Question

Докажите тождество
\frac{sina}{1+cosa} - \frac{sina}{1-cosa} = 2 ctga

Answer 1

$\frac{sina}{1+cosa} - \frac{sina}{1-cosa} = 2 ctga\\\\\frac{(sina)(1+cosa)}{(1-cosa)(1+cosa)}- \frac{(sina)(1-cosa)}{(1-cosa)(1+cosa)} = 2ctga\\\\\frac{(sina)(1+cosa)-(sina)(1-cosa)}{(1-cosa)(1+cosa)} = 2ctga\\\\\frac{(sina)(1+cosa)-(sina)(1-cosa)}{1-cos^a} = 2ctga$

Знаем, что

$sin^2a=1-cos^2a$

Решаем далее

$\frac{(sina)(1+cosa)-(sina)(1-cosa)}{sin^2a}=2ctga\\\\\frac{(1+cosa)(1-cosa)}{sin^2a}=2ctga\\\\\frac{1+cosa-1+cosa}{sina} = 2ctga\\\\\frac{2cosa}{sina} = 2ctga\\\\2ctga=2ctga$