Q113 / 160 · TS EAMCET 2010 · Mathematics

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

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

Solution

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

Hence, the correct answer is option: 'B'.

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