#### Answer

$\sec\theta-\cos\theta=\sec\theta-\cos\theta$

#### Work Step by Step

Work on the left hand side:
Step 1: Replace $\tan\theta$ with $\frac{\sin\theta}{\cos\theta}$ by definition.
Hence,
$\sin\theta\times \tan\theta= \sin\theta\times\frac{\sin\theta}{cos\theta}$
$\sin\theta\times \tan\theta= \frac{\sin^{2}\theta}{\cos\theta}$
Step 2: By identity, $\sin^{2}\theta+\cos^{2}\theta=1$.
Replace $\sin^{2}\theta$ by $1-\cos^{2}\theta$
Hence,
$\sin\theta\times \tan\theta = \frac{1-\cos^{2}\theta}{\cos\theta}$
Step 3: Split the denominator across two terms
$\sin\theta\times \tan\theta$= $\frac{1}{\cos\theta} -
\frac{\cos^{2}\theta}{\cos\theta}$
Step 4: Cancel 1 $cos\theta$ in the 2nd term and replace the $\frac{1}{\cos\theta}$ in the first term by $\sec\theta$ by definition
Hence,
$\sin\theta\times \tan\theta = \sec\theta-\cos\theta$