# Thread: Need help once again

1. ## Need help once again

I'm pretty sure I'm suppose to take tan on both sides but then I'm kinda stuck, assistance will be appreciated!

Much thanks

2. ## Re: Need help once again

Originally Posted by mkv_86
I'm pretty sure I'm suppose to take tan on both sides but then I'm kinda stuck, assistance will be appreciated!

Much thanks
I looked at it quickly and you can probably use the tan special angle formula with tan(a+b) = [tana+tanb]/(1-tanatanb). Can't see a faster method for now.

3. ## Re: Need help once again

Originally Posted by pikachu975
I looked at it quickly and you can probably use the tan special angle formula with tan(a+b) = [tana+tanb]/(1-tanatanb). Can't see a faster method for now.
That's correct. You use the formula repeatedly.

$tan(LHS) = \frac {tan(arctan{\frac{1}{3}}+arctan{\frac {1}{5}}) + tan(arctan{\frac{1}{7}} + arctan{\frac{1}{8}})}{1 - tan(arctan {\frac {1}{3}} + arctan {\frac{1}{5}})tan(arctan{\frac {1}{7}} + arctan {\frac {1}{8}})} \\ \\ = \frac {\frac {4}{7} + \frac {3} {11}}{1 - \frac {4}{7}\times \frac {3}{11}} = \frac {44+21}{65} = 1 \\ \\ \\ Note: tan(arctan \frac{1}{3} + arctan \frac{1}{5}) = \frac {\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{3} \times \frac {1}{5}} = \frac {4}{7}$

etc etc

I have used arctan instead of tan-1 to save tedious LaTeX input.

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