Question : $\frac{\sin ^2 \theta}{\cos \theta(1+\cos \theta)}+\frac{1+\cos \theta}{\cos \theta}=$?
Option 1: $\operatorname{cosec}\theta$
Option 2: $\sec\theta$
Option 3: $2\cos\theta$
Option 4: $2\sec\theta$
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Correct Answer: $2\sec\theta$
Solution : $\frac{\sin ^2 \theta}{\cos \theta(1+\cos \theta)}+\frac{1+\cos \theta}{\cos \theta}$ $=\frac{1-\cos ^2 \theta}{\cos \theta(1+\cos \theta)}+\frac{1+\cos \theta}{\cos \theta}$ $=\frac{(1-\cos \theta)(1+\cos \theta)}{\cos \theta(1+\cos \theta)}+\frac{1+\cos \theta}{\cos \theta}$ $=\frac{1-\cos \theta}{\cos \theta}+\frac{1+\cos \theta}{\cos \theta}$ $=\frac{1-\cos \theta+1+\cos \theta}{\cos \theta}$ $=\frac{2}{\cos \theta}$ $=2\sec\theta$ Hence, the correct answer is $2\sec\theta$.
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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 : $\frac{(1+\sec \theta \operatorname{cosec} \theta)^2(\sec \theta-\tan \theta)^2(1+\sin \theta)}{(\sin \theta+\sec \theta)^2+(\cos \theta+\operatorname{cosec} \theta)^2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : Which of the following is equal to $[\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}]$?
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 : $\frac{(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)}{(\sec \theta+\tan \theta)(1-\sin \theta)}$ is equal to:
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