ત્રિકોણમિતિય વિધયો — ગણિત ધોરણ 11 સાયન્સ — Question

‎ ‎ અહી, $\triangle ABC$ માં,

$(sin A + sin B + sin C) (sin A + sin B -sin C) = 3sin A sinB$

$\therefore (sin A+ sinB)^2 - sin^2C = 3 sin A sin B$

$\therefore sin^2A + sin^2B - sin^2C = sin A sin B$

$\therefore \frac{1 - cos 2A}{2} + \frac{1 - cos 2B}{2} - \frac{1 - cos 2C}{2} = sin A sin B$

$\therefore \frac{1}{2} - \frac{1}{2} (cos2A + cos2B - cos2C) = sin A sin B$

$\therefore \frac{1}{2} - \frac{1}{2} [2 cos (A + B) cos (A - B) - 2 cos^2C + 1] = sin A sin B$

$\therefore \frac{1}{2}- \frac{1}{2} [(2 cos (\pi - C) cos (A - B) - 2cos^2 C + 1)] = sin A sin B$

$\therefore \frac{-1}{2} [-2 cos C cos (A - B) - 2 cos^2 C] = sinA sin B$

$\therefore cos C [cos (A - B) + cos (\pi - A + B)] = sin A sin B$

$\therefore cos C (2 sin A sin B) = sin A sin B $

$\therefore cos C = \frac{1}{2} $

$\therefore C = 60^0$