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If $a \sin ^2 \theta+b \cos ^2 \theta=c$, then $\tan ^2 \theta$ is equal to
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The correct answer is:
$\frac{c-b}{a-c}$
$a \sin ^2 \theta+b \cos ^2 \theta=c$
On dividing both sides by $\cos ^2 \theta$
$a \tan ^2 \theta+b=c \sec ^2 \theta$
$\Rightarrow \quad a \tan ^2 \theta+b=c\left(1+\tan ^2 \theta\right)$
$\Rightarrow \quad a \tan ^2 \theta+b=c+c \tan ^2 \theta$
$\Rightarrow \quad b=c+c \tan ^2 \theta-a \tan ^2 \theta$
$\Rightarrow \quad(c-a) \tan ^2 \theta=(b-c)$
$\Rightarrow \quad \tan ^2 \theta=\frac{b-c}{c-a}$ or $\frac{c-b}{a-c}$
On dividing both sides by $\cos ^2 \theta$
$a \tan ^2 \theta+b=c \sec ^2 \theta$
$\Rightarrow \quad a \tan ^2 \theta+b=c\left(1+\tan ^2 \theta\right)$
$\Rightarrow \quad a \tan ^2 \theta+b=c+c \tan ^2 \theta$
$\Rightarrow \quad b=c+c \tan ^2 \theta-a \tan ^2 \theta$
$\Rightarrow \quad(c-a) \tan ^2 \theta=(b-c)$
$\Rightarrow \quad \tan ^2 \theta=\frac{b-c}{c-a}$ or $\frac{c-b}{a-c}$
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