Prove that $\frac{cot(x)}{1-tan(x)} + \frac{tan(x)}{1 - cot(x)}=1 + tan(x) + cot(x)$

Q. Prove: $\frac{cot(x)}{1-tan(x)} + \frac{tan(x)}{1 - cot(x)}=1 + tan(x) + cot(x)$}


A. $$\begin{array}{lll}\\
\frac{cot(x)}{1-tan(x)} + \frac{tan(x)}{1 - cot(x)} &= \frac{\frac{1}{tan(x)}}{1 - tan(x)} + \frac{tan(x)}{1 - \frac{1}{tan(x)}}&\mbox{Expressing } cot(x)=1/tan(x)\\
&= \frac{1}{tan(x)(1-tan(x))} + \frac{\tan^2(x)}{tan(x) - 1}& \mbox{Simplifying fractions.}\\
&= \frac{1}{tan(x)(1-tan(x))} - \frac{\tan^2(x)}{1- tan(x)}& \mbox{Last term as a subtraction }\\
&= \frac{1 - tan^3(x)}{tan(x)(1-tan(x)}& \mbox{Rewriting as a single fraction}\\
&= \frac{(1 - tan(x)){(1 + tan(x)} + \tan^2(x)}{tan(x)(1-tan(x))}& \mbox{Factoring a difference of cubes}\\
&= \frac{(1 + tan(x) +\tan^2(x)}{tan(x)}&\mbox{Cancellation by division}\\
&= cot(x) + 1 + tan(x)& \mbox{Division}\\
&= 1 + tan(x) + cot(x)& \mbox{Rearranging terms.}\\
\end{array}
$$

Derivation of the formula for cos(alpha - beta)

Derivation of the length of a chord

Let an arc on the unit circle with starting point at $ P_0 = (1, 0)$ subtend an angle of $\beta$ radians. Recall that its endpoint will be
$P_1 = (\cos(\beta), \sin(\beta))$. The length of the chord connecting the point (1,0) to the terminal point can be determined using the
distance formula $$d = \sqrt{(\Delta x)^2 + (\Delta y)^2}=\sqrt{(\cos(\beta) - 1)^2 + (\sin(\beta) - 0) ^2}$$. Simplifying we will have
$d = \sqrt{\cos^2(\beta) + \sin^2 (\beta) - 2\cos(\beta) + 1}= \sqrt{2- 2\cos(\beta)}.$

The fundamental formula for the cosine of the difference in angles: $\cos(\alpha - \beta)$

Now let $\alpha$ be an angle greater than $\beta$. The new endpoint will be at $P_2 =(\cos(\alpha), sin(\alpha))$. Consider the chord $P_1P_2$.
By the distance formula, the length of this chord will be given by $\sqrt{(\cos (\alpha) - \cos(\beta))^2 +(\sin(\alpha)-\sin(\beta))^2}$
and expanding we obtain
$\sqrt{(\cos^2(\alpha) - 2 cos(\alpha)cos(\beta) + cos^2(\beta) + (\sin^2(\alpha) - 2 sin(\beta)\sin(\alpha)+\sin^2(\beta))} $

Further simplifying we obtain $\sqrt{(\cos^2(\alpha) + \sin^2(\alpha) + \cos^2(\beta) + \sin^2(\beta) - 2\cos(\alpha) cos\beta - \sin(\alpha) \sin(\beta)}$.
and finally we have $d = \sqrt{2 -2 (\cos(\alpha) cos\beta + \sin(\alpha) \sin(\beta)}$
But this chord lengthis also given by $\sqrt{ 2 - 2 cos(\alpha - \beta)}$.

Equating the two expressions , we obtain the formula

$$\cos(\alpha - \beta) = \cos(\alpha) cos(\beta) + \sin(\alpha) \sin(\beta)$$

This is an important result. Let us try to derive further formulas from this.
Substiture $-\beta$ for $\beta$. We sould then have $\cos(\alpha - (-\beta)) = \cos(\alpha) cos(-\beta) + \sin(\alpha) \sin(-\beta)$.
or $$\cos(\alpha + \beta)) = \cos(\alpha) cos(\beta) - \sin(\alpha) \sin(\beta)$$ using the facts that the cosine function is even $\cos(-alpha) = \cos(alpha)$
that the sine function is odd $sin(-\beta) = -sin(beta)$.We can combine the two identities as $$\cos(\alpha \pm \beta) = cos(\alpha) cos(\beta) \mp \sin(\beta) sin(\alpha)$$.