Question

sinα__ _sinα
-------- + ---------
1+cosα 1-cosα

Answer 1

$\frac{Sin \alpha }{1+Cos \alpha }+ \frac{Sin \alpha }{1-Cos \alpha }= \frac{Sin \alpha -Sin \alpha Cos \alpha +Sin \alpha +Sin \alpha Cos \alpha }{(1+Cos \alpha )(1-Cos \alpha )} = \frac{2Sin \alpha }{1-Cos ^{2} \alpha } = \frac{2Sin \alpha }{Sin ^{2} \alpha }$$= \frac{2}{Sin \alpha }$

Answer 2

$\frac{sin(a)}{1+cos(a)} +\frac{sin(a)}{1-cos(a)} =\frac{(1-cos(a))sin(a)+(1+cos(a))sin(a)}{(1+cos(a))(1-cos(a))} =\\=\frac{sin(a)-cos(a)sin(a)+sin(a)+cos(a)sin(a)}{1-cos^2(a)} =\frac{2sin(a)}{sin^2(a)} =\frac{2}{sin(a)}$