Question

Express ${\tan }^{-1}\left(\frac{\cos x}{1-\sin x}\right),-\frac{\pi }{2}<x<\frac{\pi }{2}$ in the simplest form.

Answer 1

$ta{n}^{-1}\frac{cosx}{1-sinx}$

Calculate cos x & 1-sin x .

As $cos2x=co{s}^{2}x-si{n}^{2}x.$

$cos2\left(\frac{x}{2}\right)=co{s}^2\frac{x}{2}-si{n}^2\frac{x}{2}$

$cosx=co{s}^2\frac{x}{2}-si{n}^2\frac{x}{2}$

Similarly $sinx=2sin\frac{x}{2}cos\frac{x}{2}$

Now

$ta{n}^{-1}\left[\frac{co{s}^2\frac{x}{2}-si{n}^2\frac{x}{2}}{1-\left(2sin\frac{x}{2}cos\frac{x}{2}\right)}\right]$

$ta{n}^{-1}\left[\frac{co{s}^2\frac{x}{2}-si{n}^2\frac{x}{2}}{co{s}^2\frac{x}{2}+si{n}^2\frac{x}{2}-2sin\frac{x}{2}cos\frac{x}{2}}\right]$

$ta{n}^{-1}\left[\frac{(cos\frac{x}{2}+sin\frac{x}{2})(cos\frac{x}{2}-sin\frac{x}{2})}{(cos\frac{x}{2}-sin\frac{x}{2}{)}^2}\right]$

$ta{n}^{-1}\left[\frac{cos\frac{x}{2}+sin\frac{x}{2}}{cos\frac{x}{2}-sin\frac{x}{2}}\right]$

$ta{n}^{-1}\left[\frac{cos\frac{x}{2}+sin\frac{x}{2}/cos\frac{x}{2}}{cos\frac{x}{2}-sin\frac{x}{2}/cos\frac{x}{2}}\right]$

$ta{n}^{-1}\left[\frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}\right]$

$ta{n}^{-1}\left(\frac{tan\frac{\pi }{4}+tan\frac{x}{2}}{1-tan\frac{\pi }{4}tan\frac{x}{2}}\right)$

$=ta{n}^{-1}\left[tan\right(\frac{\pi }{4}+\frac{x}{2}\left)\right]$

$=\frac{\pi }{4}+\frac{x}{2}.$