Question
Prove the following trigonometric identities.
$\frac{1-\cos\text{A}}{1+\cos\text{A}}=(\cot\text{A}-\text{cosec A})^2$
$\frac{1-\cos\text{A}}{1+\cos\text{A}}=(\cot\text{A}-\text{cosec A})^2$
✓
Answer
$\text{L.H.S}=\frac{(1-\cos\text{A})}{(1+\cos\text{A})}\times\frac{(1-\cos\text{A})}{(1-\cos\text{A})}$
$=\frac{(1-\cos\text{A})^2}{(1-\cos^2\text{A})}$
$=\frac{(1-\cos\text{A})^2}{\sin^2\text{A}}$
$=\Big(\frac{1}{\sin\text{A}}-\frac{\cos\text{A}}{\sin\text{A}}\Big)^2$
$=(\text{cosec A}-\cot\text{A})^2$
$=(\cot\text{A}-\text{cosec A})^2$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
$=\frac{(1-\cos\text{A})^2}{(1-\cos^2\text{A})}$
$=\frac{(1-\cos\text{A})^2}{\sin^2\text{A}}$
$=\Big(\frac{1}{\sin\text{A}}-\frac{\cos\text{A}}{\sin\text{A}}\Big)^2$
$=(\text{cosec A}-\cot\text{A})^2$
$=(\cot\text{A}-\text{cosec A})^2$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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