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Trigonometric Identities — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsTrigonometric Identities3 Marks
Question
Prove the following trigonometric identities.
$\sqrt{\frac{\sec\theta-1}{\sec\theta+1}}+\sqrt{\frac{\sec\theta+1}{\sec\theta-1}}=2 \text{cosec }\theta$
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Answer

We have,
$\text{L.H.S}=\sqrt{\frac{\sec\theta-1}{\sec\theta+1}}+\sqrt{\frac{\sec\theta+1}{\sec\theta-1}}=\frac{\sqrt{\sec\theta-1}}{\sqrt{\sec\theta+1}}+\frac{\sqrt{\sec\theta+1}}{\sqrt{\sec\theta-1}}$
$=\frac{\sqrt{\sec\theta-1}\sqrt{\sec\theta-1}+\sqrt{\sec\theta+1}\sqrt{\sec\theta+1}}{\sqrt{\sec\theta+1}\sqrt{\sec\theta-1}}$
$=\frac{(\sqrt{\sec\theta-1})^2+(\sqrt{\sec\theta+1})^2}{\sqrt{(\sec\theta-1)(\sec\theta+1)}}$
$=\frac{\sec\theta-1+\sec\theta+1}{\sqrt{\sec^2\theta-1}}$
$=\frac{2\sec\theta}{\sqrt{\tan^2\theta}}$
$=\frac{2\sec\theta}{\tan\theta}$
$=\frac{2\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}}$
$=2\frac{1}{\sin\theta}$
$=2\text{cosec }\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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