Question : The value of $\frac{\operatorname{sin} 58^{\circ}}{\cos 32^{\circ}}+\frac{\sin 55^{\circ} \sec 35^{\circ}}{\tan 5^{\circ} \tan 45^{\circ} \tan 85^{\circ}}$ is equal to:
Option 1: 2
Option 2: 1
Option 3: 0
Option 4: 3
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Correct Answer: 2
Solution :
Given:
$\frac{\operatorname{sin} 58^{\circ}}{\cos 32^{\circ}}+\frac{\sin 55^{\circ} \sec 35^{\circ}}{\tan 5^{\circ} \tan 45^{\circ} \tan 85^{\circ}}$
$=\frac{\operatorname{sin} 58^{\circ}}{\cos (90^{\circ}-58^{\circ})}+\frac{\sin 55^{\circ} \cot 5^{\circ}}{\cos 35^{\circ} \tan 45^{\circ} \tan 85^{\circ}}$
$=\frac{\operatorname{sin} 58^{\circ}}{\operatorname{sin} 58^{\circ}}+\frac{\sin 55^{\circ} \cot (90^{\circ}-85^{\circ})}{\cos (90^{\circ}-55^{\circ}) \tan 45^{\circ} \tan 85^{\circ}}$
$= 1+\frac{\sin 55^{\circ} \tan 85^{\circ}}{\sin 55^{\circ} \tan 45^{\circ} \tan 85^{\circ}}$
$=1+\frac{1}{ \tan 45^{\circ}}$
$= 1+1$
$=2$
Hence, the correct answer is 2.
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