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# Prove that $2sin^{-1}\frac{3}{5}-tan^{-1}\frac{17}{31}=\frac{\pi}{4}$

Toolbox:
• $sin^{-1}x=tan^{-1}\large\frac{x}{\sqrt{1-x^2}}$
• $2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2}\: if\: -1 \leq x \leq 1$
• $tan^{-1}x-tan^{-1}y=tan^{-1}\large\frac{x-y}{1+xy} xy > -1$
By taking $x=\frac{3}{5},\:\frac{x}{\sqrt{1-x^2}}=\frac{\frac{3}{5}}{\sqrt{1-\frac{9}{25}}}$
$=\large\frac{3}{5}\times\large\frac{5}{4}=\large\frac{3}{4}$
We know that $sin^{-1}x=tan^{-1}\frac{x}{\sqrt{1-x^2}}$
$\Rightarrow\:\large\:sin^{-1}\large\frac{3}{5}=tan^{-1}\large\frac{3}{4}$
$L.H.S= \: 2tan^{-1}\large\frac{3}{4}-tan^{-1}\large\frac{17}{31}$
We know that $2tan^{-1}x=tan^{-1}\frac{2x}{1-x^2}$
By taking $x=\frac{3}{4},\: \frac{2x}{1-x^2}=\frac{2.\frac{3}{4}}{1-\frac{9}{16}}=\frac{3}{2}\times\frac{16}{7}=\frac{24}{7}$
$\Rightarrow\:\large\:2tan^{-1}\large\frac{3}{4}=\large\:tan^{-1}\large\frac{24}{7}$
Substituting the value in L.H.S.
$\Rightarrow tan^{-1}\large\frac{24}{7}-tan^{-1}\large\frac{17}{31}$
By taking $x=\frac{24}{7}\:and\:y=\frac{17}{31}$
$\frac{x-y}{1+xy}=\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7}.\frac{17}{31}}=1$
$\Rightarrow tan^{-1}\large\frac{24}{7}-tan^{-1}\large\frac{17}{31} = tan^{-1}1=\large\frac{\pi}{4}$
=R.H.S.
edited Mar 20, 2013