# How do you prove  (cosx - sinx)/(cos^3x - sin^3x) = 1/(tanxcos^2x + 1)?

Mar 29, 2018

Verified below...

#### Explanation:

Given:
$\frac{\cos x - \sin x}{{\cos}^{3} x - {\sin}^{3} x} = \frac{1}{\tan x {\cos}^{2} x + 1}$

When working with the denominator: $\left({a}^{3} - {b}^{3}\right) = \left(a - b\right) \left({a}^{2} + b a + {b}^{2}\right)$

$\frac{\cos x - \sin x}{\left(\cos x - \sin x\right) \left({\cos}^{2} x + \cos x \sin x + {\sin}^{2} x\right)} = \frac{1}{\tan x {\cos}^{2} x + 1}$

$\frac{\cancel{\cos x - \sin x}}{\cancel{\cos x - \sin x} \left({\cos}^{2} x + \cos x \sin x + {\sin}^{2} x\right)} = \frac{1}{\tan x {\cos}^{2} x + 1}$

$\frac{1}{{\cos}^{2} x + \cos x \sin x + {\sin}^{2} x} = \frac{1}{\tan x {\cos}^{2} x + 1}$

Remember: ${\sin}^{2} x + {\cos}^{2} x = 1$

$\frac{1}{1 + \cos x \sin x} = \frac{1}{\tan x {\cos}^{2} x + 1}$

Remember $\tan x = \sin \frac{x}{\cos} x$ therefore: $\sin x = \tan x \cdot \cos x$:
$\frac{1}{1 + \cos x \left(\tan x \cdot \cos x\right)} = \frac{1}{\tan x {\cos}^{2} x + 1}$

$\frac{1}{\tan x {\cos}^{2} x + 1} = \frac{1}{\tan x {\cos}^{2} x + 1}$

Mar 29, 2018

Proved below.

#### Explanation:

lhs=

$\frac{\cos x - \sin x}{{\cos}^{3} x - {\sin}^{3} x}$

Now,
Let
$a = \cos x , b = \sin x$

$\frac{\cos x - \sin x}{{\cos}^{3} x - {\sin}^{3} x} = \frac{a - b}{{a}^{3} - {b}^{3}}$

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + {b}^{2} + a b\right)$

$\frac{a - b}{{a}^{3} - {b}^{3}} = \frac{\left(a - b\right)}{\left(a - b\right) \left({a}^{2} + {b}^{2} + a b\right)}$

Cancelling a-b

$\frac{a - b}{{a}^{3} - {b}^{3}} = \frac{1}{{a}^{2} + {b}^{2} + a b}$

${a}^{2} + {b}^{2} + a b = {\left(\cos x\right)}^{2} + {\left(\sin x\right)}^{2} + \left(\cos x\right) \left(\sin x\right)$

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

${\cos}^{2} x + {\sin}^{2} x = 1$

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

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

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

${\left(\cos x\right)}^{2} + {\left(\sin x\right)}^{2} + \left(\cos x\right) \left(\sin x\right) = 1 + {\cos}^{2} x \tan x$

$\frac{a - b}{{a}^{3} - {b}^{3}} = \frac{1}{{a}^{2} + {b}^{2} + a b}$

$\frac{\cos x - \sin x}{{\cos}^{3} x - {\sin}^{3} x} = \frac{1}{1 + {\cos}^{2} x \tan x}$

$= r h s$

Mar 29, 2018

factor cos ${\cos}^{3} x - {\sin}^{3} x$=$\left(\cos x - \sin x\right) \left({\cos}^{2} x + \cos x \cdot \sin x + {\sin}^{2} x\right)$

#### Explanation:

(cosx-sinx)/(cos^3x-sin^3x)=(cosx-sinx)/((cosx-sinx)(cos^2x+cosxsinx+sin^2x)
[cancelling the $\left(\cos x - \sin x\right)$ both in numerator and denominator]

$= \frac{1}{{\cos}^{2} x + \cos x \sin x + {\sin}^{2} x}$
[using trignometric identities ${\cos}^{2} x + {\sin}^{2} x = 1$]
$= \frac{1}{1 + \cos x \sin x}$

multiplying and dividing by $\cos x$

$= \frac{1}{1 + \cos x \sin x \left(\cos \frac{x}{\cos} x\right)}$
$= \frac{1}{1 + \tan x {\cos}^{2} x}$