Question : The expression $\frac{\tan ^6 \theta-\sec ^6 \theta+3 \sec ^2 \theta \tan ^2 \theta}{\tan ^2 \theta+\cot ^2 \theta+2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Option 1: $\sec ^2 \theta \operatorname{cosec}^2 \theta$
Option 2: $-\sec ^2 \theta \operatorname{cosec}^2 \theta$
Option 3: $\cos ^2 \theta \sin ^2 \theta$
Option 4: $-\cos ^2 \theta \sin ^2 \theta$
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Correct Answer: $-\cos ^2 \theta \sin ^2 \theta$
Solution :
$\frac{\tan ^6 \theta-\sec ^6 \theta+3 \sec ^2 \theta \tan ^2 \theta}{\tan ^2 \theta+\cot ^2 \theta+2}$
Since $0^{\circ}<\theta<90^{\circ}$, putting $\theta=45^{\circ}$
$\frac{\tan ^6 45^{\circ}-\sec ^6 45^{\circ}+3 \sec ^2 45^{\circ} \tan ^2 45^{\circ}}{\tan ^2 45^{\circ}+\cot ^245^{\circ}+2}$
$=\frac{1-(\sqrt2)^6+ 3\times (\sqrt2)^2\times 1}{1+1+2}$
$=\frac{1-8+6}{1+1+2}=-\frac{1}{4}$
From option (iv),
$-\cos ^2 \theta \sin ^2 \theta$
Putting $\theta=45^{\circ}$,
$=-(\frac{1}{\sqrt{2}} )^2(\frac{1}{\sqrt{2}} )^2=-\frac{1}{4}$
Hence, the correct answer is $-\cos ^2 \theta \sin ^2 \theta$.
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