Question

The number of solutions of ${\tan }^{2}x-{\sec }^{8}x+1=0$ in $(0,12)$ is:

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

The equation is ${\tan }^{2}x-{\sec }^{8}x+1=0$
Now,
${\tan }^{2}x={\sec }^{2}x-l$
On substituting
${\sec }^{2}x-l-{\sec }^{8}x+1=0$
${\sec }^{2}x(1-{\sec }^{6}x)=0$
${\sec }^{6}x=1$
$({\sec }^{3}x{)}^2=1$
$\sec x=\pm 1$
Solutions are $\pi ,2\pi ,3\pi$
$0$ is not included in the range and
$4\pi >12$
Therefore there are only $3$ solutions