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This question belongs to Arithmetic Ability Trigonometry
Trigonometry
?

The value of $$\frac{1}{{\sqrt 2 }}{\text{sin}}\frac{\pi }{6}$$ . $${\text{cos}}\frac{\pi }{4}$$ - $$\cot \frac{\pi }{3}$$ . $${\text{sec}}\frac{\pi }{6}$$ + $$\frac{{5\tan \frac{\pi }{4}}}{{12\sin \frac{\pi }{2}}}$$   is equal to?

Answer & Solution
Correct Answer: Option A
$$\eqalign{ & \frac{1}{{\sqrt 2 }}{\text{sin}}\frac{\pi }{6}{\text{.cos}}\frac{\pi }{4} - \cot \frac{\pi }{3}{\text{.sec}}\frac{\pi }{6}{\text{ + }}\frac{{5\tan \frac{\pi }{4}}}{{12\sin \frac{\pi }{2}}} \cr & \Rightarrow \frac{1}{{\sqrt 2 }} \times \frac{1}{2} \times \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 3 }} \times \frac{2}{{\sqrt 3 }} + \frac{{5 \times 1}}{{12 \times 1}} \cr & \Rightarrow \frac{1}{4} - \frac{2}{3} + \frac{5}{{12}} \cr & \Rightarrow \frac{{3 - 8 + 5}}{{12}} \cr & \Rightarrow 0 \cr & {\text{ }} \cr} $$
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