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 $$\cos(\alpha \pm \beta) = cos(\alpha) cos(\beta) \mp \sin(\beta) sin(\alpha)$$ .

The angle between the hour and minute hands of a clock.

Q. The time shown in the clock is 7:35 what is the angle between the hour and minute hands of a clock?

A. Consider the simpler problem if the time is 7:00. As there are 12 hours in a complete revolution of the hour hand, one hour subtends an angle
of 360/12 = 30 degrees. Thus, if H is the number of hours, the angle made by the hour hand from the noon positioon would be 30H.

Now let us consider the minute hand. if M is the number of minutes, consider that if M is 15 minutes , the angle which the minute hand makes is
90 degrees. This means that each minute ssubends an angle of 90/15 = 6 degrees or 6M.

But each revolution of the minute hand imparts 1/12 revolution of the hour hand. In other words, in the time covered by the minute hand, the hour hand will cover an
addittional 6M/12 or M/2 degrees.

The angle between the hour hand and minute hand for a time reading of H:M will then be |30H + M/2 - 6M|. If this angle is greater than 180 degrees,
we take the difference from 360 degrees.

Getting back to our original problem, for a time of 7:35, the angle beween the hour and minute hands is |30(7) + 35/2 - 6(35)| = 17.5 degrees.