Question : If $\tan\theta =\frac{3}{4}$, then the value of $\frac{4\sin^{2}\theta–2\cos^{2}\theta}{4\sin^{2}\theta+3\cos^{2}\theta}$ is equal to:
Option 1: $\frac{1}{21}$
Option 2: $\frac{2}{21}$
Option 3: $\frac{4}{21}$
Option 4: $\frac{8}{21}$
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Correct Answer: $\frac{1}{21}$
Solution :
$\frac{4\sin^{2}\theta–2\cos^{2}\theta}{4\sin^{2}\theta+3\cos^{2}\theta}$
Dividing this numerator and denominator by $\cos\theta$ we have,
$\frac{4\sin^{2}\theta–2\cos^{2}\theta}{4\sin^{2}\theta+3\cos^{2}\theta} = \frac{4\tan^{2}\theta–2}{4\tan^{2}\theta+3}$
Putting the value of $\tan\theta =\frac{3}{4}$ in this expression, we have,
= $\frac{4×\frac{9}{16}–2}{4×\frac{9}{16}+3}$
= $\frac{\frac{9}{4}–2}{\frac{9}{4}+3}$
= $\frac{\frac{1}{4}}{\frac{21}{4}}$
= $\frac{1}{21}$
Hence, the correct answer is $\frac{1}{21}$.
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