Question

Number of ordered pairs $(a,x)$ satisfying the equation ${\sec }^2(a+2)x+a^2-1=0;-\pi <x<\pi$ is

• A. 2
• B. 1
• C. 3
• D. infinite

Answer 1

The correct option is C 3

using
$se{c}^{2}x=1+ta{n}^{2}x$
we get
$ta{n}^2(a+2)x+a^2=0$
this is possible only when a = 0 and $ta{n}^2(a+2)x=0$
so
$ta{n}^{2}2x=0$
$2x=-\pi ,0,\pi$