# Thread: A problem involving Inverse trignometry.

1. ## A problem involving Inverse trignometry.

Solve the equation:
Tan-1 [(x-1)/(x-2)] + Tan-1 [(x+1)/(x+2)] = π/4

2. ## Re: A problem involving Inverse trignometry.

We have
$\tan^{-1} \left( \frac{x-1}{x-2}\right)+\tan^{-1} \left( \frac{x+1}{x+2}\right)=\frac{\pi}{4}$

By using the famous identity, $\arctan(x)+\arctan(y)= \arctan \left ( \frac{x+y}{1-xy}\right)$ we obtain

$\tan^{-1} \left( \frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{x+1}{x+2} \frac{x-1}{x-2}}\right)=\frac{\pi}{4}$

$\frac{(x-1)(x+2)+(x+1)(x-2)}{x^2-2-(x^2-1)} = \tan \left( \frac{\pi}{4}\right)$

$x^2+2x-x-2+x^2-2x+x-2=-1$

$2x^2-4=-1$

$2x^2=3$

$x= \pm \sqrt{\frac{3}{2}}$

3. ## Re: A problem involving Inverse trignometry.

I had tried it the same way, but I had a few queries regarding it as thus.

1)This identity is valid when the following conditions are satisfied:
a) x>0
b) y>0
c) xy<1

I was apprehensive about how I could ascertain the fact that the question satisfies these conditions. Or is it any other vital piece of concept I am lacking?

2)Secondly could you please tell me if the site "www.wolframalpha.com", which is a computational knowledge engine, gives the same answer as the above(ie: root of 1.5), or first of all is it correct to tally this answer to the one that computational engine is producing?

Well, I thank you sbhatnagar.

And I would be glad if you, or anyone could help me with my doubts as stated above.