# $cot^{-1}\left [ (cos\, \alpha )^{\frac{1}{2}} \right ]+tan^{-1}\left [ (cos\, \alpha )^{\frac{1}{2}} \right ]=x\; \; then\; sinx=$  Option 1) 1 Option 2) $\dpi{100} cot^{2}(\alpha /2)\;$ Option 3) $\dpi{100} \; \; tan\alpha \;$ Option 4) $\dpi{100} \; \; cot(\alpha /2)$

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)

$\dpi{100} cot^{2}(\alpha /2)\;$

This is incorrect option

Option 3)

$\dpi{100} \; \; tan\alpha \;$

This is incorrect option

Option 4)

$\dpi{100} \; \; cot(\alpha /2)$

This is incorrect option

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