Simplify :
`( tan(alpha/2 ) + tan^3( alpha/2 ) )/ ( 1 + tan^2(alpha/2 ) )`

1+tan alpha*tan((alpha)/(2))

If tan (alpha + beta) = 2 and tan (alpha - beta) = 1, then tan (2 alpha ) is equal to:

If tan x = tan y cos2 alpha prove that: tan (yx) = (tan ^ (2) alpha * sin2y) / (1 + tan ^ (2) alpha * cos2y)

(1-2sin^(2)alpha)/(1+sin2 alpha)=(1-tan alpha)/(1+tan alpha)

(1-2sin^(2)alpha)/(1+sin2 alpha)=(1-tan alpha)/(1+tan alpha)

If tan 2 alpha=(3)/(4) , then find tan alpha .

If tan alpha = 1/3 and tan beta = 1/2 , then: tan(alpha + beta) =

[[1,-tan((alpha)/(2))tan((alpha)/(2)),1]][[1,-tan((alpha)/(2))tan((alpha)/(2)),1]]