Question

Show that $\frac{1}{\cos \theta }-\cos \theta =\tan \theta .\sin \theta$.

Answer 1

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

$=\frac{1-co{s}^2\theta }{cos\theta }$

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

$=\frac{sin\theta }{cos\theta }.sin\theta$

$=tan\theta .sin\theta$