Question
Prove the following trigonometric identities.
$\frac{1+\sin\theta}{\cos\theta}+\frac{\cos\theta}{1+\sin\theta}=2\sec\theta$
$\frac{1+\sin\theta}{\cos\theta}+\frac{\cos\theta}{1+\sin\theta}=2\sec\theta$
✓
Answer
$\text{L.H.S}=\frac{1+\sin\theta}{\cos\theta}+\frac{\cos\theta}{1+\sin\theta}$
$=\frac{(1+\sin\theta)^2+\cos^2\theta}{\cos\theta(1+\sin\theta)}$
$=\frac{1+\sin^2\theta+2\sin\theta+\cos^2\theta}{\cos\theta(1+\sin\theta)}$
$=\frac{1+(\sin^2\theta+\cos^2\theta)+2\sin\theta}{\cos\theta(1+\sin\theta)}$
$=\frac{1+1+2\sin\theta}{\cos\theta(1+\sin\theta)}$
$=\frac{2+2\sin\theta}{\cos\theta(1+\sin\theta)}$
$=\frac{2(1+\sin\theta)}{\cos\theta(1+\sin\theta)}$
$=\frac{2}{\cos\theta}$
$=2\sec\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
$=\frac{(1+\sin\theta)^2+\cos^2\theta}{\cos\theta(1+\sin\theta)}$
$=\frac{1+\sin^2\theta+2\sin\theta+\cos^2\theta}{\cos\theta(1+\sin\theta)}$
$=\frac{1+(\sin^2\theta+\cos^2\theta)+2\sin\theta}{\cos\theta(1+\sin\theta)}$
$=\frac{1+1+2\sin\theta}{\cos\theta(1+\sin\theta)}$
$=\frac{2+2\sin\theta}{\cos\theta(1+\sin\theta)}$
$=\frac{2(1+\sin\theta)}{\cos\theta(1+\sin\theta)}$
$=\frac{2}{\cos\theta}$
$=2\sec\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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