Question

Tan inverse ( cos x/ 1- sin x)

Answer 1

Question :

$\sf \tan {}^{ - 1} (\frac{cos x}{1 - sin x} )$

Formula's used :

• Trignometric Formulas

$1) \sin( \frac{\pi}{2} - x ) = \cos(x)$

$2) \cos( \frac{\pi}{2} - x ) = \sin(x)$

$3) \sin(2x) = 2 \sin(x) \cos(x)$

$4) \cos(2x) = 1 - 2 \sin {}^{2} (x)$

$5) \tan(\frac{\pi}{2} - x) = \cot(x)$

•Properties of inverse Trignometric Functions

$1) \sin {}^{ - 1} (sin x) = x$

$2) \cos {}^{ - 1} ( \cos x) = x$

$3) \tan {}^{ - 1} (tan x) = x$

Note :

These inverse properties are within the range of inverse Trignometric Functions.

Solution:

$\bf \tan{}^{ - 1}( \dfrac{cosx}{1 - sin x} )$

$= \sf tan {}^{ - 1} ( \frac{sin( \frac{\pi}{2} - x)}{1 - cos( \frac{\pi}{2} - x)} )$

We know that sin2x = 2sinxcosx

⇒sinx = 2sin(x/2) cos (x/2)

and cos2x= 1-2sin²x

⇒cosx = 1-2sin²(x/2)

_______________________________

$= \tan {}^{ - 1} ( \frac{2 \sin( \frac{\pi}{4} - \frac{x}{2} ) \sin( \frac{\pi}{4} - \frac{x}{2} ) }{2 \sin {}^{2} ( \frac{\pi}{4} - \frac{x}{2}) } )$

$= \tan {}^{ - 1}( \frac{\cancel 2 \cancel sin( \frac{\pi}{4} - \frac{x}{2}) \times \cos( \frac{\pi}{4} - \frac{x}{2} )}{ \cancel 2 \cancel sin( \frac{\pi}{4} - \frac{x}{2} ) \times sin( \frac{\pi}{4} - \frac{x}{2} )} )$

$= \tan {}^{ - 1} ( \cot( \frac{\pi}{4} - \frac{x}{2} ))$

$= \tan {}^{ - 1} ( \tan( \frac{\pi}{2} - ( \frac{\pi}{4} - \frac{x}{2} ))$

we know that tan⁻¹(tanx) = x

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

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

Therefore,

$\bf \tan {}^{ - 1} ( \dfrac{cosx}{1 - sinx} ) = \dfrac{\pi}{4} + \dfrac{x}{2}$