Question

$arc{\tan }^{}\frac{x}{\sqrt{(a²-x²)}}=?$

Answer 1

Here the basic trigonometric function of $Sin\theta =y$ can be changed to $\theta ={\sin }^{-1}y$

Let $x=a\sin \theta$

$\Rightarrow \theta ={\sin }^{-1}\left(\frac{x}{a}\right)$

Now, Using trigonometric identity we get

$\begin{array}{rcl}{\tan }^{-1}\frac{x}{\sqrt{\left(a^2-x^2\right)}}& =& {\tan }^{-1}\left(\frac{a\sin \theta }{\sqrt{(a²-{\left(a\sin \theta \right)}^2}}\right)\\ & =& {\tan }^{-1}\left(\frac{a\sin \theta }{\sqrt{(a²-a^2{\sin }^2\theta }}\right)\\ & =& {\tan }^{-1}\left(\frac{a\sin \theta }{\sqrt{a^2(1-{\sin }^2\theta )}}\right)\\ & =& {\tan }^{-1}\left(\frac{a\sin \theta }{a\cos \theta }\right)\\ & =& {\tan }^{-1}(\tan \theta )\\ & =& \theta \\ & =& {\sin }^{-1}\left(\frac{x}{a}\right)\end{array}$

Thus, ${\tan }^{-1}\frac{x}{\sqrt{(a²-x²)}}$$={\sin }^{-1}\left(\frac{x}{a}\right)$