Question

me pueden ayudar demostrando las siguientes indentidades xfavor

Answer 1

$\frac{1-cost}{tan^2t} = \frac{1-cost}{sec^2t-1}= \frac{1-cost}{(sect+1)(sect-1)} \\=\frac{1-cost}{(sect+1)(sect-1)} \cdot \frac{cost+1}{cost+1} = \frac{1-cos^2t}{(sect+1)(1-cost+sect-1)} \\ \frac{1-cos^2t}{(sect+1)( \frac{1}{cost}-cost) } =\frac{1-cos^2t}{(sect+1)( \frac{1-cos^2t}{cost}) } \\=\frac{(1-cos^2t)(cost)}{(sect+1)( 1-cos^2t) } = \frac{cost}{sect+1}$

Saludos!