Question

Prove that:

$\frac{cos\theta }{(1-tan\theta )}-\frac{si{n}^2\theta }{(cos\theta -sin\theta )}=(cos\theta +sin\theta )$

Answer 1

$\frac{cos\theta }{(1-tan\theta )}-\frac{si{n}^2\theta }{(cos\theta -sin\theta )}$

$=\frac{cos\theta }{1-\frac{sin\theta }{cos\theta }}-\frac{si{n}^2\theta }{cos\theta -sin\theta }$

$=\frac{co{s}^2\theta }{cos\theta -sin\theta }-\frac{si{n}^2\theta }{cos\theta -sin\theta }$

$=\frac{co{s}^2\theta -si{n}^2\theta }{cos\theta -sin\theta }$

$=\frac{(cos\theta -sin\theta )(cos\theta +sin\theta )}{cos\theta -sin\theta }$

$=cos\theta +sin\theta$