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 $\cos ^{2} \theta$
$a \tan ^{2} \theta +b=c \sec ^{2} \theta$
$\Rightarrow a \tan ^{2} \theta +b=c\left(1+\tan ^{2} \theta\right)$
$\rightarrow a \tan ^{2} \theta +b=c +c \tan ^{2} \theta$
$\Rightarrow b=c+ c \tan ^{2} \theta-a \tan ^{2} \theta$
$\Rightarrow (c-a) \tan ^{2} \theta=(b-c)$
$\Rightarrow \tan ^{2} \theta=\frac{b-c}{c-a}$ or $\frac{c-b}{a-c}$