Evaluate the value of sin -6-cos -6

Sin π6 = 12 and Cos π6 = √(3)2 So Given expression = 1+√(3)2 ...

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Evaluate the value of sin $\frac{π}{6}$ +cos $\frac{π}{6}$

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(x + \frac{π}{6})

(x + \frac{π}{6})

π

sin [\frac{π}{6} + A] . cos [\frac{π}{3} - B] + sin [\frac{π}{3} - B] . cos [\frac{π}{6} + A] = cos (A - B)

cos [{cos}^{- 1} (- \frac{\sqrt{3}}{2}) + \frac{π}{6}]

cos [{cos}^{- 1} (- \frac{\sqrt{3}}{2}) + \frac{π}{6}]

{sin}^{- 1} (\frac{2 π}{4})

{cos}^{- 1} (cos \frac{7 π}{6})

{tan}^{- 1} (tan \frac{2 π}{3})

cos ({cos}^{- 1} (\frac{\sqrt{3}}{2}) + \frac{π}{6})

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