Question

Find the min. value of $\frac{a^2}{{\cos }^{2}x}+\frac{b^2}{{\sin }^{2}x}$

• A. $(a+b{)}^2$
• B. $(a-b{)}^2$
• C. $a^2+b^2$
• D. $a^2-b^2$

Answer 1

The correct option is A $(a+b{)}^2$

Let $f\left(x\right)=\frac{a^2}{{\cos }^{2}x}+\frac{b^2}{{\sin }^{2}x}=a^2{\sec }^{2}x+b^{2}co{\sec }^{2}x=(a^2+b^2)+(a^2{\tan }^{2}x+b^2{\cot }^{2}x)$
$\Rightarrow f\left(x\right)=(a^2+b^2)+(a\tan x-b\cot x{)}^2+2ab=(a+b{)}^2+(a\tan x-b\cot x{)}^2$
Hence minimum value of $f\left(x\right)$ is $(a+b{)}^2$