Question : The value of $\frac{\sec 54^{\circ}}{\operatorname{cosec} 36^{\circ}}+\frac{\tan 70^{\circ}}{\cot 20^{\circ}}-2 \tan 45^{\circ}$ is equal to:
Option 1: 2
Option 2: 0
Option 3: 1
Option 4: 3
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Correct Answer: 0
Solution :
$\frac{\sec 54^{\circ}}{\operatorname{cosec} 36^{\circ}}+\frac{\tan 70^{\circ}}{\cot 20^{\circ}}-2 \tan 45^{\circ}$
$= \frac{\sec 54^{\circ}}{\operatorname{cosec} (90^\circ-54^{\circ})}+\frac{\tan 70^{\circ}}{\cot (90^\circ-70^{\circ})}-2 \tan 45^{\circ}$
$=\frac{\sec 54^{\circ}}{\sec 54^{\circ}}+\frac{\tan 70^{\circ}}{\tan 70^{\circ}}-2×1$
= 1 + 1 – 2
= 0
Hence, the correct answer is 0.
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