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# Could someone help me out with this?

Let $$a$$, $$b$$ and $$c$$ be positive real numbers such that $\theta=\tan^{-1} \sqrt{\frac {a(a+b+c)}{bc}} + \tan^{-1} \sqrt{\frac {b(a+b+c)}{ca}} + \tan^{-1} \sqrt{\frac {c(a+b+c)}{ab}}$ Then find the value of $$\tan \theta$$.

I'm thinking of using that identity $$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\frac{xy}{1-xy}\right)$$, but I can't get anywhere with it. Is there any way of solving this without that identity?

Note by Omkar Kulkarni
2 years, 3 months ago

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A possible way is :

Let $$\displaystyle \alpha =\arctan \left( \sqrt{\frac{a(a+b+c)}{abc}}\right) = \arctan \left(a \sqrt{\frac{a+b+c}{abc}}\right) =\arctan ( ak)$$. Similarly let $$\displaystyle \beta = \arctan (bk)$$ and $$\gamma = \arctan(ck)$$ , where $$\displaystyle k = \sqrt{\frac{(a+b+c)}{abc}}$$.

So,$$\displaystyle \theta = \alpha + \beta + \gamma \Rightarrow \tan \theta = \tan (\alpha + \beta + \gamma)$$.

$$\displaystyle \begin{array} \\ & = \frac{\tan \alpha + \tan \beta + \tan \gamma - \tan \alpha \tan \beta \tan \gamma}{\tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \gamma \tan \alpha } \\ & = \frac{(a+b+c)k -abck^3}{(ab+bc+ca)k^2} \\ & = \frac{k\left( (a+b+c) - abc \left(\frac{a+b+c}{abc} \right) \right) }{(ab+bc+ca)k^2} \\ & = 0 \\ \end{array}$$.

$$\displaystyle \therefore\boxed{ \tan \theta = 0}$$. · 2 years, 3 months ago

Ohh okay. Is there no way other than using the $$\tan(\alpha+\beta+\gamma)$$ identity? · 2 years, 3 months ago

I cannot of think about such a method now. I'll post if I get one. · 2 years, 3 months ago