If \(\tan\alpha=2\), then what is the value of \(\displaystyle\frac{\mathrm{cosec}^2\alpha-\sec^2\alpha}{\mathrm{cosec}^2\alpha+\sec^2\alpha}+\frac{3}{5}\)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0
From \(\tan\alpha\), the identities \(\sec^2\alpha=1+\tan^2\alpha\) and \(\mathrm{cosec}^2\alpha=1+\cot^2\alpha\) give both squares directly, with no sign question since only squares appear.
\(\sec^2\alpha=1+4=5\)
\(\cot\alpha=\displaystyle\frac{1}{2}\), so \(\mathrm{cosec}^2\alpha=1+\frac{1}{4}=\frac{5}{4}\)
\(\displaystyle\frac{\frac{5}{4}-5}{\frac{5}{4}+5}=\frac{-\frac{15}{4}}{\frac{25}{4}}=-\frac{3}{5}\)
\(\displaystyle-\frac{3}{5}+\frac{3}{5}=0\)