Question

:1) sina/1+cosa - sina/1-cosa 2) sin^2a + 2cos^2a - 1

Answer 1

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

$sin^2 \alpha + 2cos^2 \alpha - 1 =1-cos^2 \alpha + 2cos^2 \alpha - 1 =cos^2 \alpha$