Question

$ta{n}^{-1}\sqrt{\frac{a-x}{a+x}}$

Answer 1

$y=ta{n}^{-1}\left[\frac{1+x^2}{x}\right]=ta{n}^{-1}\left[cot\frac{\theta }{2}\right]$
$\therefore y=\frac{\pi }{2}-\frac{\theta }{2}$
$\therefore y=\frac{\pi }{2}-\frac{1}{2}ta{n}^{-1}x$
$y=ta{n}^{-1}\sqrt{\frac{a-x}{a+x}}$
$Letx=acos\theta \theta =co{s}^{-1}\left(\frac{x}{a}\right)$
$\sqrt{\frac{a-x}{a+x}}=\sqrt{\frac{a-acos\theta }{a+acos\theta }}=\sqrt{\frac{1-cos\theta }{1+cos\theta }}=\sqrt{\frac{sin\frac{\theta }{2}+co{s}^2\frac{\theta }{2}-co{s}^2\frac{\theta }{2}+si{n}^2\frac{\theta }{2}}{co{s}^2\frac{\theta }{2}+si{n}^2\frac{\theta }{2}+co{s}^2\frac{\theta }{2}-si{n}^2\frac{\theta }{2}}}$
$\sqrt{\frac{a-x}{a+x}}=\sqrt{\frac{2si{n}^2\frac{\theta }{2}}{2co{s}^2\frac{\theta }{2}}}=tan\frac{\theta }{2}$
$\therefore y=ta{n}^{-1}\left[tan\frac{\theta }{2}\right]=\frac{\theta }{2}$
$\therefore y=\frac{1}{2}co{s}^{-1}\left(\frac{x}{a}\right)$