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

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