Question

The value of $si{n}^2\alpha +cose{c}^2\alpha$ is always

• A. greater than 1
• B. less than 1
• C. greater than or equal to 2
• D. equal to 2

Answer 1

The correct option is C greater than or equal to 2

Given $si{n}^2\alpha +cose{c}^2\alpha$

$=si{n}^2\alpha +\frac{1}{si{n}^2\alpha }$

we know that $A.M\ge G.M$

apply the above relation to $si{n}^2\alpha +\frac{1}{si{n}^2\alpha }$

$\Rightarrow \frac{si{n}^2\alpha +\frac{1}{si{n}^2\alpha }}{2}\ge \sqrt{si{n}^2\alpha \times \frac{1}{{\sin }^2\alpha }}$

$\Rightarrow \frac{si{n}^2\alpha +\frac{1}{si{n}^2\alpha }}{2}\ge \sqrt{1}$

$\Rightarrow si{n}^2\alpha +\frac{1}{si{n}^2\alpha }\ge 2$

$\Rightarrow si{n}^2\alpha +cose{c}^2\alpha \ge 2$

$Opt:\left[C\right]$