tan75°/(1-tan^2 75°)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to calculate `tan 75^o/(1-tan^2 75^o).`

Write `tan 75^o = tan(30^o + 45^o)`

You should evaluate`tan 75^o`  using the formula`tan (a+b) = (tan a + tan b)/(1 - tan a * tan b).`

`` `tan(30^o + 45^o) = (tan30 ^o+ tan 45^o)/(1 - tan 30^o* tan 45^o).`

`` `tan 75^o = (sqrt3/3 + 1)/(1 - sqrt3/3) =gt tan 75^o = (3 + sqrt3)/(3 - sqrt3)`

Calculate `1 - tan^2 75^o = 1 - ((3 + sqrt3)/(3 - sqrt3))^2`

`1 - tan^2 75^o = ((3 - sqrt3)^2 - (3+sqrt3)^2)/(3 - sqrt3)^2`

Use difference of squares: `(3 - sqrt3)^2 - (3+sqrt3)^2 = ((3 - sqrt3 - 3 - sqrt3)(3 - sqrt3 + 3+sqrt3))`

Use difference of squares: `(3 - sqrt3)^2 - (3+sqrt3)^2 = (-2sqrt3)*6`

Use difference of squares: `(3 - sqrt3)^2 - (3+sqrt3)^2 = -12 sqrt3.`

`` `tan 75^o/(1-tan^2 75^o) = (3 + sqrt3)/(-12 sqrt3*(3 - sqrt3))`

Multiply the fraction by the conjugate of denominator:

`tan 75^o/(1-tan^2 75^o) = (3 + sqrt3)^2/(-12 sqrt3*(3 - sqrt3)(3 + sqrt3))`

Expand the binomial:

`tan 75^o/(1-tan^2 75^o) = (9 + 6sqrt3 + 3)^2/(-12 sqrt3*(9 - 3))`

`` `tan 75^o/(1-tan^2 75^o) = (12 + 6sqrt3)/(-12*6*sqrt3)`

`tan 75^o/(1-tan^2 75^o) = 6*(2 + sqrt3)/(-12*6*sqrt3)`

`tan 75^o/(1-tan^2 75^o) = (2 + sqrt3)/(-12*sqrt3)`

Multiply the fraction by `-sqrt3` :

`tan 75^o/(1-tan^2 75^o) = -sqrt3*(2 + sqrt3)/36`

Evaluating the given fraction yields: `tan 75^o/(1-tan^2 75^o) =-sqrt3*(2 + sqrt3)/36.`

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