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