Question

Prove that : $2{\sec }^2\theta -{\sec }^4\theta -co{\sec }^2\theta -co{\sec }^4\theta ={\cot }^4\theta -{\tan }^4\theta$

Answer 1

Consider LHS

$2{\sec }^2\theta -{\sec }^4\theta -2{\csc }^2\theta +{\csc }^4\theta$

$=2{\sec }^2\theta -2{\csc }^2\theta -{\sec }^4\theta +{\csc }^4\theta$

$=2({\sec }^2\theta -{\csc }^2\theta )-({\sec }^4\theta -{\csc }^4\theta )$

$=2({\sec }^2\theta -{\csc }^2\theta )-({\sec }^2\theta -{\csc }^2\theta )({\sec }^2\theta +{\csc }^2\theta )$

$=({\sec }^2\theta -{\csc }^2\theta )[2-(1+{\tan }^2\theta +1+{\cot }^2\theta \left)\right]$

$=(1+{\tan }^2\theta -1-{\cot }^2\theta )[2-(2+{\tan }^2\theta +{\cot }^2\theta \left)\right]$

$=({\tan }^2\theta -{\cot }^2\theta )[2-2-{\tan }^2\theta -{\cot }^2\theta ]$

$=({\tan }^2\theta -{\cot }^2\theta )[-({\tan }^2\theta +{\cot }^2\theta \left)\right]$

$=({\cot }^2\theta -{\tan }^2\theta )({\cot }^2\theta +{\tan }^2\theta )$

$={\cot }^4\theta -{\tan }^4\theta$

$=RHS$