Question : If $\tan\alpha=2$, then the value of $\frac{\operatorname{cosec}^{2}\alpha-\sec^{2}\alpha}{\operatorname{cosec}^{2}\alpha+\sec^{2}\alpha}$ is:
Option 1: $-\frac{15}{9}$
Option 2: $-\frac{3}{5}$
Option 3: $\frac{3}{5}$
Option 4: $\frac{17}{5}$
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Correct Answer: $-\frac{3}{5}$
Solution : Given: $\tan\alpha=2$ We know that $\sec^{2}\alpha = 1 + \tan^{2}\alpha$ and $\operatorname{cosec}^{2}\alpha = 1 + \cot^{2}\alpha$ Since $\tan\alpha = 2$, we have $\sec^{2}\alpha = 1 + (2)^{2} = 5$ Also, $\cot\alpha = \frac{1}{\tan\alpha} = \frac{1}{2}$ So $\operatorname{cosec}^{2}\alpha = 1 + (\frac{1}{2})^{2} = \frac{5}{4}$ Putting the values, we get: $\frac{\operatorname{cosec}^{2}\alpha-\sec^{2}\alpha}{\operatorname{cosec}^{2}\alpha+\sec^{2}\alpha} = \frac{\frac{5}{4}-5}{\frac{5}{4}+5} = \frac{\frac{-15}{4}}{\frac{25}{4}} = -\frac{15}{25} = -\frac{3}{5}$ Hence, the correct answer is $-\frac{3}{5}$.
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