`285^0 = 270^0 + 15^0 `
`sin (270^0 + 15^0) = -cos 15^0 `
`= -cos(45^0 - 30^0) `
`= -[cos 45^0 * cos 30^0 + sin 45^0 * sin 30^0]`
`= -[1/sqrt2 * sqrt3/2 + 1/sqrt2 * 1/2] `
`=-[(sqrt3 + 1)/(2sqrt2)]`
`cos (270^0 + 15^0) = sin 15^0 `
`= sin(45^0 -30^0) `
`= sin45^0 * cos 30^0 - cos 45^0 * sin 30^0 `
`= 1/sqrt2 * sqrt3/2 - 1/sqrt2 *1/2 `
Now `tan(285^0)` . this can be evaluated by using `tan(A+B)` identity or
`tan A = (sin A)/(cos A)`
`tan(285^0) = [sin(285^0)]/[cos(285^0)] `
`= -[(sqrt3 + 1)/(2sqrt2)]/[(sqrt3 - 1)/(2sqrt2)] `
`=-[(sqrt3 + 1)]/[(sqrt3 - 1)]`
by rationalizing the denominator we get
`tan(285^0) = -(2+sqrt3) `
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