Question

Let $x$ and $y$ be $2$ real numbers which satisfy the equations $({\tan }^{2}x-{\sec }^{2}y)=\frac{5a}{6}-3$ and $(-{\sec }^{2}x+{\tan }^{2}y)=a^2$, then the value of a can be equal to

• A. $\frac{2}{3}$
• B. $\frac{-2}{3}$
• C. $\frac{3}{2}$
• D. $\frac{-3}{2}$

Answer 1

Answer: (A), (D)

The correct options are
A $\frac{2}{3}$
D $\frac{-3}{2}$