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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Question : $\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)} \times \frac{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}}{\tan \theta+\cot \theta}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : The expression $\frac{(1-\sin \theta+\cos \theta)^2(1-\cos \theta) \sec ^3 \theta\; {\operatorname{cosec}}^2 \theta}{(\sec \theta-\tan \theta)(\tan \theta+\cot \theta)}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : The expression $\frac{\cos ^4 \theta-\sin ^4 \theta+2 \sin ^2 \theta+3}{(\operatorname{cosec} \theta+\cot \theta+1)(\operatorname{cosec} \theta-\cot \theta+1)-2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $\frac{(1+\sec \theta \operatorname{cosec} \theta)^2(\sec \theta-\tan \theta)^2(1+\sin \theta)}{(\sin \theta+\sec \theta)^2+(\cos \theta+\operatorname{cosec} \theta)^2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
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