Inverse Trigonometric Functions Question 157
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
Options:
A) 1
B) $ \frac{1}{\sqrt{2}} $
C) $ \frac{1}{3} $
D) $ \frac{1}{2} $
Show Answer
Answer:
Correct 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},\cos 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 $ a+b+c=\frac{3\pi }{4}\Rightarrow a+b=\frac{3\pi }{4}-c $
Take cos on each side, we get $ \cos (a+b)=cos( \frac{3\pi }{4}-c ) $
$ \Rightarrow \cos a\cos b-\sin a\sin b $
$ =\cos { \pi -( \frac{\pi }{4}+c ) }=-\cos ( \frac{\pi }{4}+c ) $ Put values of $ \cos a,cosb $ and $ \sin a,\sin b, $ we get $ \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 \sqrt{1-q}=\sqrt{q} $ Squaring on both side; $ \Rightarrow 1-q=q $
$ \Rightarrow 1=2q\Rightarrow q=\frac{1}{2} $
BETA
