Question

$\frac{{\sec }^4\theta }{a}+\frac{{\tan }^4\theta }{b}=\frac{1}{a+b}$ , then

• A. $|a|=|b|$
• B. $b<0,a>0,|b|\le a$
• C. $|b|\ge |a|$
• D. $b>0,a<0,b\le |a|$

Answer 1

Answer: (B), (D)

$b<0,a>0,|b|\le a$
$b>0,a<0,b\le |a|$

${\sec }^2\theta =1+{\tan }^2\theta$
Substituting this value in the equation in question, we get a quadratic equation in ${\tan }^2\theta$ as $(a+b){\tan }^4\theta +2b{\tan }^2\theta +\frac{b^2}{a+b}=0$
The roots of this equation have to be positive and so we get the condition that $\frac{-b}{a+b}\ge 0.$

Hence, option B and D.