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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

 

Answers (1)

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.

Posted by

Vakul

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