Prove the following

LHS $=\frac{1+\sec \theta-\tan \theta}{1+\sec \theta+\tan \theta}$

$=\frac{1+1 / \cos \theta-\sin \theta / \cos \theta}{1+1 / \cos \theta+\sin \theta / \cos \theta}$ $\left[\because \sec \theta=\frac{1}{\cos \theta}\right.$ and $\left.\tan \theta=\frac{\sin \theta}{\cos \theta}\right]$

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

$\left[\because 1+\cos \theta=2 \cos ^{2} \frac{\theta}{2}\right.$ and $\left.\sin \theta=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right]$

$=\frac{2 \cos ^{2} \frac{\theta}{2}-2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}+2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}}=\frac{2 \cos \frac{\theta}{2}\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)}{2 \cos \frac{\theta}{2}\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)}$

$=\frac{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}} \times \frac{\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)}{\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)}$ [ by rationalisation]

$=\frac{\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)^{2}}{\left(\cos ^{2} \frac{\theta}{2}-\sin ^{2} \frac{\theta}{2}\right)}$ $\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right.$ and $\left.(a-b)(a+b)=\left(a^{2}-b^{2}\right)\right]$

$=\frac{\left(\cos ^{2} \frac{\theta}{2}+\sin ^{2} \frac{\theta}{2}\right)-\left(2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}\right)}{\cos \theta}$ $\left[\because \cos ^{2} \frac{\theta}{2}-\sin ^{2} \frac{\theta}{2}=\cos \theta\right]$

$=\frac{1-\sin \theta}{\cos \theta}$ $\left[\because \sin ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}=1\right]$

$=\mathrm{RHS}$ 

Hence proved.