The value of  tan^{-1}\left ( 1 \right )+tan^{-1}\left ( 0 \right )+tan^{-1}\left ( 2 \right )+tan^{-1}\left ( 3 \right )  is equal to

  • Option 1)

    \pi

  • Option 2)

    \frac{5\pi}{4}

  • Option 3)

    \frac{\pi}{2}

  • Option 4)

    None of these

 

Answers (1)
V Vakul

As we learnt in

Formulae of Inverse Trigonometric Functions -

\sin ^{-1}x + \sin ^{-1}y= \sin ^{-1}\left [ x\sqrt{1-y^{2}}+ y \sqrt{1-x^{2}} \right ]

\sin ^{-1}x + \sin ^{-1}y=\pi - \sin ^{-1}\left [ x\sqrt{1-y^{2}}+ y \sqrt{1-x^{2}} \right ]

- wherein

x\geqslant 0, y\geqslant 0, x^{2}+ y^{2}\leqslant 1

 

x\geqslant 0, y\geqslant 0, x^{2}+y^{2}> 1

 

 or

 

Formulae of Inverse Trigonometric Functions -

\cos ^{-1}x + \cos ^{-1}y = \cos ^{-1}\left \left ( xy- \sqrt{1-x^{2}}\sqrt{1-y^{2}} \right )

- wherein

x\geqslant 0, y\geqslant 0

 

 \tan ^{-1}(1)+\tan ^{-1}(0)+ \tan ^{-1} (2)+ \tan ^{-1}(3)

\Rightarrow \frac{\pi }{4}+0+ \tan ^{-1}\frac{s}{16}

\Rightarrow \frac{\pi }{4}+ \tan ^{-1}(-1)=\frac{\pi }{4}+\pi -\frac{\pi }{4}=\pi

 

 


Option 1)

\pi

Correct option

Option 2)

\frac{5\pi}{4}

Incorrect option

Option 3)

\frac{\pi}{2}

Incorrect option

Option 4)

None of these

Incorrect option

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