Question
Prove the following trigonometric identities.
$\frac{1}{\sec\text{A}-1}+\frac{1}{\sec\text{A}+1}=2\text{cosec A}\cot\text{A}$
$\frac{1}{\sec\text{A}-1}+\frac{1}{\sec\text{A}+1}=2\text{cosec A}\cot\text{A}$
✓
Answer
We need to prove $\frac{1}{\sec\text{A}-1}+\frac{1}{\sec\text{A}+1}=2\text{cosec A}\cot\text{A}$
Solving the L.H.S, we get
$\text{L.H.S}=\frac{1}{\sec\text{A}-1}+\frac{1}{\sec\text{A}+1}=\frac{\sec\text{A}+1+\sec \text{A}-1}{(\sec\text{A}-1)(\sec\text{A}+1)}$
$=\frac{2\sec\text{A}}{\sec^2\text{A}-1}$
Further using the property $1+\tan^2\theta=\sec^2\theta,$ we get
So,
$\frac{2\sec\text{A}}{\sec^2\text{A}-1}=\frac{2\sec\text{A}}{\tan^2\text{A}}$
$=\frac{2\Big(\frac{1}{\cos\text{A}}\Big)}{\frac{\sin^2\text{A}}{\cos^2\text{A}}}$
$=2\frac{1}{\cos\text{A}}\times\frac{\cos^2\text{A}}{\sin^2\text{A}}$
$=2\Big(\frac{\cos\text{A}}{\sin\text{A}}\Big)\times\frac{1}{\sin\text{A}}$
$=2\text{cosec A}\cot\text{A}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Solving the L.H.S, we get
$\text{L.H.S}=\frac{1}{\sec\text{A}-1}+\frac{1}{\sec\text{A}+1}=\frac{\sec\text{A}+1+\sec \text{A}-1}{(\sec\text{A}-1)(\sec\text{A}+1)}$
$=\frac{2\sec\text{A}}{\sec^2\text{A}-1}$
Further using the property $1+\tan^2\theta=\sec^2\theta,$ we get
So,
$\frac{2\sec\text{A}}{\sec^2\text{A}-1}=\frac{2\sec\text{A}}{\tan^2\text{A}}$
$=\frac{2\Big(\frac{1}{\cos\text{A}}\Big)}{\frac{\sin^2\text{A}}{\cos^2\text{A}}}$
$=2\frac{1}{\cos\text{A}}\times\frac{\cos^2\text{A}}{\sin^2\text{A}}$
$=2\Big(\frac{\cos\text{A}}{\sin\text{A}}\Big)\times\frac{1}{\sin\text{A}}$
$=2\text{cosec A}\cot\text{A}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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