Question

If $y=co{s}^{2}x+se{c}^{2}x$, then

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

Answer 1

Answer: (C)

Given $y=co{s}^{2}x+se{c}^{2}x$
$\Rightarrow y=co{s}^{2}x+\frac{1}{co{s}^{2}x}\left(\because cosx=\frac{1}{secx}\right)$
$\Rightarrow y=co{s}^{2}x+\frac{1}{co{s}^{2}x}+2-2$
$\Rightarrow y={\left(cosx-\frac{1}{cosx}\right)}^2+2$
$\Rightarrow y={\left(cosx-secx\right)}^2+2$
As ${\left(cosx-secx\right)}^2=0$ or positive
$\therefore y=2$ or $y\ge 2$