Question : The value of $\frac{\cot30^{\circ}-\cot75^{\circ}}{\tan15^{\circ}-\tan60^{\circ}}$ is equal to:
Option 1: –1
Option 2: 0
Option 3: 1
Option 4: 2
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Correct Answer: –1
Solution :
We know that $\tan(90-\theta)=\cot \theta$
$\frac{\cot30^{\circ}-\cot75^{\circ}}{\tan15^{\circ}-\tan60^{\circ}}$
= $\frac{\tan(90^{\circ}-30^{\circ})-\tan(90^{\circ}-75^{\circ})}{\tan15^{\circ}-\tan60^{\circ}}$
= $\frac{\tan60^{\circ}-\tan15^{\circ}}{\tan15^{\circ}-\tan60^{\circ}}$
= $\frac{-1(\tan15^{\circ}-\tan60^{\circ})}{\tan15^{\circ}-\tan60^{\circ}}$
= $-1$
Hence, the correct answer is –1.
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