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The value of \small \tan ^{-1}\left [ \frac{\sqrt{1+x^{2}}+\sqrt{1 -x^{2}}}{\sqrt{1 +x^{2}}-\sqrt{1 -x^{2}}} \right ],

\small \left | x \right |< \frac{1}{2},x\neq 0 is equal to:

  • Option 1)

    \frac{\pi }{4} +\frac{1}{2} \cos ^{-1} x^{2}

  • Option 2)

    \frac{\pi }{4} +\cos ^{-1} x^{2}

  • Option 3)

    \frac{\pi }{4} -\frac{1}{2} \cos ^{-1} x^{2}

  • Option 4)

    \frac{\pi }{4} -\cos ^{-1} x^{2}

 

Answers (1)

As we learnt in 

Trigonometric Ratios of Submultiples of an Angle -

Trigonometric ratios of submultiples of an angle 1

- wherein

This shows the formulae for half angles and their doubles.

 

 \tan^{-1} \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}\:-\sqrt{1-x^{2}}}

Let x^{2}=\cos 2t

   As, x^{2}>0, \:\:\:2t \varepsilon(0.\frac{\pi}{2})\:\Rightarrow t \varepsilon(0.\frac{\pi}{4})                     ..............( i )

Then, the expression becomes =\tan ^{-1} \left ( \frac{\sqrt{1+\cos2t}\:+\sqrt{1-\cos2t}}{\sqrt{1+\cos2t}\:-\sqrt{1-\cos 2t}} \right )

=\tan ^{-1} \frac{\sqrt{2\cos^{2}t}\:+\sqrt{2\sin^{2}t}}{\sqrt{2\cos^{2}t}\:-\sqrt{2\sin^{2}t}}

=\tan^{-1}\frac{\left| \cos t\right|\:+\left| \sin t\right|}{\left| \cos t\right|\:-\left| \sin t\right|}

=\tan^{-1}\frac{1+\left| \tan t\right|}{1-\left| \tan t\right|}                 -   {Dividing numerator and denominator by |cos t|}

=\tan^{-1}\frac{1+ \tan t}{1-\tan t}                    - {From ( i )}

=\tan^{-1} \tan(\frac{\pi}{4}+\frac{1}{2}\:\cos^{-1}\:x^{2})

=\frac{\pi}{4}+\frac{1}{2}\:\cos^{-1}\:x^{2}

 


Option 1)

\frac{\pi }{4} +\frac{1}{2} \cos ^{-1} x^{2}

This option is correct.

Option 2)

\frac{\pi }{4} +\cos ^{-1} x^{2}

This option is incorrect.

Option 3)

\frac{\pi }{4} -\frac{1}{2} \cos ^{-1} x^{2}

This option is incorrect.

Option 4)

\frac{\pi }{4} -\cos ^{-1} x^{2}

This option is incorrect.

Posted by

Vakul

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