Question : $(\sin \theta+\operatorname{cosec} \theta)^2+(\cos \theta+\sec \theta)^2=$?
Option 1: $5+\tan ^2 \theta+\cot ^2 \theta$
Option 2: $7+\tan ^2 \theta-\cot ^2 \theta$
Option 3: $7+\tan ^2 \theta+\cot ^2 \theta$
Option 4: $5+\tan ^2 \theta-\cot ^2 \theta$
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Correct Answer: $7+\tan ^2 \theta+\cot ^2 \theta$
Solution : $(\sin \theta+\operatorname{cosec} \theta)^2+(\cos \theta+\sec \theta)^2$ = $(\sin^2 \theta+\operatorname{cosec}^2 \theta+2\sin \theta\operatorname{cosec} \theta)+(\cos^2 \theta+\sec^2 \theta+2\cos \theta\sec \theta)$ = $\sin^2 \theta+\cos^2 \theta+\operatorname{cosec}^2 \theta+2\sin \theta\operatorname{cosec} \theta+\sec^2 \theta+2\cos \theta\sec \theta$ = $1+1+\cot^2 \theta+2+1+\tan^2 \theta+2$ = $7+\cot^2 \theta+\tan^2 \theta$ Hence, the correct answer is $7+\tan^2 \theta+\cot^2 \theta$.
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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 : $\frac{(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)}{(\sec \theta+\tan \theta)(1-\sin \theta)}$ is equal to:
Question : Which of the following is equal to $[\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}]$?
Question : The value of $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$ is:
Question : Let $0^{\circ}<\theta<90^{\circ}$, $\left(1+\cot ^2 \theta\right)\left(1+\tan ^2 \theta\right) × (\sin \theta-\operatorname{cosec} \theta)(\cos \theta-\sec \theta)$ is equal to:
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