To evaluate $\frac{\cot25+\cot55}{\tan25+\tan55}+ \frac{\cot55+\cot100}{\tan55+\tan100}+\frac{\cot100+\cot25}{\tan100+\tan25}$ To evaluate $$\frac{\cot25^{\circ}+\cot55^{\circ}}{\tan25^{\circ}+\tan55^{\circ}}+ \frac{\cot55^{\circ}+\cot100^{\circ}}{\tan55^{\circ}+\tan100^{\circ}}+\frac{\cot100^{\circ}+\cot25^{\circ}}{\tan100^{\circ}+\tan25^{\circ}}$$
i took lcm and after usual trigonometric identities i have reduced above expression to $$\cot25^{\circ}\cot55^{\circ}+\cot55^{\circ}\cot100^{\circ}+\cot100^{\circ}\cot25^{\circ}$$
How do i proceed from here?
Thanks
 A: Note that $25 + 55 + 100 = 180$ so $100 = 180 - (25 + 55)$ and $\cot(180 - \theta) = -\cot(\theta)$ so you can rewrite your equation as:
$$\cot(\alpha)\cot(\beta) - \cot(\alpha+\beta)(\cot(\alpha) + \cot(\beta))$$
Where $\alpha = 25$ and $\beta = 55$. Turns out, regardless of the values of $\alpha$ and $\beta$ (as long as $\cot$ is defined) the value of the above equation is always $1$. Let's try to prove that.
$$\cot(\alpha)\cot(\beta) = 1 + \cot(\alpha+\beta)(\cot(\alpha) + \cot(\beta))$$
Multiplying the whole thing by $\sin^2(\alpha)\sin(\beta)\cos(\beta) + \sin(\alpha)\sin^2(\beta)\cos(\alpha)$ yields (after expanding and simplifying):
$$\sin^2(\alpha)\sin(\beta)\cos(\beta) + \sin(\alpha)\sin^2(\beta)\cos(\alpha) = \sin^2(\alpha)\sin(\beta)\cos(\beta) + \sin(\alpha)\sin^2(\beta)\cos(\alpha)$$
Which is indeed true.
A: We know that $$\cot A + \cot B = \frac {\cot A \cot B -1}{\cot (A+B)} $$
Now using this identity, we get, $$P =\cot 55^\circ [\cot 25^\circ + \cot 100^\circ] + \cot 25^\circ \cot 100^\circ $$ $$P =\cot 55^\circ [\frac {\cot 25^\circ \cot 100^\circ -1}{\cot 125^\circ}] + \cot 25^\circ \cot 100^\circ $$ $$P = \cot 25^\circ \cot 100^\circ  [1 + \frac {\cot 55^\circ}{\cot 125^\circ}] - \frac {\cot 55^\circ}{\cot 125^\circ} $$
Now notice $\cot (180 - \alpha) = -\cot \alpha $. When $\alpha =55^\circ $, then $\cot 125^\circ = - \cot 55^\circ $.
Thus, $$P = \cot 25^\circ \cot 100^\circ [1-1] - \frac {\cot 55^\circ}{-\cot 55^\circ} =1$$ Hope it helps. 
