Prove that tan A + tan÷ tan a - tan b= sin a + b ÷ Sin A - b

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Solution

(sinA/cosA + sinB/cosB) / (sinA/cosA - sinB/cosB)

multiply everything by cosAcosB / cosAcosB

(sinA cosB + sinB cosA) / (sinA cosB - sinB cosA)

which is recognizably the formula

sin(A+B)/sin(A-B)

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Similar questions

Prove that:
$\frac{t a n A + t a n B}{t a n A - t a n B} = \frac{s i n (A + B)}{s i n (A - B)}$

\frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\sin (A + B)}{\sin (A - B)}

tan A + tan B + tan C - tan A tan B tan C = \frac{sin (A + B + C)}{cos A cos B cos C}

\frac{sin (A + B)}{cos A cos B} = tan A + tan B

If $\frac{m}{n} = \frac{t a n A}{t a n B}$ , find the value of $\frac{m + n}{m - n}$

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