Question

Let $x$ and $y$ be 2 real numbers which satisfy the equations $\left(\tan ^2 x-\sec ^2 y\right)=\frac{5 a}{6}-3$ 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 $6 a^2+5 a-6=0 \Rightarrow 6 a^2+9 a-4 a-6=0 \Rightarrow(2 a+3)(3 a-2)=0$
Hence $a =\frac{-3}{2}$ or $\left.\frac{2}{3}\right] $