Question

How
$\frac{\sec \theta -1}{\tan \theta }=\frac{1-\cos \theta }{\frac{\cos \theta }{\frac{\sin \theta }{\cos \theta }}}$

Answer 1

$\frac{sec\theta -1}{tan\theta }$

$(tan\theta =\frac{sin\theta }{cos\theta };sec\theta =\frac{1}{cos\theta })$

$\therefore \frac{\frac{1}{cos\theta }-1}{\frac{sin\theta }{cos\theta }}$

$\Rightarrow \frac{\left(\frac{1-cos\theta }{cos\theta }\right)}{\left(\frac{sin\theta }{cos\theta }\right)}$

= RHS

Hence verified.