We consider tan(3A) as tan(2A+A) and then we use the formula for tan(a+b) which is this:and we substitute 2A for a and A for b: equation 1: 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 substituting A for a and also A for b: Next we must substitute for tan(2A) in equation 1: To simplify that compound fraction multiply top and bottom by (1-tan2A) Edwin