Question

${\cot }^2\theta -{\tan }^2\theta$ can also be written as $cose{c}^2\theta -{\sec }^2\theta$.

Answer 1

Now,

${\cot }^2\theta -{\tan }^2\theta$

$=(cose{c}^2\theta -1)-({\sec }^2\theta -1)$ [ Since $cose{c}^2\theta -{\cot }^2\theta =1$ and ${\sec }^2\theta -{\tan }^2\theta =1$]

$=cose{c}^2\theta -{\sec }^2\theta$.