Question

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Answer 1

`$ (√(3) - 1)/(2√(2)) $`

`$ (-(√(3) + 1))/(2√(2)) $`

`$ - (√(3) - 1)/(√(3) + 1) $`

Explanation:

Given
`$ (11 \pi)/(12) = (3 \pi)/(4) + (\pi)/(6) $`

(A)
`$ sin((11\pi)/(12)) = sin ((3 \pi)/(4)  + (\pi)/(6)) $`

We know that Sin(A + B) = SinA cosB + cosAsinB

Substituting in the above formula we get:

`$ sin ((3\pi)/(4) + (\pi)/(6)) = (1)/(√(2)) . (√(3))/(2) + (-1)/(√(2)). (1)/(2) $`

`$ \implies (1)/(√(2)) ((√(3) - 1)/(2)) = (√(3) - 1)/(2√(2))`

(B) Cos(A + B) = CosAcosB - SinASinB

`$ cos((11\pi)/(12)) = cos((3\pi)/(4) + (\pi)/(6)}) $`

`$ \implies (-1)/(√(2)). (√(3))/(2) - (1)/(√(2)) . (1)/(2) $`

`$ \implies cos((11\pi)/(12)) = cos((3\pi)/(4) + (\pi)/(6)) $`

`$ = (-(√(3) + 1))/(2√(2))`

(C) Tan(A + B) =
`$ (Sin(A +B))/(Cos(A + B)) $`

From the above obtained values this can be calculated and the value is
`$ - (√(3) - 1)/(√(3) + 1) $`.