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
Correct Answer: A
Solution :
[a] Let \[{{\sin }^{-1}}\frac{1}{a}=\theta ;{{\sin }^{-1}}\frac{1}{b}=\phi \] then \[{{\sin }^{-1}}\frac{1}{x}=\theta +\phi \] \[\Rightarrow \sin {{\sin }^{-1}}\frac{1}{x}=\sin (\theta +\phi )\] \[\Rightarrow \frac{1}{x}=\sin \theta cos\phi +cos\theta sin\phi \] \[=\frac{1}{a}\sqrt{1-\frac{1}{{{b}^{2}}}}+\sqrt{1-\frac{1}{{{a}^{2}}}.}\frac{1}{b}=\frac{\sqrt{{{b}^{2}}-1}}{ab}+\frac{\sqrt{{{a}^{2}}-1}}{ab}\] \[\Rightarrow x=\frac{ab}{\sqrt{{{a}^{2}}-1}+\sqrt{{{b}^{2}}-1}}\]
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