Question
Prove the following trigonometric identities.
$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$
$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$
✓
Answer
We have to prove the following identity-
$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$
Consider the L.H.S
$\text{L.H.S}=\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}$
$=\Big(\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}\Big)\Big(\frac{1+\cos\theta+\sin\theta}{1+\cos\theta+\sin\theta}\Big)$
$=\frac{(1+\cos\theta+\sin\theta)^2}{(1+\cos\theta)^2-\sin^2\theta}$
$=\frac{2+2(\cos\theta+\sin\theta+\sin\theta\cos\theta)}{2\cos^2\theta+2\cos\theta}$
$=\frac{2(1+\cos\theta)(1+\sin\theta)}{2\cos\theta(1+\cos\theta)}$
$=\frac{1+\sin\theta}{\cos\theta}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$
Consider the L.H.S
$\text{L.H.S}=\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}$
$=\Big(\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}\Big)\Big(\frac{1+\cos\theta+\sin\theta}{1+\cos\theta+\sin\theta}\Big)$
$=\frac{(1+\cos\theta+\sin\theta)^2}{(1+\cos\theta)^2-\sin^2\theta}$
$=\frac{2+2(\cos\theta+\sin\theta+\sin\theta\cos\theta)}{2\cos^2\theta+2\cos\theta}$
$=\frac{2(1+\cos\theta)(1+\sin\theta)}{2\cos\theta(1+\cos\theta)}$
$=\frac{1+\sin\theta}{\cos\theta}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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