Home  >>  CBSE XII  >>  Math

# Prove that $4\tan^{-1}\large\frac{1}{5}-\tan^{-1}\frac{1}{70}+\tan^{-1}\frac{1}{99}=\large\frac{\pi}{4}$

Toolbox:
• $$2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2}$$, |x|<1
• $$tan^{-1}x-tan^{-1}y=tan^{-1}\large\frac{x-y}{1+xy}$$ , xy>-1
• $$tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy},\:\:\:xy<1$$
L.H.S.
By taking $$x=\large\frac{1}{5},\:\large\frac{2x}{1-x^2}=\large\frac{2\frac{1}{5}}{1-\large\frac{1}{25}}=\large\frac{2}{5}.\large\frac{25}{24}=\large\frac{5}{12}$$

By substituting $$x=\large\frac{1}{5}$$ in the above formula of $$2tan^{-1}x$$ we get
$$\Rightarrow\: 2tan^{-1}\large\frac{1}{5}=tan^{-1}\large\frac{5}{12}$$
By taking $$x=\large\frac{5}{12},\frac{2x}{1-x^2}=\large\frac{2\frac{5}{12}}{1-\large\frac{25}{144}}=\large\frac{10}{12}.\large\frac{144}{119}=\large\frac{120}{119}$$

Apply once again the formula of $$2tan^{-1}x$$ by taking $$x=\large\frac{5}{12}$$ we get
$$\Rightarrow\: 4tan^{-1}\large\frac{1}{5}=2tan^{-1}\large\frac{5}{12}=tan^{-1}\large\frac{120}{119}$$
By taking $$x=\large\frac{1}{99}\:and\:y=\large\frac{1}{70},$$
$$\large\frac{x-y}{1+xy}=\large\frac{\large\frac{1}{99}-\large\frac{1}{70}}{1+\large\frac{1}{99}.\large\frac{1}{70}}=-\large\frac{29}{6930}.\large\frac{6930}{6931}=-\large\frac{1}{239}$$
$$tan^{-1}\large\frac{1}{99}-tan^{-1}\large\frac{1}{70}=-tan^{-1}\large\frac{1}{239}$$

Substituting the values L.H.S. becomes,
$$\Rightarrow\: 4tan^{-1}\large\frac{1}{5}+(tan^{-1}\large\frac{1}{99}-tan^{-1}\large\frac{1}{70})=$$
$$tan^{-1}\large\frac{120}{119}-tan^{-1}\large\frac{1}{239}$$

Apply $$tan^{-1}x-tan^{-1}y$$ formula by taking $$x=\large\frac{120}{119}\:and\:y=\large\frac{1}{239}$$
$$tan^{-1}\large\frac{120}{119}-tan^{-1}\large\frac{1}{239} =tan^{-1}\large\frac{\large\frac{120}{119}-\large\frac{1}{239}}{1-\large\large\frac{120}{119}.\large\frac{1}{239}}$$
$$=tan^{-1}1=\large\frac{\pi}{4}$$
=R.H.S.

edited Mar 19, 2013