Explanation

(d)
Let $\alpha = \cos^{- 1}{\sqrt{P},\beta = \cos^{- 1}\sqrt{1 - P}}$
And $\gamma = \cos^{- 1}\sqrt{1 - q}$
$\Rightarrow \cos{\alpha = \sqrt{p,}\ \cos\beta = \sqrt{1 - p}}$
And $\cos{\gamma = \sqrt{1 - q}}$
Therefore, $\sin{\alpha = \sqrt{1 - p,}\sin{\beta = \sqrt{p}\ and\sin{\gamma = \sqrt{q}}}}$
The given equation may be written as
$\alpha + \beta + \gamma = \frac{3\pi}{4}$
$\Rightarrow \alpha + \beta = \frac{3\pi}{4} - \gamma$
$\Rightarrow \cos{(\alpha + \beta) = \cos\left( \frac{3\pi}{4} - \gamma \right)}$
$\Rightarrow \cos{\alpha\cos{\beta - \sin{\alpha\sin\beta}}}$
$= \cos{\left\{ \pi - \left( \frac{\pi}{4} + \gamma \right) \right\} = - \cos\left( \frac{\pi}{4} + \gamma \right)}$
$\Rightarrow \sqrt{p}\sqrt{1 - p} - \sqrt{1 - p}\sqrt{p} = - \left( \frac{1}{\sqrt{2}}\sqrt{1 - q} - \frac{1}{\sqrt{2}}\sqrt{q} \right)$
$\Rightarrow 0 = \sqrt{1 - q} - \sqrt{q} \Rightarrow 1 - q = q$
$\Rightarrow q = \frac{1}{2}$