Question

If $co{s}^{-1}\sqrt{p}+co{s}^{-1}\sqrt{1-p}+co{s}^{-1}\sqrt{1-q}=\frac{3\pi }{4}$, then the value of q is

• A. $1$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

Let $\alpha =co{s}^{-1}\sqrt{p}:\beta =co{s}^{-1}\sqrt{1-p}$
and $\gamma =co{s}^{-1}\sqrt{1-q}$
or
$cos\alpha =\sqrt{p}:cos\beta =\sqrt{1-p}$
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 $\alpha +\beta +\gamma =\frac{3\pi }{4}$
$or,\alpha +\beta =\frac{3\pi }{4}-\gamma$
or, $cos(\alpha +\beta )=cos(\frac{3\pi }{4}-\gamma )$
$\Rightarrow cos\alpha cos\beta -sin\alpha sin\beta$ = $cos\{\pi -(\frac{\pi }{4}+\gamma )\}=-cos(\frac{\pi }{4}+\gamma )$
$\Rightarrow \sqrt{p}\sqrt{1-p}-\sqrt{1-p}\sqrt{p}=-(\frac{1}{\sqrt{2}}\sqrt{1-q}-\frac{1}{\sqrt{2}}.\sqrt{q})$
$\Rightarrow 0=\sqrt{1-q}-\sqrt{q}\Rightarrow 1-q=q\Rightarrow q=\frac{1}{2}$