Question

If TanA=ptanB then prove that Sin(A+B)=(p+1)(sin(A-B))/(p-1)

Answer 1

Answer on Question#39018 – Math – Trigonometry

If $\mathrm{TanA} = \mathrm{atanB}$ then prove that $\sin (A + B) = (p + 1)(\sin (A - B)) / (p - 1)$

Solution:

Initial condition:

$$\tan A = p \cdot \tan B$$

We can write the tangent as $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$.

$$\frac{\sin A}{\cos A} = p \cdot \frac{\sin B}{\cos B}$$ $$p \cos A \cdot \sin B = \sin A \cdot \cos B$$

Since of the sum and difference:

$$\sin (A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B$$ $$\sin (A - B) = \sin A \cdot \cos B - \cos A \cdot \sin B$$

(1) in (2) and (3):

$$\sin (A + B) = p \cos A \cdot \sin B + \cos A \cdot \sin B = \cos A \cdot \sin B (p + 1)$$ $$\sin (A - B) = p \cos A \cdot \sin B - \cos A \cdot \sin B = \cos A \cdot \sin B (p - 1)$$

From (5):

$$\cos A \cdot \sin B = \frac{\sin (A - B)}{p - 1}$$

(6) in (4):

$$\sin (A + B) = \cos A \cdot \sin B (p + 1) = \frac{(p + 1)(\sin (A - B))}{p - 1}$$