Question

if tan( a+b) =2 and tan b = 1/3 find tan a

Answer 1

\(\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}\)

Answer 2

so

\(\frac{\tan(a)+\frac{1}{3}}{1-\tan(a)\frac{1}{3}}=2\)


solve for \(\tan(a)\)

Answer 3

i don/t know how to solve for a

Answer 4

Here: step by step, ok?

Answer 5

\[\frac{ \tan(a)+\frac{ 1 }{3 } }{1-\tan(a)(\frac{ 1 }{ 3 }) }=2\]We are going to get rid of the denominator by multiplying it by both sides like this:

Answer 6

\[\tan(a)+\frac{ 1 }{ 3 }=2(1-\tan(a)(\frac{ 1 }{ 3 }))\]

Answer 7

now we will multiply both sides by 3 to get of the 3 in the denominator of the 1/3and at the same time distribute the 2 into the parenthesis:\[3\tan(a)+1=2-2\tan(a)\]

Answer 8

now move the -2tan(a) over to the other side:\[3\tan(a)-2\tan(a)+1=2\]and move the +1 over to the other side:\[\tan(a)=1\]and from here you use the inverse tangent button on your calculator to find that a = 45

Answer 9

we dont want to solve for a, we want to solve for tan(a)