Prove the following trigonometric identities.

Solution:

(i) In the given question, we need to prove $\left(\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}\right)^{2}=\frac{1-\cos \theta}{1+\cos \theta}$

Taking $\sin \theta$ common from the numerator and the denominator of the L.H.S, we get

$\left(\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}\right)^{2}=\left(\frac{(\sin \theta)(\operatorname{cosec} \theta+1-\cot \theta)}{(\sin \theta)(\operatorname{cosec} \theta+1+\cot \theta)}\right)^{2}$

$=\left(\frac{1+\operatorname{cosec} \theta-\cot \theta}{1+\operatorname{cosec} \theta+\cot \theta}\right)^{2}$

Now, using the property $1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta$, we get

$\left(\frac{1+\operatorname{cosec} \theta-\cot \theta}{1+\operatorname{cosec} \theta+\cot \theta}\right)^{2}=\left(\frac{\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)+\operatorname{cosec} \theta-\cot \theta}{1+\operatorname{cosec} \theta+\cot \theta}\right)^{2}$

Using $a^{2}-b^{2}=(a+b)(a-b)$, we get

$\left(\frac{\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)+\operatorname{cosec} \theta-\cot \theta}{1+\operatorname{cosec} \theta+\cot \theta}\right)^{2}=\left(\frac{(\operatorname{cosec} \theta+\cot \theta)(\operatorname{cosec} \theta-\cot \theta)+(\operatorname{cosec} \theta-\cot \theta)}{1+\operatorname{cosec} \theta+\cot \theta}\right)^{2}$

Taking $\operatorname{cosec} \theta-\cot \theta \operatorname{common}$ from the numerator, we get

$\left(\frac{(\operatorname{cosec} \theta+\cot \theta)(\operatorname{cosec} \theta-\cot \theta)+\operatorname{cosec} \theta-\cot \theta}{1+\operatorname{cosec} \theta+\cot \theta}\right)^{2}=\left(\frac{(\operatorname{cosec} \theta-\cot \theta)(\operatorname{cosec} \theta+\cot \theta+1)}{1+\operatorname{cosec} \theta+\cot \theta}\right)^{2}$

$=(\operatorname{cosec} \theta-\cot \theta)^{2}$

Using $\cot \theta=\frac{\cos \theta}{\sin \theta}$ and $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$, we get

$(\operatorname{cosec} \theta-\cot \theta)^{2}=\left(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right)^{2}$

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

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

Now, using the property $\sin ^{2} \theta+\cos ^{2} \theta=1$, we get

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

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

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

Hence proved.

(ii)
Consider the LHS.

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

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

$=\frac{(\sec \theta-\tan \theta)+\left(\sec ^{2} \theta-\tan ^{2} \theta\right)}{1+\sec \theta}$ $\left(\sec ^{2} \theta-\tan ^{2} \theta=1\right)$

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

$=\sec \theta-\tan \theta$

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

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

= RHS
Hence proved.