Question : $\frac{\cos 20^{\circ}}{\sin 70^{\circ}}+\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}=$_________,
Option 1: $-2$
Option 2: $\frac{1}{2}$
Option 3: $-\frac{1}{2}$
Option 4: $2$
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Correct Answer: $2$
Solution : We know that $\text{sin}(90^{\circ} - \theta)= \cos \theta$ $\frac{\cos 20^{\circ}}{\sin 70^{\circ}}+\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}$ = $\frac{\cos 20^{\circ}}{\sin (90^{\circ}-20^{\circ})}+\frac{\cos \theta}{\cos \theta}$ = $\frac{\cos 20^{\circ}}{\cos (20^{\circ})}+\frac{\cos \theta}{\cos \theta}$ = 1 + 1 = 2 Hence, the correct answer is $2$.
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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:
Option 1: $\cos ^2 \theta$
Option 2: $-\sin ^2 \theta$
Option 3: $\sec ^2 \theta$
Option 4: $-\sec ^2 \theta$
Question : If $\left(\frac{\cos A}{1-\sin A}\right)+\left(\frac{\cos A}{1+\sin A}\right)=4$, then what will be the value of $A$? $\left(0^{\circ}<\theta<90^{\circ}\right)$
Option 1: $90^{\circ}$
Option 2: $45^{\circ}$
Option 3: $60^{\circ}$
Option 4: $30^{\circ}$
Question : $\cos \left(30^{\circ}+\theta\right)-\sin \left(60^{\circ}-\theta\right)=$ _____________.
Option 1: $\frac{\sqrt{3}}{2}$
Option 2: $0$
Option 3: $\frac{1}{2}$
Option 4: $\frac{1}{\sqrt{2}}$
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:
Option 1: $\operatorname{cosec} \theta \sec \theta$
Option 2: $\operatorname{cosec} \theta$
Option 3: $\sin \theta \cos \theta$
Option 4: $\sec \theta$
Question : What is $\tan \frac{\theta}{2}$?
Option 1: $\frac{\cos \theta}{1-\sin \theta}$
Option 2: $\frac{\sin \theta}{1-\cos \theta}$
Option 3: $\frac{\cos \theta}{1-\cos \theta}$
Option 4: $\frac{\sin \theta}{1+\cos \theta}$
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