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Sum of Trigonometric Ratios in Terms of Their Product

If A+B+C=π ...

Question

If $A + B + C = π$ then the expression $\frac{sin 2 A + sin 2 B - sin 2 C}{sin 2 A + sin 2 B + sin 2 C}$ reduces to

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Solution

Given that
$A + B + C = π$
Now
$\frac{sin 2 A + sin 2 B - s i n 2 C}{sin 2 A + sin 2 B + s i n 2 C}$

$\frac{2 sin \frac{2 A + 2 B}{2} cos \frac{2 A - 2 B}{2} - sin 2 C}{2 sin \frac{2 A + 2 B}{2} cos \frac{2 A - 2 B}{2} + sin 2 C}$ [using the formula of $s i n 2 A + s i n 2 B$ ]

$\frac{2 sin (A + B) cos (A - B) - sin 2 C}{2 sin (A + B) cos (A - B) + sin 2 C}$

$= \frac{2 sin C cos (A - B) - 2 sin C cos C}{2 sin C cos (A - B) + 2 sin C cos C}$ $∵ A + B = π - C$

$= \frac{2 sin C [cos (A - B) - cos C]}{2 sin C [cos (A - B) + cos C]}$

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

$= \frac{2 cos \frac{(A - B + A + B)}{2} cos \frac{(A - B - A + B)}{2}}{2 sin \frac{(A - B + A + B)}{2} sin \frac{(A - B - A + B)}{2}}$ [using the formula of $cos (A - B) \pm cos (A + B)$ ]

$= \frac{2 cos (\frac{2 A}{2}) cos (\frac{2 B}{2})}{2 sin (\frac{2 A}{2}) s i n (\frac{2 B}{2})}$

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

$= cot A cot B$

Hence, it is the reduces from of the given equations.

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Sum of Trigonometric Ratios in Terms of Their Product

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