Question : The value of $\frac{\left(\cos 9^{\circ}+\sin 81^{\circ}\right)\left(\sec 9^{\circ}+{\operatorname{cosec}} \;81^{\circ}\right)}{{\operatorname{cosec}}^2 \;71^{\circ}+\cos ^2 15^{\circ}-\tan ^2 19^{\circ}+\cos ^2 75^{\circ}} $ is:
Option 1: 1
Option 2: 4
Option 3: 5
Option 4: 2
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Correct Answer: 2
Solution :
Given: $\frac{\left(\cos 9^{\circ}+\sin 81^{\circ}\right)\left(\sec 9^{\circ}+{\operatorname{cosec}} \;81^{\circ}\right)}{{\operatorname{cosec}}^2 \;71^{\circ}+\cos ^2 15^{\circ}-\tan ^2 19^{\circ}+\cos ^2 75^{\circ}} $
$= \frac{\left(\cos 9^{\circ}+\sin (90^\circ - 9^{\circ})\right)\left(\sec 9^{\circ}+\operatorname{cosec}(90^\circ - 9^{\circ})\right)}{{\operatorname{cosec}}^2 \;(90^\circ- 19^{\circ})+\cos ^2 15^{\circ}-\tan ^2 19^{\circ}+\cos ^2 (90^\circ - 75^{\circ})} $
$= \frac{\left(\cos 9^{\circ}+\cos 9^{\circ}\right)\left(\sec 9^{\circ}+\sec 9^{\circ}\right)}{\sec^2 19^{\circ}+\cos ^2 15^{\circ}-\tan ^2 19^{\circ}+\sin ^2 15^\circ} $
$= \frac{2 \cos 9^{\circ} \times 2 \sec 9^{\circ}}{2}$
$= 2$
Hence, the correct answer is 2.
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