Question

70. If arccos(sqrt(cos alpha)) * arctan(sqrt(cos alpha)) = x then the value of sin x is

Answer 1

sin x = sin{arccos(sqrt(cos alpha)) * arctan(sqrt(cos alpha))}

Answer 2

We start with

$$x=\arccos\Bigl(\sqrt{\cos\alpha}\Bigr)\cdot \arctan\Bigl(\sqrt{\cos\alpha}\Bigr).$$

There is no standard trigonometric identity that “combines” a product of an inverse‐cosine and an inverse‐tangent into a simpler function. In other words, without additional restrictions on the parameter $\alpha$ (or further information) the expression for $x$ cannot be simplified further so that $\sin x$ can be written in a “nice” form.

Thus one acceptable answer is to simply write

$$\sin x=\sin\Bigl\{\arccos\Bigl(\sqrt{\cos\alpha}\Bigr)\cdot \arctan\Bigl(\sqrt{\cos\alpha}\Bigr)\Bigr\}.$$

Any answer equivalent to the above is correct.