Browse Questions

# Evaluate :$\large\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}$

$\begin{array}{1 1}(a)\;\large\frac{\sin\theta}{\cos\theta}&(b)\;\tan\large\frac{\theta}{2}\\(c)\;\cot\large\frac{\theta}{2}&(d)\;0\end{array}$

We have $\large\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}$
$\Rightarrow \large\frac{(1-\cos \theta)+\sin\theta}{(1+\cos\theta)+\sin\theta}$
$\Rightarrow \large\frac{2\sin^2\Large\frac{\theta}{2}+2\sin\Large\frac{\theta}{2}\cos\Large\frac{\theta}{2}}{2\cos^2\Large\frac{\theta}{2}+2\sin\Large\frac{\theta}{2}\cos\Large\frac{\theta}{2}}$
$\Rightarrow \large\frac{2\sin\Large\frac{\theta}{2}\big(\sin\Large\frac{\theta}{2}\cos\Large\frac{\theta}{2}\big)}{2\cos\Large\frac{\theta}{2}\big(\sin\Large\frac{\theta}{2}\cos\Large\frac{\theta}{2}\big)}$
$\Rightarrow \large\frac{\sin\Large\frac{\theta}{2}}{\cos\Large\frac{\theta}{2}}$
$\Rightarrow \tan\large\frac{\theta}{2}$
Hence (b) is the correct answer.