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?

## Comments

Sort by:

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

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

\(\displaystyle \therefore\boxed{ \tan \theta = 0} \). – Sudeep Salgia · 2 years, 1 month ago

Log in to reply

– Omkar Kulkarni · 2 years, 1 month ago

Ohh okay. Is there no way other than using the \(\tan(\alpha+\beta+\gamma)\) identity?Log in to reply

– Sudeep Salgia · 2 years, 1 month ago

I cannot of think about such a method now. I'll post if I get one.Log in to reply

– Omkar Kulkarni · 2 years, 1 month ago

Okay thanks!Log in to reply