Question : $\frac{1}{1+\cos(90^{\circ}- \theta)}$ + $\frac{1}{1-\cos(90^{\circ}- \theta)}$ =?
Option 1: $2\sec^2 \theta$
Option 2: $1$
Option 3: $2$
Option 4: $2 \tan^2 \theta$
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Correct Answer: $2\sec^2 \theta$
Solution : $\frac{1}{1+\cos(90^{\circ}- \theta)}$ + $\frac{1}{1-\cos(90^{\circ}- \theta)}$ = $\frac{1}{1+\sin \theta}$ + $\frac{1}{1-\sin \theta}$ = $\frac{(1-\sin \theta)+(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}$ = $\frac{2}{1-\sin ^2 \theta}$ = $\frac{2}{\cos ^2 \theta}$ = $2\sec^2 \theta$ Hence, the correct answer is $2\sec^2 \theta$.
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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 : $\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $\frac{(1+\sec \theta \operatorname{cosec} \theta)^2(\sec \theta-\tan \theta)^2(1+\sin \theta)}{(\sin \theta+\sec \theta)^2+(\cos \theta+\operatorname{cosec} \theta)^2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : The expression $\frac{\left(1-2 \sin ^2 \theta \cos ^2 \theta\right)(\cot \theta+1) \cos \theta}{\left(\sin ^4 \theta+\cos ^4 \theta\right)(1+\tan \theta) \operatorname{cosec} \theta}-1,0^{\circ}<\theta<90^{\circ}$, equals:
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:
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