Question

sec⁴θ – tan⁴θ = 1 + 2 tan²θ

Answer 1

$$\begin{gathered} {\sec }^4\theta -{\tan }^4\theta \\ ={\left({\sec }^2\theta \right)}^2-{\left({\tan }^2\theta \right)}^2 \\ =\left({\sec }^2\theta -{\tan }^2\theta \right)\left({\sec }^2\theta +{\tan }^2\theta \right)\left[a^2-b^2=\left(a-b\right)\left(a+b\right)\right] \end{gathered}$$
$$\begin{gathered} =1\times \left(1+{\tan }^2\theta +{\tan }^2\theta \right)\left(1+{\tan }^2\theta ={\sec }^2\theta \right) \\ =1+2{\tan }^2\theta \end{gathered}$$