Question : $\cos ^2 35^{\circ}+\cos 55^{\circ} \cdot \sin 35^{\circ}+\frac{\tan 34^{\circ}}{\cot 56^{\circ}}=$________.
Option 1: 2
Option 2: 3
Option 3: 4
Option 4: 1
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Correct Answer: 2
Solution :
We know that $\sin \theta = \cos (90°-\theta)$
And, $\tan \theta = \cot (90°-\theta)$
So, $\cos ^2 35^{\circ}+\cos 55^{\circ} \cdot \sin 35^{\circ}+\frac{\tan 34^{\circ}}{\cot 56^{\circ}}$
= $\cos ^2 35^{\circ}+\cos (90^{\circ}-35^{\circ}) \cdot\sin 35^{\circ}+\frac{\tan(90^{\circ}- 56^{\circ})}{\cot 56^{\circ}}$
= $\cos ^2 35^{\circ}+\sin 35^{\circ} \cdot \sin 35^{\circ}+\frac{\cot 56^{\circ}}{\cot 56^{\circ}}$
= $\cos ^2 35^{\circ}+\sin^2 35^{\circ}+\frac{\cot 56^{\circ}}{\cot 56^{\circ}}$
= $1+1$
= $2$
Hence, the correct answer is 2.
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