Question

Prove that$\frac{\cos \theta }{1-\tan \theta }+\frac{\sin \theta }{1-\cot \theta }=\cos \theta +\sin \theta$

Answer 1

Taking LHS we get

$\frac{cos\theta }{1-tan\theta }+\frac{sin\theta }{1-cot\theta }$

$=\frac{cos\theta }{1-\frac{sin\theta }{cos\theta }}+\frac{sin\theta }{1-\frac{cos\theta }{sin\theta }}$

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

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

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

BY using $a^2-b^2=(a+b)(a-b)$

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

$=cos\theta +sin\theta$

=RHS proved