cot^{-1}\left [ (cos\, \alpha )^{\frac{1}{2}} \right ]+tan^{-1}\left [ (cos\, \alpha )^{\frac{1}{2}} \right ]=x\; \; then\; sinx= 

  • Option 1)

    1

  • Option 2)

    cot^{2}(\alpha /2)\;

  • Option 3)

    \; \; tan\alpha \;

  • Option 4)

    \; \; cot(\alpha /2)

 

Answers (1)

As we learnt in 

Important Results of Inverse Trigonometric Functions -

\tan ^{-1}x + \cot ^{-1}x = \frac{\pi }{2}

- wherein

When x\epsilon R

 

 \cot ^{-1}(cos\ a)^\frac{1}{2}+tan ^{-1}(cos\ a)^\frac{1}{2}=x

let (\cos a)^\frac{1}{2}=t

Then \cot ^{-1}t+\tan ^{^{-1}}t=\frac{\pi}{2} [identity]

Thus, x=\frac{\pi}{2}

\sin x=1


Option 1)

1

This is correct option

Option 2)

cot^{2}(\alpha /2)\;

This is incorrect option

Option 3)

\; \; tan\alpha \;

This is incorrect option

Option 4)

\; \; cot(\alpha /2)

This is incorrect option

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