If $\sec \mathrm{x}+{\sec }^2\mathrm{x}=1$ then the value of ${\tan }^8\mathrm{x}-{\tan }^4\mathrm{x}- 2{\tan }^2\mathrm{x}+1$will be equal to

Solution:

$\begin{array}{l}\sec \mathrm{x}=1-{\sec }^2\mathrm{x}\\ \Rightarrow \sec \mathrm{x}=-{\tan }^2\mathrm{x}\end{array}$

Squaring both side, we get

$\begin{array}{r}{\sec }^2\mathrm{x}={\tan }^4\mathrm{x}\\ \Rightarrow 1+{\tan }^2\mathrm{x}={\tan }^4\mathrm{x}\end{array}$

Squaring both side, we get

$\begin{array}{l}{\tan }^8\mathrm{x}={\tan }^4\mathrm{x}+2{\tan }^2\mathrm{x}+1\\ \Rightarrow {\tan }^8\mathrm{x}-{\tan }^4\mathrm{x}-2{\tan }^2\mathrm{x}=1\\ \Rightarrow {\tan }^8\mathrm{x}-{\tan }^4\mathrm{x}-2{\tan }^2\mathrm{x}+1=2\end{array}$