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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


Team Careers360 24th Jan, 2024
Answer (1)
Team Careers360 25th Jan, 2024

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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