Question

If $y = cos^2x + sec^2x$, then

• A. $y \le 2$
• B. $y \le 1$
• C. $y \ge 2$
• D. $1 < y < 2$

Answer 1

Answer: (C)

Given $y = cos^2 \,x + sec^2\, x$
$\Rightarrow \quad y = cos^{2}\,x + \frac{1}{cos^{2}\,x} \left(\because cos\,x = \frac{1}{sec\,x}\right)$
$\Rightarrow \quad y = cos^{2}\,x + \frac{1}{cos^{2}\,x} +2 - 2$
$\Rightarrow \quad y = \left(cos\,x - \frac{1}{cos\,x}\right)^2 + 2$
$\Rightarrow \quad y = \left(cos\,x-sec\,x\right)^{2} + 2$
As $\left(cos\, x - sec \,x\right)^{2} = 0$ or positive
$\therefore \quad y = 2$ or $y \ge 2$