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# Prove that $$tan^{-1} \bigg(\frac{\large 1}{\large 4} \bigg) + tan^{-1} \bigg( \frac{\large 2}{\large 9} \bigg ) = \frac{\large 1}{\large 2} cos^{-1} \bigg( \frac{3}{5} \bigg).$$

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• $$tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}\:\:\:xy<1$$
• $$tan^{-1}x=cos^{-1}\large\frac{1}{\sqrt{1+x^2}}$$
• $$2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2}$$
L.H.S.=
We shall rewrie the question as prove $$2.(tan^{-1}\large\frac{1}{4}+tan^{-1}\large\frac{2}{9}=cos^{-1}\large\frac{3}{5}$$

By taking $$x=\large\frac{1}{4}\:and\:y=\large\frac{2}{9}\:we\:get$$
$$\large\frac{x+y}{1-xy}=\large\frac{\large\frac{1}{4}+\large\frac{2}{9}}{1-\large\frac{1}{4}.\large\frac{2}{9}}=\large\frac{17}{36}.\large\frac{36}{34}=\large\frac{17}{34}=\large\frac{1}{2}$$

Substituting in the above formula we get L.H.S.=
$$tan^{-1}\large\frac{1}{4}+tan^{-1}\large\frac{2}{9}=tan^{-1}\large\frac{1}{2}$$
By taking $$x=\large\frac{1}{2},\:\large\frac{2x}{1-x^2}=\frac{2\times\frac{1}{2}}{1-\large\frac{1}{4}}=\large\frac{4}{3}$$

Substituting in the above formula of $$2tan^{-1}x$$ we get
$$2tan^{-1}\large\frac{1}{2}= tan^{-1}\large\frac{4}{3}$$
By taking $$x=\large\frac{4}{3}\:we\:get\:\large\frac{1}{\sqrt{1+x^2}}=\large\frac{1}{\sqrt{1+\large\frac{16}{9}}}=\large\frac{3}{5}$$

By substituting in the formula of $$tan{-1}x=cos^{-1}\large\frac{1}{\sqrt{1+x^2}}$$ we get
$$tan{-1}\large\frac{4}{3}=cos^{-1}\large\frac{3}{5}$$
Then L.H.S. becomes
$$\Rightarrow\:2tan^{-1}\large\frac{1}{2}=tan^{-1}\large\frac{4}{3}=cos^{-1}\large\frac{3}{5}$$
=R.H.S.