Question

If \[{{\cos }^{-1}}\sqrt{p}+{{\cos }^{-1}}\sqrt{1-p}+{{\cos }^{-1}}\sqrt{1-q}=\dfrac{3\pi }{4}\]then the value of q is
A. 1
B. \[\dfrac{1}{\sqrt{2}}\]
C. \[\dfrac{1}{3}\]
D. \[\dfrac{1}{2}\]

Answer 1

Hint: Here in this problem assume the ideal case conditions that the three each cos terms are equal and each cos term is equal to \[\dfrac{\pi }{4}\]. We know that when an inverse cos term moves from left hand side to right hand side or vice versa characterized by equation symbol then it becomes cos there by we will get the value of p and then the value of q.

Complete step-by-step answer:
Given that \[{{\cos }^{-1}}\sqrt{p}+{{\cos }^{-1}}\sqrt{1-p}+{{\cos }^{-1}}\sqrt{1-q}=\dfrac{3\pi }{4}\]
Under ideal case conditions,
\[{{\cos }^{-1}}\sqrt{p}={{\cos }^{-1}}\sqrt{1-p}={{\cos }^{-1}}\sqrt{1-q}=\dfrac{\pi }{4}\]
Hence \[p=\dfrac{1}{2}\]because the value of \[\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\]. . . . . . . . . . . . . . . . . . . . . . . . (1)
Similarly \[\sqrt{1-p}=\dfrac{1}{\sqrt{2}}\]since the value of \[\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\]
\[\sqrt{1-p}=\dfrac{1}{\sqrt{2}}\]. . . . . . . . . . . . . . . . . . . . . . . . . (2)
Now squaring on both the left hand side and right hand side then we will get,
\[1-p=\dfrac{1}{2}\]
\[p=1-\dfrac{1}{2}\]
\[p=\dfrac{1}{2}\]
Similarly \[\sqrt{1-q}=\dfrac{1}{\sqrt{2}}\]. . . . . . . . . . . . . . . . . . . . . (3)
Now squaring on both the left hand side and right hand side then we will get,

\[1-q=\dfrac{1}{2}\]
\[q=1-\dfrac{1}{2}\]
\[q=\dfrac{1}{2}\]
So the obtained value of q is \[\dfrac{1}{2}\]
So the correct option for above question is option (D)

Note: We know the basic trigonometric formula that is the value of \[\cos \dfrac{\pi }{4} = \dfrac{1}{\sqrt{2}}\]. In some problems we have to use the ideal case conditions and then proceed so that we will get the required parameters. Carefully do the basic mathematical operations and find the value of required parameters asked in question.