The value of frac{1}{sqrt{2}} sin frac{pi}{6} | vedclass

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(A) Given expression: $\frac{1}{\sqrt{2}} \sin \frac{\pi}{6} \cos \frac{\pi}{4} - \cot \frac{\pi}{3} \sec \frac{\pi}{6} + \frac{5 	an \frac{\pi}{4}}{

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(A) Given expression: $\frac{1}{\sqrt{2}} \sin \frac{\pi}{6} \cos \frac{\pi}{4} - \cot \frac{\pi}{3} \sec \frac{\pi}{6} + \frac{5 	an \frac{\pi}{4}}{12 \sin \frac{\pi}{2}}$Substitute the standard trigonometric values:$\sin \frac{\pi}{6} = \frac{1}{2}$,$\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,$\cot \frac{\pi}{3} = \frac{1}{\sqrt{3}}$,$\sec \frac{\pi}{6} = \frac{2}{\sqrt{3}}$,$	an \frac{\pi}{4} = 1$,$\sin \frac{\pi}{2} = 1$Substituting these values into the expression:$= \frac{1}{\sqrt{2}} \left( \frac{1}{2} 	imes \frac{1}{\sqrt{2}} ight) - \left( \frac{1}{\sqrt{3}} 	imes \frac{2}{\sqrt{3}} ight) + \frac{5 	imes 1}{12 	imes 1}$$= \left( \frac{1}{\sqrt{2}} 	imes \frac{1}{2\sqrt{2}} ight) - \frac{2}{3} + \frac{5}{12}$$= \frac{1}{4} - \frac{2}{3} + \frac{5}{12}$$= \frac{3 - 8 + 5}{12} = \frac{0}{12} = 0$

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

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