Question

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

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

Answer 1

Answer: (D)

$\frac{1}{2}$

${\cos }^{-1}\left(\sqrt{p}\right)+{\cos }^{-1}\left(\sqrt{1-p}\right)+{\cos }^{-1}\left(\sqrt{1-q}\right)=\frac{3\pi }{4}$
Under ideal case
${\cos }^{-1}\left(\sqrt{p}\right)={\cos }^{-1}\left(\sqrt{1-p}\right)={\cos }^{-1}\left(\sqrt{1-q}\right)=\frac{\pi }{4}$
Hence
$p=\frac{1}{2}$ since $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$
Similarly,
$\sqrt{1-p}=\frac{1}{\sqrt{2}}$ or $1-p=\frac{1}{2}$ $⟹p=1-\frac{1}{2}$$=\frac{1}{2}$.

$\sqrt{1-q}=\frac{1}{\sqrt{2}}$ or $1-q=\frac{1}{2}$$⟹q=1-\frac{1}{2}$$=\frac{1}{2}$.