Question

Write
${\tan }^{4}x$ in terms of $\cos x$

Answer 1

${\tan }^{4}x={\left({\tan }^{2}x\right)}^2({\tan }^{2}x+1={\sec }^{2}x)={({\sec }^{2}x-1)}^2={\sec }^{4}x-2{\sec }^{2}x+1=\frac{1}{{\cos }^{4}x}-\frac{2}{{\cos }^{2}x}+\frac{1}{1}=\frac{{\cos }^{4}x-2{\cos }^{2}x+1}{{\cos }^{4}x}$