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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

Michael Rozenberg
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Gathdi
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    $25+55+100=\pi$ use this formula http://math.stackexchange.com/questions/62499/proving-cota-cotb-cotb-cotc-cotc-cota-1 $$\cot(A)\cot(B)+\cot(B)\cot(C)+\cot(C)\cot(A)=1.$$ – Nosrati Feb 07 '17 at 11:53
  • @MyGlasses This should made an answer. – Jan Feb 07 '17 at 11:56
  • The proof is not too hard, take $A+B=\pi-C$ and take $\cot$ two sides then simplify it. – Nosrati Feb 07 '17 at 11:59

2 Answers2

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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.

Darth Geek
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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.