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

Step 1: Take the L.H.S of the equation and prove the R.H.S
GIVEN: \frac{1}{\cos \theta} - \cos \theta = \tan \theta \sin \theta
L.H.S : \frac{1}{\cos \theta} - \cos \theta
Step 2: Do the L.C.M to the L.H.S
EXAMPLE: = \frac{1}{\cos \theta} - \cos \theta
= \frac{1 - \cos^2 \theta}{\cos \theta}
Step 3: Use the trigonometric identities to simplify
= \frac{1 - \cos^2 \theta}{\cos \theta}
[Since, We know \sin^2 \theta + \cos^2 \theta = 1
\sin^2 \theta = 1 - \cos^2 \theta]
= \frac{\sin^2 \theta}{\cos \theta}
Rewrite this equation
= \frac{\sin \theta}{\cos \theta} \sin \theta
= \tan \theta * \sin \theta
Hence, proved