\theta}-\frac{1}{\sin\theta}=x

, then what. is

\frac{1}{\csc\theta+\cot\theta}-\frac{1}{\sin\theta}

equal to, where

0 < \theta < \frac{\pi}{2}?

$-x$
$x$
$\frac{1}{x}$
$-\frac{1}{x}$

Solution: 👈: Video Solution

1.

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

2.

\csc^2\theta-\cot^2\theta=1

3.

\frac{1}{\sin\theta}=\csc\theta

Using above formula

\frac{1}{\csc\theta-\cot\theta}-\frac{1}{\sin\theta}=X

\Rightarrow \frac{\csc\theta+\cot\theta}{(\csc\theta-\cot\theta)(\csc\theta+\cot\theta)}-\frac{1}{\sin\theta}=X

\Rightarrow \frac{\csc\theta+\cot\theta}{\csc^2\theta-\cot^2\theta}-\frac{1}{\sin\theta}=X

Since,

\csc^2\theta-\cot^2\theta=1

\Rightarrow \frac{\csc\theta+\cot\theta}{\csc^2\theta-\cot^2\theta}-\frac{1}{\sin\theta}=X

\Rightarrow \csc\theta+\cot\theta-\csc\theta=x

\Rightarrow x=\cot\theta\cdots(i)

The value of

\frac{1}{\csc\theta+\cot\theta}-\frac{1}{\sin\theta}

\Rightarrow \frac{\csc\theta-\cot\theta}{(\csc\theta+\cot\theta)(\csc\theta-\cot\theta)}-\frac{1}{\sin\theta}

\Rightarrow \frac{\csc\theta-\cot\theta}{\csc^2\theta-\cot^2\theta}-\frac{1}{\sin\theta}

\Rightarrow \csc\theta-\cot\theta-\csc\theta

\Rightarrow -\cot\theta=-x

[From equation (1)

]

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