Question

If tan-1a+tan-1b+tan-1c=pie,then prove that a+b+c=abc

Answer 1

We are asked to prove:

$If \, \, \tan^{-1} a + \tan^{-1}b + \tan^{-1}c = \pi \, \, then \, \, a+b+c=abc$


We will use the following concepts:

$\boxed{\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)} \, \, \, if \, xy<1\\ \\ \\ \\ \boxed{\tan^{-1}x + \tan^{-1}y =\pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)} \, \, \, if xy>1 \\ \\ \\ \boxed{\tan(\pi - \theta) = -\tan \theta} \\ \\ \\ \boxed{\tan (-\theta) = - \tan \theta}$

Now we can solve the question:


PART 1: Assuming ab<1


$\tan^{-1}a+\tan^{-1}b+\tan^{-1}c = \pi \\ \\ \\ \implies \tan^{-1}a+\tan^{-1}b = \pi - \tan^{-1}c \\ \\ \\ \implies \tan^{-1}\left(\frac{a+b}{1-ab}\right) = \pi - \tan^{-1}c \\ \\ \\ \text{Taking tan on both sides} \\ \\ \\ \implies \tan \left(\tan^{-1}\left(\frac{a+b}{1-ab} \right) \right) = \tan (\pi - \tan^{-1}c) \\ \\ \\ \implies \frac{a+b}{1-ab} = -\tan(\tan^{-1}c) \\ \\ \\ \implies \frac{a+b}{1-ab} = -c \\ \\ \\ \implies a+b = -c(1-ab) \\ \\ \\ \implies a+b = -c+abc \\ \\ \\ \implies \boxed{a+b+c=abc} \\ \\ \\ \text{Hence Proved}$

PART 2: Assuming ab>1


[tex]\tan^{-1}a+\tan^{-1}b+\tan^{-1}c = \pi \\ \\ \\ \implies \tan^{-1}a+\tan^{-1}b = \pi - \tan^{-1}c \\ \\ \\ \implies \pi + \tan^{-1}\left(\frac{a+b}{1-ab}\right) = \pi - \tan^{-1}c \\ \\ \\ \tan^{-1}\left(\frac{a+b}{1-ab}\right) = -\tan^{-1}c \\ \\ \\ \text{Taking tan on both sides} \\ \\ \\ \implies \tan \left(\tan^{-1}\left(\frac{a+b}{1-ab} \right) \right) = \tan (-\tan^{-1}c) \\ \\ \\ \implies \frac{a+b}{1-ab} = -\tan(\tan^{-1}c) \\ \\ \\ \implies \frac{a+b}{1-ab} = -c \\ \\ \\ \implies a+b = -c(1-ab) \\ \\ \\ \implies a+b = -c+abc \\ \\ \\ \implies \boxed{a+b+c=abc} \\ \\ \\ \text{Hence Proved}[/tex]

Hope it helps
Purva
Brainly Community

Answer 2

for solution refer attachment