1. ## Prove this

$tan^{-1}\left(\frac{tan2\theta + tanh2\phi}{tan2\theta - tanh2\phi}\right)+tan^{-1}\left(\frac{tan\theta - tanh\phi}{tan\theta + tanh\phi}\right)=tan^{-1}\left(cot\theta coth\phi \right)$

Thank You Very Much....

2. ## Re: Prove this

Are you sure you typed the formula correctly? I don't believe the two sides are equal. I tried values of theta = 0.1 and phi = 0.2 and I get a result of -1.59 for the left side and +1.55 for the right side.

3. ## Re: Prove this

Yes, I typed correctly as given in the book, may be it was printed wrong in book, thanks ..

4. ## Re: Prove this

Originally Posted by kjchauhan

$tan^{-1}\left(\frac{tan2\theta + tanh2\phi}{tan2\theta - tanh2\phi}\right)+tan^{-1}\left(\frac{tan\theta - tanh\phi}{tan\theta + tanh\phi}\right)=tan^{-1}\left(cot\theta coth\phi \right)$

Thank You Very Much....
take the tan for both sides
left side
$tan \left(tan^{-1}\left(\frac{tan2\theta + tanh2\phi}{tan2\theta - tanh2\phi}\right)+tan^{-1}\left(\frac{tan\theta - tanh\phi}{tan\theta + tanh\phi}\right)\right)$

then use the identity $\tan (a+b) = \frac{\tan a + \tan b }{1 - \tan a \tan b }$
resulting

$\dfrac{\dfrac{tan2\theta + tanh2\phi}{tan2\theta - tanh2\phi} + \dfrac{tan\theta - tanh\phi}{tan\theta + tanh\phi}}{1 - \left(\dfrac{tan2\theta + tanh2\phi}{tan2\theta - tanh2\phi}\right)\left(\dfrac{tan\theta - tanh\phi}{tan\theta + tanh\phi}\right)}$

we have to prove that the previous equal $\cot \theta \coth \phi$

You can make the nominator and denominator with common denominator then cancel it

5. ## Re: Prove this

Thank you very much for giving such an important hint..

6. ## Re: Prove this

Originally Posted by ebaines
Are you sure you typed the formula correctly? I don't believe the two sides are equal. I tried values of theta = 0.1 and phi = 0.2 and I get a result of -1.59 for the left side and +1.55 for the right side.
Update - the formula is indeed correct. My mistake was in relying on my PC to provide the arctan calculation, and forgetting that it can be plus or minus pi. With that correction the formula does indeed work. Sorry for the earlier post.