Q

# Help me understand! - Trigonometry - BITSAT-2

If    $2 tan^{-1}\left ( cosx \right )=tan^{-1}\left ( 2cosec\, \, x \right )$   then the value of x is

• Option 1)

$\frac{3\pi}{4}$

• Option 2)

$\frac{\pi}{4}$

• Option 3)

$\frac{\pi}{3}$

• Option 4)

None of these

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As we discussed in concept

Relation between all the Inverse Trigonometric Functions -

$\sin ^{-1}x = \tan ^{-1}\frac{x}{\sqrt{1-x^{2}}} = cosec ^{-1}\frac{1}{x} = \cos ^{-1}\sqrt{1-x^{2}} = \cot ^{-1}\frac{\sqrt{1-x^{2}}}{x} = \sec ^{-1}\frac{1}{\sqrt{1-x^{2}}}$

- wherein

This happens with domain of functions, x$>$0

$2tan^{-1}(cosx)=tan^{-1}(2cosecx)$

$\therefore 2tan^{-1}x=tan^{-1}\: \frac{2x}{1-x^{2}}$

=> $tan^{-1}\: \frac{2cosx}{1-cos^{2}x}=tan^{-1}(2cosecx)$

$\therefore \frac{2cosx}{sin^{2}x}\:=\: \frac{2}{sinx}$

=> $\frac{1}{sinx}[cotn-1]\:=\:0$

$\therefore cotn-1=0$                            $n=\frac{\pi}{4}$

Option 1)

$\frac{3\pi}{4}$

This option is incorrect.

Option 2)

$\frac{\pi}{4}$

This option is correct.

Option 3)

$\frac{\pi}{3}$

This option is incorrect.

Option 4)

None of these

This option is incorrect.

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