Question
If $${\sin ^{ - 1}}\frac{1}{x} = {\sin ^{ - 1}}\frac{1}{a} + {\sin ^{ - 1}}\frac{1}{b},$$ then the value of $$x$$ is
A.
$$\frac{{ab}}{{\sqrt {{a^2} - 1} + \sqrt {{b^2} - 1} }}$$
B.
$$\frac{{ab}}{{\sqrt {{a^2} - 1} - \sqrt {{b^2} - 1} }}$$
C.
$$\frac{{2ab}}{{\sqrt {{a^2} - 1} + \sqrt {{b^2} - 1} }}$$
D.
None of these
Answer :
$$\frac{{ab}}{{\sqrt {{a^2} - 1} + \sqrt {{b^2} - 1} }}$$
Solution :
$$\eqalign{
& {\text{Let, }}{\sin ^{ - 1}}\frac{1}{a} = \theta ;{\sin ^{ - 1}}\frac{1}{b} = \phi {\text{ then }}{\sin ^{ - 1}}\frac{1}{x} = \theta + \phi \cr
& \Rightarrow \sin {\sin ^{ - 1}}\frac{1}{x} = \sin \left( {\theta + \phi } \right) \cr
& \Rightarrow \frac{1}{x} = \sin \theta \cos \phi + \cos \theta \sin \phi \cr
& = \frac{1}{a}\sqrt {1 - \frac{1}{{{b^2}}}} + \sqrt {1 - \frac{1}{{{a^2}}}} \cdot \frac{1}{b} = \frac{{\sqrt {{b^2} - 1} }}{{ab}} + \frac{{\sqrt {{a^2} - 1} }}{{ab}} \cr
& \Rightarrow x = \frac{{ab}}{{\sqrt {{a^2} - 1} + \sqrt {{b^2} - 1} }} \cr} $$