Question

Solve ${\sin }^{-1}\left(\frac{x}{\sqrt{x^2+a^2}}\right)$

• A. ${\cos }^{-1}\left(\frac{x}{a}\right)$
• B. ${\csc }^{-1}\left(\frac{x}{a}\right)$
• C. ${\tan }^{-1}\left(\frac{x}{a}\right)$
• D. None of these

Answer 1

Answer: (C)

${\tan }^{-1}\left(\frac{x}{a}\right)$

We have,

${\sin }^{-1}\left(\frac{x}{\sqrt{x^2+a^2}}\right)$

Since,

${\sin }^{-1}x={\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$

Therefore,

$={\tan }^{-1}\left(\frac{\frac{x}{\sqrt{x^2+a^2}}}{\sqrt{1-{\left(\frac{x}{\sqrt{x^2+a^2}}\right)}^2}}\right)$

$={\tan }^{-1}\left(\frac{\frac{x}{\sqrt{x^2+a^2}}}{\sqrt{1-\frac{x^2}{(x^2+a^2)}}}\right)$

$={\tan }^{-1}\left(\frac{\frac{x}{\sqrt{x^2+a^2}}}{\sqrt{\frac{x^2+a^2-x^2}{(x^2+a^2)}}}\right)$

$={\tan }^{-1}\left(\frac{\frac{x}{\sqrt{x^2+a^2}}}{\sqrt{\frac{a^2}{(x^2+a^2)}}}\right)$

$={\tan }^{-1}\left(\frac{\frac{x}{\sqrt{x^2+a^2}}}{\frac{a}{\sqrt{x^2+a^2}}}\right)$

$={\tan }^{-1}\left(\frac{x}{a}\right)$

Hence, this is the answer.