L.H.S. = \(\frac{1+cos\theta+sin\theta}{1+cos\theta-sin\theta}\) = \(\frac{(1+cos\theta+sin\theta)(1+cos\theta-sin\theta)}{(1+cos\theta+sin\theta)-(1+cos\theta-sin\theta)}\)

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

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

= \(\frac{1-cos^2\theta}{sin\theta(1-cos\theta)}\) = \(\frac{sin^2\theta}{sin\theta(1-cos\theta)}\)

= \(\frac{sin\theta}{1-cos\theta}\) = \(\cfrac{2sin\frac{\theta}2cos\frac{\theta}2}{2sin^2\frac{\theta}2}\) 

 L.H.S. = cot θ/2

R.H.S. = \(\frac{1+sin\theta}{cos\theta}\)

\(=\frac{sin^2\theta/2+cos^2\theta/2+2sin\theta/2cos\theta/2}{cos^2\theta/2-sin^2\theta/2}\)

\(=\frac{(cos\theta/2+sin\theta/2)^2}{(cos\theta/2+sin\theta/2)(cos\theta/2-sin\theta/2)}\) 

\(=\frac{cos\theta/2+sin\theta/2}{cos\theta/2 - sin\theta/2}\)

\(= \frac{(cos\theta/2+sin\theta/2)+(cos\theta/2-sin\theta/2)}{(cos\theta/2+sin\theta/2)-(cos\theta/2-sin\theta/2)}\)

\(=\frac{2cos\theta/2}{2sin\theta/2}=cot\theta/2\)

Hence, L.H.S = R.H.S.

∴ \(\frac{1+cos\theta+sin\theta}{1+cos\theta-sin\theta}\) = \(\frac{1+sin\theta}{cos\theta}\)

Hence Proved