Inverse Trigonometric Functions Question 67
Question: If $ {{\sin }^{-1}}\frac{1}{x}={{\sin }^{-1}}\frac{1}{a}+{{\sin }^{-1}}\frac{1}{b}, $ then the value of x is
Options:
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
Show Answer
Answer:
Correct Answer: A
Let $ {{\sin }^{-1}}\frac{1}{a}=\theta ;{{\sin }^{-1}}\frac{1}{b}=\phi $ then $ {{\sin }^{-1}}\frac{1}{x}=\theta +\phi $
$ \Rightarrow \sin {{\sin }^{-1}}\frac{1}{x}=\sin (\theta +\phi ) $
$ \Rightarrow \frac{1}{x}=\sin \theta cos\phi +cos\theta sin\phi $
$ =\frac{1}{a}\sqrt{1-\frac{1}{b^{2}}}+\sqrt{1-\frac{1}{a^{2}}.}\frac{1}{b}=\frac{\sqrt{b^{2}-1}}{ab}+\frac{\sqrt{a^{2}-1}}{ab} $
$ \Rightarrow x=\frac{ab}{\sqrt{a^{2}-1}+\sqrt{b^{2}-1}} $
BETA
