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

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. 
