# How do you verify (tanx) /(1-cotx) + (cotx) /(1-tanx) = 1 +secx cscx?

Nov 26, 2015

Carefully.

#### Explanation:

I find that its easier to convert directly to sine and cosine. I recognize the trig identities better.

$\frac{\sin \frac{X}{\cos} X}{1 - \cos \frac{X}{\sin} X} + \frac{\cos \frac{X}{\sin} X}{1 - \sin \frac{X}{\cos} X}$

There are too many fractions. Turn the $1$'s into $\sin X \text{/} \sin X$ and $\cos X \text{/} \cos X$, then combine the denominators into fractions over $\sin X$ and $\cos X$.

$\frac{\sin \frac{X}{\cos} X}{\frac{\sin X - \cos X}{\sin} X} + \frac{\cos \frac{X}{\sin} X}{\frac{\cos X - \sin X}{\cos} X}$

Now we can get rid of these fractions of fractions by flipping the denominators and multiplying them by the numerators.

${\sin}^{2} \frac{X}{\cos X \left(\sin X - \cos X\right)} + {\cos}^{2} \frac{X}{\sin X \left(\cos X - \sin X\right)}$

Cross multiply the denominators to get a common denominator.

$\frac{{\sin}^{3} X \left(\cos X - \sin X\right) + {\cos}^{3} \left(\sin X - \cos X\right)}{\sin X \cos X \left(\sin X - \cos X\right) \left(\cos X - \sin X\right)}$

Multiply through the parenthesis.

$\frac{{\sin}^{3} X \cos X - {\sin}^{4} X - {\cos}^{4} X + \sin X {\cos}^{3} X}{\sin X \cos X \left(- {\sin}^{2} X + 2 \sin X \cos X - {\cos}^{2} X\right)}$

Pull a factor of $\sin X \cos X$ out of two of the terms in the numerator. There is also a $- \left({\sin}^{2} X + {\cos}^{2} X\right)$ in the denominator, which by the Pythagorean theorem is equal to $- 1$.

$\frac{\left({\sin}^{2} X + {\cos}^{2} X\right) \sin X \cos X - {\sin}^{4} X - {\cos}^{4} X}{\sin X \cos X \left(2 \sin X \cos X - 1\right)}$

We can use the Pythagorean theorem again on the top. Also, pull out a $- 1$ from the ""^4 terms in the numerator.

$\frac{\sin X \cos X - \left({\sin}^{4} X + {\cos}^{4} X\right)}{\sin X \cos X \left(2 \sin X \cos X - 1\right)}$

Split the ${\sin}^{4} X$ term into two ${\sin}^{2} X$ terms. Then we can use the Pythagorean theorem, ${\sin}^{2} X = 1 - {\cos}^{2} X$ to replace one of the ${\sin}^{2}$ terms. Do a similar action with the $\cos {X}^{4}$ term.

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

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

Multiply through the terms in the top parenthesis.

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

Once again Pythagorean theorem out the ${\sin}^{2} X$ and ${\cos}^{2} X$ terms.

$\frac{\sin X \cos X - \left(1 - 2 {\sin}^{2} X {\cos}^{2} X\right)}{\sin X \cos X \left(2 \sin X \cos X - 1\right)}$

Multiply the $- 1$ through the parenthesis in the numerator.

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

Pull out a factor of $\sin X \cos X$ from two of the terms in the numerator.

$\frac{\sin X \cos X \left(1 + 2 \sin X \cos X\right) - 1}{\sin X \cos X \left(2 \sin X \cos X - 1\right)}$

Adding and subtracting a $1$ inside the parenthesis is the same as adding $0$.

$\frac{\sin X \cos X \left(1 + 2 \sin X \cos X + \left(1 - 1\right)\right) - 1}{\sin X \cos X \left(2 \sin X \cos X - 1\right)}$

Combine the two $+ 1$ terms in the parenthesis.

$\frac{\sin X \cos X \left(2 \sin X \cos X - 1 + 2\right) - 1}{\sin X \cos X \left(2 \sin X \cos X - 1\right)}$

Now pull the $2$ out. Remember to multiply by $\sin X \cos X$.

$\frac{\sin X \cos X \left(2 \sin X \cos X - 1\right) + 2 \sin X \cos X - 1}{\sin X \cos X \left(2 \sin X \cos X - 1\right)}$

Finally we have a term in the numerator that is the same as the denominator. Split the addition terms in the numerator to get;

$\frac{\sin X \cos X \left(2 \sin X \cos X - 1\right)}{\sin X \cos X \left(2 \sin X \cos X - 1\right)} + \frac{2 \sin X \cos X - 1}{\sin X \cos X \left(2 \sin X \cos X - 1\right)}$

The first term simplifies to $1$ and in the second term the $\left(2 \sin X \cos X - 1\right)$s cancel out.

$1 + \frac{1}{\sin X \cos X}$

Now convert to $\sec X$and $\csc X$

$1 + \sec X \csc X$

Sep 22, 2016

$L H S = \tan \frac{x}{1 - \cot x} + \cot \frac{x}{1 - \tan x}$

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

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

$= \cot \frac{x}{1 - \tan x} - {\tan}^{2} \frac{x}{1 - \tan x}$

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

$= \frac{1 - {\tan}^{3} x}{\tan x \left(1 - \tan x\right)}$

=(cancel((1-tanx))(1+tanx+tan^2x))/(tanxcancel((1 - tanx))

=(tanx+sec^2x)/(tanx

$= \tan \frac{x}{\tan} x + {\sec}^{2} \frac{x}{\tan} x$

$= 1 + \sec x \cdot \sec x \cdot \left(\cos \frac{x}{\sin} x\right)$

$= 1 + \sec x \cdot \left(\frac{1}{\sin} x\right)$

$= 1 + \sec x \cdot \csc x = R H S$