Question

The minimum value of $a{\tan }^{2}x+b{\cot }^{2}x$ equals the maximum value of $a{\sin }^2\theta +b{\cos }^2\theta$ where $a>b>0$, when

• A. $a=b$
• B. $a=2b$
• C. $a=3b$
• D. $a=4b$

Answer 1

Answer: (D)

$a=4b$

Minimum value of $a{\tan }^{2}x+b{\cot }^{2}x$ is $2\sqrt{ab}$

and maximum value of $a{\sin }^2\theta +b{\cos }^2\theta$is $a$

$\because a{\sin }^2\theta +b{\cos }^2\theta =(a-b){\sin }^2\theta +b$

and $0\le {\sin }^2\theta \le 1$

Given; $a=2\sqrt{ab}.$

$\therefore a=4b$