# 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${}{}{}$ – Ali Caglayan Nov 25 '13 at 17:35

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ \begin{align} &\tan\pars{\overbrace{\cos^{-1}\left(\frac{3}{5}\right)}^{\ds{\equiv\ x}}} = \tan\pars{\tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{1}{9}\right)} = {1/4 + 1/9 \over 1 - \pars{1/4}\pars{1/9}} = {13 \over 35} \end{align} $$\tan\pars{x} = {\root{1 - \cos^{2}\pars{x}} \over \cos\pars{x}} = {\root{1 - \pars{3/5}^{2}} \over 3/5} = {4/5 \over 3/5} = {4 \over 3} \color{#0000ff}{\Huge\not=} {13 \over 35}$$

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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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