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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