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: $2$
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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