Inverse Trigonometric Functions Question 7
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
[Karnataka CET 2002; Pb. CET 2000]
Options:
A) 1
B) $ \frac{1}{\sqrt{2}} $
C) $ \frac{1}{3} $
D) $ \frac{1}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ \alpha ={{\cos }^{-1}}\sqrt{p}; $
$ \beta ={{\cos }^{-1}}\sqrt{1-p} $ and $ \gamma ={{\cos }^{-1}}\sqrt{1-q}or\cos \alpha =\sqrt{p};\cos \beta =\sqrt{1-p} $ and $ \cos \gamma =\sqrt{1-q.} $
Therefore $ \sin \alpha =\sqrt{1-p}, $ $ \sin \beta =\sqrt{p} $ and $ \sin \gamma =\sqrt{q} $ .
The given equation may be written as $ \alpha +\beta +\gamma =\frac{3\pi }{4} $
Or $ \alpha +\beta =\frac{3\pi }{4}-\gamma $
Or $ \cos (\alpha +\beta )=\cos ( \frac{3\pi }{4}-\gamma ) $
Therefore $ \cos \alpha \cos \beta -\sin \alpha \sin \beta = $ $ \cos { \pi -( \frac{\pi }{4}+\gamma ) }=-\cos ( \frac{\pi }{4}+\gamma ) $
Therefore $ \sqrt{p}\sqrt{1-p}-\sqrt{1-p}\sqrt{p} $ $ =-( \frac{1}{\sqrt{2}}\sqrt{1-q}-\frac{1}{\sqrt{2}}.\sqrt{q} ) $
Therefore $ 0=\sqrt{1-q}-\sqrt{q} $
Therefore $ 1-q=q $
Therefore $ q=\frac{1}{2}. $
BETA
