Question

Let $x$ and $y$ be 2 real numbers which satisfy the equations $\left({\tan }^{2}x-{\sec }^{2}y\right)=\frac{5a}{6}-3$ and $\left(-{\sec }^{2}x+{\tan }^{2}y\right)=a^2$, 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)

On adding these two equations, we get
$\left({\tan }^{2}x-{\sec }^{2}y\right)+\left(-{\sec }^{2}x+{\tan }^{2}y\right)=\frac{5}{6}a-3+a^2\Rightarrow -2=\frac{5}{6}a-3+a^2$
Hence $6a^2+5a-6=0\Rightarrow 6a^2+9a-4a-6=0\Rightarrow (2a+3)(3a-2)=0$
Hence $a=\frac{-3}{2}$ or $\frac{2}{3}]$