Question

Tan 3A in terms of tan​

Answer 1

Here's the sum rule for the tangent function:

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

In the special case a=b, this becomes the double angle formula:

$\tan(a+a)=\tan(2a)=\dfrac{\tan(a)+\tan(a)}{1-\tan(a)\tan(a)}=\dfrac{2\tan(a)}{1-\tan^2(a)}$

In your case, you case use the sum rule once:

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

And use it again, in the special case of the double angle:

$\dfrac{\dfrac{2\tan(a)}{1-\tan^2(a)}+\tan(a)}{1-\dfrac{2\tan(a)}{1-\tan^2(a)}\tan(a)}$

We can obvisouly simplify this expression a lot: let's deal with the numerator and denominator separately: the numerator is

$\dfrac{2\tan(a)}{1-\tan^2(a)}+\tan(a) = \dfrac{2\tan(a)+\tan(a)-\tan^3(a)}{1-\tan^2(a)}$

and the denominator is

$1-\dfrac{2\tan(a)}{1-\tan^2(a)}\tan(a) = \dfrac{1-\tan^2(a)-2\tan^2(a)}{1-\tan^2(a)} = \dfrac{1-3\tan^2(a)}{1-\tan^2(a)}$

So, the fraction is

$\dfrac{2\tan(a)+\tan(a)-\tan^3(a)}{1-\tan^2(a)}\cdot \dfrac{1-\tan^2(a)}{1-3\tan^2(a)} = \dfrac{2\tan(a)+\tan(a)-\tan^3(a)}{1-3\tan^2(a)}$