Question

$\frac{cos\theta }{1-sin\theta }=\frac{1+cos\theta +sin\theta }{1+cos\theta -sin\theta }$

Answer 1

$RHS=\frac{1+cos\theta +sin\theta }{1+cos\theta -sin\theta }$

$=\frac{\left[\right(1+cos\theta )+sin\theta ]}{(1+cos\theta )-sin\theta }\times \frac{\left[\right(1+cos\theta )+sin\theta ]}{(1+cos\theta +sin\theta )}$

$=\frac{\left(\right(1+cos\theta )+sin\theta {)}^2}{(1+cos\theta {)}^2-si{n}^2\theta }$ [Using $(a+b)(a+b)=(a+b{)}^2$ and $(a+b)(a-b)=a^2{b}^2]$

$=\frac{(1+cos\theta {)}^2+si{n}^2\theta +sin\theta +sin\theta (1+cos\theta )}{1+co{s}^2\theta +2cos\theta -si{n}^2\theta }$ [Using $(1+b{)}^2=a^2+b^2+2ab]$

$=\frac{1+co{s}^2+2.1cos\theta +si{n}^2\theta +2sin\theta (1+cos\theta )}{1+co{s}^2\theta +2cos\theta -(1-co{s}^2\theta )}$ [Using $si{n}^2\theta =1-co{s}^2\theta ]$

$=\frac{1+1+2cos\theta +2sin\theta (1+cos\theta )}{1-1+co{s}^2\theta +co{s}^2\theta +2cos\theta }$ [Using $si{n}^2\theta +co{s}^2\theta =1]$

$=\frac{2+2cos\theta +2sin\theta (1+cos\theta )}{2co{s}^2\theta +2cos\theta }$

$=\frac{2(1+cos\theta )+2sin\theta (1+cos\theta )}{2cos\theta (cos\theta +1)}=\frac{(1+cos\theta )(2+2sin\theta )}{2cos\theta (1+cos\theta )}$

$=\frac{1+sin\theta }{cos\theta }\times \frac{1-sin\theta }{1-sin\theta }=\frac{co{s}^2\theta }{1-sin\theta }$