Question

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

• 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)

$\frac{c-b}{a-c}$

$a{\sin }^2\theta +b{\cos }^2\theta =c$

${\cos }^2\theta (a{\tan }^2\theta +b)=c$

$a{\tan }^2\theta +b=c{\sec }^2\theta$

$a{\tan }^2\theta +b=c+c{\tan }^2\theta$

$(a-c){\tan }^2\theta =c-b$

$\therefore {\tan }^2\theta =\frac{c-b}{a-c}$