Question:
What is the value of $\frac{\tan ^{2} \theta-\sec ^{2} \theta}{\cot ^{2} \theta-\operatorname{cosec}^{2} \theta}$.
Solution:
We have,
$\frac{\tan ^{2} \theta-\sec ^{2} \theta}{\cot ^{2} \theta-\operatorname{cosec}^{2} \theta}=\frac{-1\left(\sec ^{2} \theta-\tan ^{2} \theta\right)}{-1\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)}$
$=\frac{\sec ^{2} \theta-\tan ^{2} \theta}{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}$
We know that,
$\sec ^{2} \theta-\tan ^{2} \theta=1$
$\operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1$
Therefore,
$\frac{\tan ^{2} \theta-\sec ^{2} \theta}{\cot ^{2} \theta-\operatorname{cosec}^{2} \theta}=\frac{1}{1}$
$=1$