Question

The number of solutions of the equation ${\tan }^{2}x-{\sec }^{10}x+1=0$ in $(0,10)$ is

• A. 3
• B. 6
• C. 10
• D. 11

Answer 1

Answer: (A)

3

$ta{n}^{2}x-{\sec }^{10}x+1=0$

$\Rightarrow {\sec }^{2}x-{\sec }^{10}x=0(\because 1+{\tan }^{2}x={\sec }^{2}x)$

$\Rightarrow {\sec }^{2}x(1-{\sec }^{8}x)=0$

$\Rightarrow {\sec }^{2}x(1-\sec x)(1+\sec x)(1+{\sec }^{2}x)(1+{\sec }^{4}x)=0$

$\Rightarrow$ The real roots are only for ${\sec }^{2}x=0,$

$(1-\sec x)=0,(1+\sec x)=0$

$\Rightarrow {\sec }^{2}x=0\Rightarrow \sec x=0$

$(1-\sec x)=0\Rightarrow \sec x=1$

$(1+\sec x)=0\Rightarrow \sec x=-1$

$\therefore$ only one solution $\sec x=1$ in the interval $(0,10)$