# Thread: Proving Identity of Trigonometry Ratio

1. ## Proving Identity of Trigonometry Ratio

Given that :
$sinB/sinA=cos(A+B)$

To prove :
$tan(A+B)=2tanA$

How to do this prove ? thank you very much

2. Hello mandyn318

Welcome to Math Help Forum!
Originally Posted by mandyn318
Given that :
$sinB/sinA=cos(A+B)$

To prove :
$tan(A+B)=2tanA$

How to do this prove ? thank you very much
This looks easy, but I think it's surprisingly tricky. I have a proof that assumes $\cos A > 0$ and $\cos B>0$. I'll leave it to you to check what happens if you take the negative values of the square roots.

To make it easier to read, then, I'm going to let:
$\sin A = a$ and $\sin B = b$
So:
$\cos A = \sqrt{1-a^2}$ and $\cos B = \sqrt{1-b^2}$, taking the positive square roots
Therefore the original given equation:
$\frac{\sin B}{\sin A} = \cos(A+B)$
$=\cos A \cos B - \sin A \sin B$
can be written as:
$\frac ba=\sqrt{(1-a^2)(1-b^2)}-ab$

$\Rightarrow b(1+a^2)=a\sqrt{(1-a^2)(1-b^2)}$
...(1)
Now
$\tan(A+B)=\sin(A+B)\cdot\frac{1}{\cos(A+B)}$
$=\sin (A+B)\cdot\frac{\sin A}{\sin B}$, from the original given
$=\frac{\sin A(\sin A \cos B + \cos A \sin B)}{\sin B}$

$=\frac{a(a\sqrt{1-b^2}+b\sqrt{1-a^2})}{b}$

$=\frac{a\Big(a\sqrt{(1-a^2)(1-b^2)}+b(1-a^2)\Big)}{b\sqrt{1-a^2}}$, multiplying top-and-bottom by $\sqrt{1-a^2}$

$=\frac{a\Big(b(1+a^2)+b(1-a^2)\Big)}{b\sqrt{1-a^2}}$ from
(1)

$=\frac{2ab}{b\sqrt{1-a^2}}$

$=2\tan A$