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# What does $\tan^{-1}{\frac{1}{4}}+\tan^{-1}{\frac{2}{9}}$ reduce to?

$\begin{array}{1 1} sin^{-1}\frac{1}{\sqrt5} \\ sin^{-1}\frac{2}{\sqrt5} \\ cos^{-1}\frac{1}{\sqrt5} \\ tan^{-1}\frac{1}{\sqrt5} \end{array}$

Can you answer this question?

Toolbox:
• $tan^{-1}x+tan^{-1}y=tan^{-1}\frac{x+y}{1-xy}\:\:\:xy<1$
• $tan^{-1}x=sin^{-1}\frac{x}{\sqrt{1+x^2}}$
By taking $x=\frac{1}{4}\:and\:y=\frac{2}{9}\:we\:get$
$\frac{x+y}{1-xy}=\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}.\frac{2}{9}}=\frac{17}{36}.\frac{36}{34}=\frac{17}{34}=\frac{1}{2}$
Substituting in the above formula we get L.H.S.=
$tan^{-1}\frac{1}{4}+tan^{-1}\frac{2}{9}=tan^{-1}\frac{1}{2}$
By taking $x=\frac{1}{2},\:\frac{x}{\sqrt{1+x^2}}=\frac{\frac{1}{2}}{\sqrt{1+\frac{1}{4}}}=\frac{1}{2}.\frac{2}{\sqrt{5}}=\frac{1}{\sqrt5}$
Substituting in the above formula of $tan^{-1}x$ we get
$tan^{-1}\frac{1}{2}= sin^{-1}\frac{1}{\sqrt 5}$
answered Feb 19, 2013
edited Jul 9, 2014