Question 1076136
<pre><b>
We consider tan(3A) as tan(2A+A) and then
we use the formula for tan(<font face="symbol">a</font>+<font face="symbol">b</font>) which is this:

{{{tan(alpha+beta) = (tan(alpha)+tan(beta))/(1-tan(alpha)tan(beta)))}}}

and we substitute 2A for <font face="symbol">a</font> and A for <font face="symbol">b</font>:

{{{tan(3A)=tan(2A+A) = (tan(2A)+tan(A))/(1-tan(2A)tan(A)))}}}

equation 1: {{{tan(3A)= (tan(2A)+tan(A))/(1-tan(2A)tan(A)))}}}

But equation 1 is not enough, for we must now find tan(2A).

We find that by considering 2A as A+A and use this formula again

{{{tan(alpha+beta) = (tan(alpha)+tan(beta))/(1-tan(alpha)tan(beta)))}}}

substituting A for <font face="symbol">a</font> and also A for <font face="symbol">b</font>:

{{{tan(2A)=tan(A+A) = (tan(A)+tan(A))/(1-tan(A)tan(B))=(2tan(A))/(1-tan^2(A))}}}

Next we must substitute for tan(2A) in equation 1:

{{{tan(3A)= (((2tan(A))/(1-tan^2(A)))+tan(A))/(1-((2tan(A))/(1-tan^2(A)))tan(A))=(((2tan(A))/(1-tan^2(A)))+tan(A))/(1-((2tan^2(A))/(1-tan^2(A)))))}}} 

To simplify that compound fraction multiply top and bottom
by (1-tan<sup>2</sup>A)

{{{tan(3A)=(2tan(A)+tan(A)(1-tan^2(A)^""))/(1*(1-tan^2(A)^"")-2tan^2(A))}}}

{{{tan(3A)=(2tan(A)+tan(A)(1-tan^2(A)^""))/(1-tan^2(A)-2tan^2(A))}}}

{{{tan(3A)=(2tan(A)+tan(A)-tan^3(A))/(1-tan^2(A)-2tan^2(A))}}}

{{{tan(3A)=(3tan(A)-tan^3(A))/(1-3tan^2(A))}}}

{{{tan(3A)=(tan(A)(3-tan^2(A)^""))/(1-3tan^2(A))}}}

Edwin</pre></b>