Question : What is the value of $\frac{\cos 50^{\circ}}{\sin 40^{\circ}}+\frac{3 \operatorname{cosec} 80^{\circ}}{\sec 10^{\circ}}–2 \cos 50^{\circ} \operatorname{cosec} 40^{\circ}$?
Option 1: 3
Option 2: 2
Option 3: 5
Option 4: 4
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Correct Answer: 2
Solution :
Given: The trigonometric expression is $\frac{\cos 50^{\circ}}{\sin 40^{\circ}}+\frac{3 \operatorname{cosec} 80^{\circ}}{\sec 10^{\circ}}-2 \cos 50^{\circ} \operatorname{cosec} 40^{\circ}$.
Use the trigonometric formulas, $\cos( 90^{\circ}- \theta) =\sin \theta$, $\operatorname{cosec}( 90^{\circ}-\theta)=\sec \theta$ and $\sin \theta=\frac{1}{\operatorname{cosec}\theta}$.
$\frac{\cos 50^{\circ}}{\sin 40^{\circ}}+\frac{3 \operatorname{cosec} 80^{\circ}}{\sec 10^{\circ}}-2 \cos 50^{\circ} \operatorname{cosec} 40^{\circ}$
$=\frac{\cos( 90^{\circ}-40^{\circ} ) }{\sin 40^{\circ}}+\frac{3 \operatorname{cosec}( 90^{\circ}-10^{\circ})}{\sec 10^{\circ}}-2 \cos( 90^{\circ}-40^{\circ}) \operatorname{cosec} 40^{\circ}$
$=\frac{\sin 40^{\circ}}{\sin 40^{\circ}}+\frac{3 \sec 10^{\circ}}{\sec 10^{\circ}}-2 \sin 40^{\circ} \frac{1}{\sin 40^{\circ}}=1+3-2=2$
Hence, the correct answer is 2.
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