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