Question : If $\tan\theta-\cot\theta=0$ and $\theta$ is positive acute angle, then the value of $\frac{\tan(\theta+15)}{\tan(\theta-15)}$ is:
Option 1: $3$
Option 2: $\frac{1}{\sqrt{3}}$
Option 3: $\frac{1}{3}$
Option 4: $\sqrt{3}$
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Correct Answer: $3$
Solution :
Given: $\tan\theta-\cot\theta=0$ and $\theta$ is a positive acute angle.
Here $\tan\theta-\cot\theta=0$
⇒ $\tan\theta=\cot\theta$
⇒ $\theta=45°$
Now, $\frac{\tan(\theta+15)}{\tan(\theta–15)}$
= $\frac{\tan(45+15)}{\tan(45–15)}$
= $\frac{\tan60°}{\tan30°}$
= $\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}$
= $\sqrt{3}×\frac{\sqrt{3}}{1}=3$
Hence, the correct answer is $3$.
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