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 equal to :

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

Answer 1

Answer: (D)

Let ${\cos }^{-1}\sqrt{p}+{\cos }^{-1}\sqrt{1-p}+{\cos }^{-1}\sqrt{1-q}=\frac{3\pi }{4}$ ......(i)
Let $a={\cos }^{-1}\sqrt{p}$
$b={\cos }^{-1}\sqrt{1-p}$ and
$c={\cos }^{-1}\sqrt{1-q}$
$\Rightarrow \cos a=\sqrt{p}$
$\cos b=\sqrt{1-p}$
$c=\sqrt{1-q}$
$\Rightarrow {\cos }^{2}a=p,$
${\cos }^{2}b=1-p{\cos }^{2}c=1-q$
Now, ${\sin }^{2}a=1-{\cos }^{2}a=1-p$
$\Rightarrow \sin a=\sqrt{1-p}$
${\sin }^{2}b=1-{\cos }^{2}b=1-1+p$
$\Rightarrow \sin b=\sqrt{p}$
${\sin }^{2}c=1-{\cos }^{2}c=1-1+q=q$
$\Rightarrow \sin c=\sqrt{q}$
$\therefore$ equation (i) can be written as
$\therefore a+b+c=\frac{3\pi }{4}$
$\Rightarrow a+b=\frac{3\pi }{4}-c$
Take cos on each side, we get
$\cos \left(a+b\right)=\cos \left(\frac{3\pi }{4}-c\right)$
$\Rightarrow \cos a\cos b-\sin a\sin b$
$=\cos \left\{\pi -\left(\frac{\pi }{4}+c\right)\right\}=-\cos \left(\frac{\pi }{4}+c\right)$
Put values of cos a, cos b and sin a, sin b,
we get
$\sqrt{p}.\sqrt{1-p}-\sqrt{1-p}\sqrt{p}$
$=-\left(\frac{1}{\sqrt{2}}\sqrt{1-q}-\frac{1}{\sqrt{2}}\sqrt{q}\right)$
$\Rightarrow 0=\sqrt{1-q}-\sqrt{q}\Rightarrow \sqrt{1-q}=\sqrt{q}$
Squaring on both side
$\Rightarrow 1-q=q$
$\Rightarrow 1=2q\Rightarrow q=\frac{1}{2}$