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)

$\frac{1}{2}$

${\cos }^{-1}\sqrt{1-p}={\sin }^{-1}\sqrt{p}$

Now, the given equation becomes:

${\cos }^{-1}\sqrt{p}+{\sin }^{-1}\sqrt{p}+{\cos }^{-1}\sqrt{1-q}=\frac{3\pi }{4}$ ...(1)

We know that,

${\sin }^{-1}A+{\cos }^{-1}A=\frac{\pi }{2}$ ...(2)

Therefore, from (1) and (2), we get

${\cos }^{-1}\sqrt{1-q}=\frac{3\pi }{4}-\frac{\pi }{2}=\frac{\pi }{4}$

$\Rightarrow \sqrt{1-q}=\frac{1}{\sqrt{2}}$

$\Rightarrow q=\frac{1}{2}$