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UPSC NDA Math Model Paper 5

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Question : 5 of 120

Marks: +1, -0

If

\cot \left[ \tan^{-1} a - \tan^{-1} \left( \frac{x-y}{x+y} \right) \right] = 1,

then

$a = \frac{y}{x}$
$a = \frac{x}{y}$
$a = \frac{x-y}{x}$
$a = \frac{x+y}{x-y}$

Solution:

S = \cot \left[ \tan^{-1} a - \tan^{-1} \left( \frac{x-y}{x+y} \right) \right]

=1

\Rightarrow \cot \left[ \tan^{-1} a - \tan^{-1} \left( \frac{x-y}{x+y} \right) \right]

= \cot 45 \; (\because \cot 45 = 1)

Taking

\cot^{-1}

on both sides

\Rightarrow \cot^{-1} \cot \left[ \tan^{-1} a - \tan^{-1} \left( \frac{x-y}{x+y} \right) \right]

= \cot^{-1}(\cot 45)

\Rightarrow \tan^{-1} a - \tan^{-1} \left( \frac{x-y}{x+y} \right) = 45

\Rightarrow \tan^{-1} \left[ \frac{ a - \frac{x-y}{x+y} }{ 1 + a \times \frac{x-y}{x+y} } \right] = 45

Taking tan on both sides

\Rightarrow \tan\tan^{-1} \left[ \frac{ a - \frac{x-y}{x+y} }{ 1 + a \frac{x-y}{x+y} } \right] = \tan 45

\Rightarrow \frac{ a - \frac{x-y}{x+y} }{ 1 + a \frac{x-y}{x+y} } = 1

\Rightarrow a - \frac{x-y}{x+y} = 1 + a \frac{x-y}{x+y}

\Rightarrow a \left( 1 - \frac{x-y}{x+y} \right) = 1 + \frac{x-y}{x+y}

\Rightarrow a (x+y-(x-y)) = x+y+x-y

\Rightarrow 2 y a = 2 x

\Rightarrow a = \frac{x}{y}

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