Question

Prove that sin(A+B)/sin(A-B)+cos(A+B) +cos(A-B)/sin(A-B)+cos(A+B)=tan(phi/4 +B)

Answer 1

Answer on Question #62878 – Math – Trigonometry

Question

Prove that

$$\frac {\sin (A + B) + \cos (A - B)}{\sin (A - B) + \cos (A + B)} = \tan \left(\frac {\pi}{4} + B\right).$$

Solution

Recall that

$$\sin (A + B) = \sin A \cos B + \cos A \sin B, \quad \sin (A - B) = \sin A \cos B - \cos A \sin B,$$ $$\cos (A + B) = \cos A \cos B - \sin A \sin B, \quad \cos (A - B) = \cos A \cos B + \sin A \sin B.$$

Transforming the left-hand side of (1) get

$$\begin{array}{l} \frac {\sin (A + B) + \cos (A - B)}{\sin (A - B) + \cos (A + B)} = \frac {(\sin A \cos B + \cos A \sin B) + (\cos A \cos B + \sin A \sin B)}{(\sin A \cos B - \cos A \sin B) + (\cos A \cos B - \sin A \sin B)} \\ = \frac {\sin A \cos B + \cos A \sin B + \cos A \cos B + \sin A \sin B}{\sin A \cos B - \cos A \sin B + \cos A \cos B - \sin A \sin B} \\ = \frac {(\sin A \cos B + \cos A \cos B) + (\cos A \sin B + \sin A \sin B)}{(\sin A \cos B + \cos A \cos B) - (\cos A \sin B + \sin A \sin B)} = \\ \end{array}$$ $$= \frac {\cos B (\sin A + \cos A) + \sin B (\cos A + \sin A)}{\cos B (\sin A + \cos A) - \sin B (\cos A + \sin A)} = \frac {(\cos B + \sin B) (\cos A + \sin A)}{(\cos A + \sin A) (\cos B - \sin B)} = \frac {\cos B + \sin B}{\cos B - \sin B}.$$

We proved that

$$\frac {\sin (A + B) + \cos (A - B)}{\sin (A - B) + \cos (A + B)} = \frac {\cos B + \sin B}{\cos B - \sin B}.$$

Transforming the right-hand side of (1) get

$$\tan \left(\frac {\pi}{4} + B\right) = \frac {\tan \left(\frac {\pi}{4}\right) + \tan B}{1 - \tan \left(\frac {\pi}{4}\right) \tan B} = \frac {1 + \tan B}{1 - \tan B} = \frac {1 + \frac {\sin B}{\cos B}}{1 - \frac {\sin B}{\cos B}} = \frac {\frac {\cos B + \sin B}{\cos B}}{\frac {\cos B - \sin B}{\cos B}} = \frac {\cos B + \sin B}{\cos B - \sin B}.$$ $$= \frac {\cos B + \sin B}{\cos B (\cos B - \sin B)} \cdot \cos B = \frac {\cos B + \sin B}{\cos B - \sin B}.$$

We proved that

$$\tan \left(\frac {\pi}{4} + B\right) = \frac {\cos B + \sin B}{\cos B - \sin B}.$$

It follows from (2) and (3) that

$$\frac {\sin (A + B) + \cos (A - B)}{\sin (A - B) + \cos (A + B)} = \frac {\cos B + \sin B}{\cos B - \sin B} = \tan \left(\frac {\pi}{4} + B\right).$$

Hence

$$\frac {\sin (A + B) + \cos (A - B)}{\sin (A - B) + \cos (A + B)} = \tan \left(\frac {\pi}{4} + B\right).$$

Q.E.D.

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