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# If ${{\cot }^{-1}}[{{\left( \cos \alpha \right)}^{\dfrac{1}{2}}}]+{{\tan }^{-1}}[{{\left( \cos \alpha \right)}^{\dfrac{1}{2}}}]=x$ . Then find the value of $\sin x$\begin{align} & \text{a) 1} \\ & \text{b) co}{{\text{t}}^{2}}\left( \dfrac{a}{2} \right) \\ & \text{c)}\text{ tan}\alpha \\ & \text{d) cot}\left( \dfrac{\alpha }{2} \right) \\ \end{align}

Hint: We know that ${{\cot }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2}$ hence we get x = $\dfrac{\pi }{2}$. Now since we know the value of x. we can easily find the value of sinx.

Now we are given that ${{\cot }^{-1}}[{{\left( \cos \alpha \right)}^{\dfrac{1}{2}}}]+{{\tan }^{-1}}[{{\left( \cos \alpha \right)}^{\dfrac{1}{2}}}]=x$ .
Let us say ${{\left( \cos \alpha \right)}^{\dfrac{1}{2}}}$ is equal to t.
Now we will note that the range of ${{\cos }^{-1}}\theta$ is $[0,\pi ]$ and hence the output of $\cos \alpha$ is also real
Hence we have $t={{(\cos \alpha )}^{\dfrac{1}{2}}}$ is a real number.
Substituting this in the above equation we get
${{\cot }^{-1}}t+{{\tan }^{-1}}t=x.................(1)$
Now we know that for all real numbers x the identity ${{\cot }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2}$ is true.
Now since this identity is true for all real numbers x and does not depend on x this identity is also true for t which is a real number.
Hence applying this to equation (1) we get $x=\dfrac{\pi }{2}$
Hence now we have the value of x is $\dfrac{\pi }{2}$.
Now we have to find the value of $\sin x$
Now since $x=\dfrac{\pi }{2}$ we have $\sin x=\sin \dfrac{\pi }{2}$.
We know that the value of $\sin \dfrac{\pi }{2}$ is 1.
Hence we get the value of $\sin x$ is equal to 1.

So, the correct answer is “Option A”.

Note: Here the equation is given in a confusing format which is nothing but a equation in form of ${{\cot }^{-1}}a+{{\tan }^{-1}}a$ and for that we know the identity = ${{\cot }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2}$ we can directly apply it to solve our question.