Question

(cosa/1-sina+cosa/1+sina)*1/1+tg²a

Answer 1

Объяснение:

$(\frac{cos\alpha }{1-sin\alpha }+\frac{cos\alpha }{1+sin\alpha })*\frac{1}{1+tg^2\alpha } .$

$1)\ \frac{cos\alpha }{1-sin\alpha }+\frac{cos\alpha }{1+sin\alpha }=\frac{cos\alpha *(1+sin\alpha )-cos\alpha *(1-sin\alpha )}{(1-sin\alpha \alpha )*(1+sin\alpha) } =\frac{cos\alpha +sin\alpha *cos\alpha -cos\alpha +sin\alpha *cos\alpha }{1-sin^2\alpha }=\\=\frac{2*sin\alpha *cos\alpha }{cos^2\alpha }=\frac{2*sin\alpha }{cos\alpha } .$$2)\ \frac{1}{1+tg^2\alpha } =\frac{1}{1+\frac{sin^2\alpha }{cos^2\alpha } } =\frac{1}{\frac{sin^2\alpha +cos^2\alpha }{cos^2\alpha } } =\frac{cos^2\alpha }{1} =cos^2\alpha.$

$3)\ \frac{2*sin\alpha }{cos\alpha } *cos^2\alpha =2*sin\alpha *cos\alpha =sin2\alpha .$