Prove that secx-cosx=tanxsinx?

Oct 8, 2016

Explanation:

$\sec x - \cos x$

= $\frac{1}{\cos} x - \cos x$

= $\frac{1 - {\cos}^{2} x}{\cos} x$

= ${\sin}^{2} \frac{x}{\cos} x$

= $\sin \frac{x}{\cos} x \times \sin x$

= $\tan x \sin x$

Oct 8, 2016

I believe you have made an error in the question as:

$\sec \left(x\right) - \cos \left(x\right) \equiv \frac{1}{\cos} x - \cos x$

$\therefore \sec \left(x\right) - \cos \left(x\right) \equiv \frac{1}{\cos} x - \cos x \cdot \cos \frac{x}{\cos} x$

$\therefore \sec \left(x\right) - \cos \left(x\right) \equiv \frac{1}{\cos} x - {\cos}^{2} \frac{x}{\cos} x$

$\therefore \sec \left(x\right) - \cos \left(x\right) \equiv \frac{1 - {\cos}^{2} x}{\cos} x$

$\therefore \sec \left(x\right) - \cos \left(x\right) \equiv \frac{{\sin}^{2} x}{\cos} x$ (using ${\sin}^{2} x + {\cos}^{2} x \equiv 1$)

$\therefore \sec \left(x\right) - \cos \left(x\right) \equiv \sin \frac{x}{\cos} x \cdot \sin x$

$\therefore \sec \left(x\right) - \cos \left(x\right) \equiv \tan x \cdot \sin x$