Question

If $3{\cos }^{2}A=\cos {60}^{\circ }+{\sin }^2{45}^{\circ }$ then find the value of ${\sec }^{2}A$.

• A. $1$
• B. $0$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

$3{\cos }^{2}A=\cos {60}^0+{\sin }^{2}45$

$\Rightarrow 3{\cos }^{2}A=\frac{1}{2}+(\frac{1}{\sqrt{2}}{)}^2$

$\Rightarrow 3{\cos }^{2}A=\frac{1}{2}+\frac{1}{2}=1$

$\Rightarrow {\cos }^{2}A=\frac{1}{3}$
Or ${\sec }^{2}A=\frac{1}{{\cos }^{2}A}=3$