# How to prove Trigonometry equation?

how to solve following equation

$$\tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{1}{9}\right) = \cos^{-1}\left(\frac{3}{5}\right)$$ How to prove the above equation?

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Have you attempted anything? –  Zhoe Nov 25 '13 at 17:04
I was trying with tana+tanb formula but not able to convert it –  subodh joshi Nov 25 '13 at 17:05
One problem is of course that the equation is wrong. –  Daniel Fischer Nov 25 '13 at 17:10
There may be a mistake in the equation –  Dutta Nov 25 '13 at 17:10
*Disprove${}{}{}$ –  Alizter Nov 25 '13 at 17:35


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U mean above equation wrong? –  user110715 Nov 25 '13 at 17:28
@user110715 Yes. –  Felix Marin Nov 25 '13 at 17:28

From this or Ex$\#5$ of Page $\#276$ of this $$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$$ if $xy<1$

Now, as the principal value of $\tan$ lies $\in\left[-\frac\pi2,\frac\pi2\right],$

If $\displaystyle \tan^{-1}z=\theta,\tan\theta=z,$ $\displaystyle\cos\theta=+\frac1{\sqrt{1+z^2}}$

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how u made cosQ like that? –  user110715 Nov 25 '13 at 17:14
@user110715, $\sec^2\theta=1+\tan^2\theta,$ right? –  lab bhattacharjee Nov 25 '13 at 17:16

After you apply the formula, let $cos \theta= \frac{3}{5}$. Now convert this to $tan \theta$, which is trivial, and compare both sides.

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