If tan(A+B)=x and tanB=1/2prove that tanA=2x-1/x+2and obtain an expression for tan(A-B)in terms of x.?

Answer 1

By the tan addition identity,

tan(A + B) = [tan(A) + tan(B)] / [1 - tan(A)tan(B)]

But, tan(A + B) = x, so replacing tan(A + B) with x, we get

x = [tan(A) + tan(B)] / [1 - tan(A)tan(B)]

But we're given tan(B) = 1/2, so replacing every occurrence of tan(B) with (1/2), we get

x = [tan(A) + (1/2)] / [1 - tan(A)(1/2)]

Let's simplify this complex fraction by multiplying top and bottom by 2.

x = [2tan(A) + 1] / [2 - tan(A)]

Now, we solve for tan(A) algebraically. Multiply both sides by
[2 - tan(A)], to obtain

x[2 - tan(A)] = 2tan(A) + 1. Expanding the left hand side,

2x - xtan(A) = 2tan(A) + 1

Now, move everything with a tan(A) to the right hand side; everything else, to the left hand side.

2x - 1 = 2tan(A) + xtan(A)

Factor tan(A) on the right hand side.

2x - 1 = tan(A) [2 + x]

Divide both sides by (2 + x)

tan(A) = (2x - 1) / (2 + x)

2) Find tan(A - B) in terms of x.

tan(A - B) = [tan(A) - tan(B)] / [1 + tan(A)tan(B)]

Substituting tan(A) = (2x - 1) / (2 + x) and tan(B) = (1/2)

= [(2x - 1)/(2 + x) - (1/2)] / [1 + (1/2)(2x - 1)/(2 + x)]

Multiply top and bottom by 2(2 + x), to get rid of all fractions.

= [(2x - 1)(2) - (2 + x)] / [2(2 + x) + (2x - 1)(1/2)(2)]
= [4x - 2 - 2 - x] / [4 + 2x + 2x - 1]
= [3x - 4] / [4x + 3]

Edit: Corrected from a previously wrong answer.

Answer 2

by formula, tan(A+B) = (tanA + tanB)/(1 - tanAtanB)
given, tan(A+B) = x & tanB = 1/2. hence,
x = (tanA + 1/2)/(1 - (tanA)/2)
x - (x/2)tanA = tanA + 1/2
tanA + (x/2)tanA = x - 1/2
tanA (1 + x/2) = x - 1/2
tanA = ((2x - 1)/2)(2/(x + 2))
tanA = (2x - 1)(x + 2)

expression for tan(A-B) :
by formula, tan(A - B) = (tanA - tanB)/(1 + tanAtanB)
we know that , tanA= (2x - 1)(x + 2)
tanB = 1/2 hence,
tan(A - B) =((2x - 1)(x + 2) - (1/2))/(1 + (2x - 1)(x +2)(1/2)
=((3x - 4) / 2(x+2)) / ((2x+3) / 2(x+2))
=(3x - 4) / (2x + 3)

Answer 3

solve for tanA
x=(tanA+1/2)/(1-tanA*(1/2))

then

tan(-B)=-tanB=-1/2 and continue like above