Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

Given,

$\sec x+{\sec }^{2}x=1$

$\Rightarrow \sec x=1-{\sec }^{2}x={\tan }^{2}x$

$\Rightarrow {\sec }^{2}x={\tan }^{4}x$

$\Rightarrow 1+{\tan }^{2}x={\tan }^{4}x$

$(1+{\tan }^{2}x{)}^2=({\tan }^{4}x{)}^2$

$1+{\tan }^{4}x+2{\tan }^{2}x={\tan }^{8}x$

$\therefore {\tan }^{8}x-{\tan }^{4}x-2{\tan }^{2}x+1$

$=1+{\tan }^{4}x+2{\tan }^{2}x-{\tan }^{4}x-2{\tan }^{2}x+1$

$=2$