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.