Proving \tan A=2+\sqrt{3} using the identity \sin A=\sin(A+30^{\circ})"

[odolwa99]

In attempting this question, I decided to expand the first statement. Can anyone help me out?

Many thanks.

Homework Statement

If $\sin A=\sin(A+30^{\circ})$, show that $\tan A=2+\sqrt{3}$.

Homework Equations

The Attempt at a Solution

$\sin A=\sin A\cos30+\cos A\sin30$
$\sin A=\frac{\sin A\sqrt{3}+\cos A}{2}$
$2\sin A=\sin A\sqrt{3}+\cos A$
$\sin A(2-\sqrt{3})=\cos A$
$2-\sqrt{3}=\frac{\cos A}{\sin A}$

 

[CAF123]

In attempting this question, I decided to expand the first statement. Can anyone help me out?

Many thanks.

Homework Statement

If $\sin A=\sin(A+30^{\circ})$, show that $\tan A=2+\sqrt{3}$.

Homework Equations

The Attempt at a Solution

$\sin A=\sin A\cos30+\cos A\sin30$
$\sin A=\frac{\sin A\sqrt{3}+\cos A}{2}$
$2\sin A=\sin A\sqrt{3}+\cos A$
$\sin A(2-\sqrt{3})=\cos A$
$2-\sqrt{3}=\frac{\cos A}{\sin A}$

This implies $$\frac{\sin A}{\cos A} = \frac{1}{2 - \sqrt{3}}.$$ Now rationalise the denominator.

 

[odolwa99]

Great, thank you very much.