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

The correct option is D

$\frac{1}{2}$

Explanation for the correct option:

Step 1. Find the value of $q$:

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

$\Rightarrow$${\cos }^{-1}\sqrt{p}+{\cos }^{-1}\sqrt{(1-{\left(\sqrt{p}\right)}^2)}+{\cos }^{-1}\sqrt{(1-{\left(\sqrt{q}\right)}^2)}=\frac{3\pi }{4}$

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

Step 2. By using formula $\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{\pi }}{\mathbf{2}}$, we get

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

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

$\Rightarrow$ ${\cos }^{-1}\sqrt{(1–q)}=\frac{\pi }{4}$

$\Rightarrow$ $\sqrt{(1–q)}=\cos \left(\frac{\pi }{4}\right)$

$\Rightarrow$ $\sqrt{(1–q)}=\frac{1}{\sqrt{2}}$

Step 3. By squaring both sides, we get

$1–q=\frac{1}{2}$

$\Rightarrow$ $q=1–\frac{1}{2}$

$\mathbf{\therefore }\mathit{q}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}$

Hence, Option ‘D’ is Correct.