Question

The function $({\tan }^4\theta +{\tan }^2\theta )$ is equal to
[Hint: ${\sec }^2\theta -{\tan }^2\theta =1$]

• A. ${\sec }^4\theta -{\sec }^2\theta$
• B. ${\tan }^4\theta -{\sec }^2\theta$
• C. ${\sec }^2\theta -{\sin }^2\theta$
• D. ${\sin }^4\theta -{\cos }^2\theta$

Answer 1

The correct option is A ${\sec }^4\theta -{\sec }^2\theta$

${\tan }^4\theta +{\tan }^2\theta ={\tan }^2\theta ({\tan }^2\theta +1)$
We have,
${\sec }^2\theta -{\tan }^2\theta =1\Rightarrow ({\tan }^2\theta +1)={\sec }^2\theta$. $\Rightarrow {\tan }^2\theta ={\sec }^2\theta -1$
$\therefore {\tan }^2\theta ({\tan }^2\theta +1)=({\sec }^2\theta -1){\sec }^2\theta$$={\sec }^4\theta -{\sec }^2\theta$