Question

If $a{\sin }^2\theta +b{\cos }^2\theta =c$ then ${\tan }^2\theta$ is equal to

• A. $\frac{b-c}{a-c}$
• B. $\frac{c-b}{a-c}$
• C. $\frac{a-c}{b-c}$
• D. $\frac{a-c}{c-b}$

Answer 1

Answer: (B)

$a{\sin }^2\theta +b{\cos }^2\theta =c$
On dividing both sides by ${\cos }^2\theta$
$a{\tan }^2\theta +b=c{\sec }^2\theta$
$\Rightarrow a{\tan }^2\theta +b=c\left(1+{\tan }^2\theta \right)$
$\to a{\tan }^2\theta +b=c+c{\tan }^2\theta$
$\Rightarrow b=c+c{\tan }^2\theta -a{\tan }^2\theta$
$\Rightarrow (c-a){\tan }^2\theta =(b-c)$
$\Rightarrow {\tan }^2\theta =\frac{b-c}{c-a}$ or $\frac{c-b}{a-c}$