# Prove that (cotx-tanx)/(sinx+cosx)=cscx-secx?

May 3, 2016

#### Explanation:

$\frac{\cot x - \tan x}{\sin x + \cos x}$

= $\frac{\cos \frac{x}{\sin} x - \sin \frac{x}{\cos} x}{\sin x + \cos x}$

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

= $\frac{\left(\cos x - \sin x\right) \left(\cos x + \sin x\right)}{\sin x \cos x} \times \frac{1}{\sin x + \cos x}$

= $\frac{\left(\cos x - \sin x\right) \cancel{\left(\cos x + \sin x\right)}}{\sin x \cos x} \times \frac{1}{\cancel{\left(\sin x + \cos x\right)}}$

= $\frac{\left(\cos x - \sin x\right)}{\sin x \cos x}$

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

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

= $\csc x - \sec x$