Question : What is the value of $\frac{3 \cos 62^{\circ}}{\sin 28^{\circ}}-\frac{2 \tan 34^{\circ}}{\cot 56^{\circ}} ?$
Option 1: 3
Option 2: 1
Option 3: 5
Option 4: 4
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Correct Answer: 1
Solution :
We know that $\sin(90^{\circ} - \theta) = \cos \theta$ and $\cot(90^{\circ} - \theta) = \tan \theta$
$\frac{3 \cos 62^{\circ}}{\sin 28^{\circ}}-\frac{2 \tan 34^{\circ}}{\cot 56^{\circ}} $
$=\frac{3 \cos 62^{\circ}}{\sin (90-62)^{\circ}}-\frac{2 \tan 34^{\circ}}{\cot (90-34)^{\circ}} $
$= \frac{3 \cos 62^{\circ}}{\cos 62^{\circ}}-\frac{2 \tan 34^{\circ}}{ \tan 34^{\circ}}$
$=3 - 2 $
$= 1$
Hence, the correct answer is 1.
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