Question

If $ta{n}^{-1}2x+ta{n}^{-1}3x+\frac{11}{4}$, then find the value of ,x,.

Answer 1

${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{-11}{4},[{\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)]$

${\tan }^{-1}\left[\frac{2x+2x}{1-6x^2}\right]=\frac{-11}{4}$

${\tan }^{-1}\left(\frac{5x}{1-6x^2}\right)=\frac{-11}{4}$

$\frac{5x}{1-6x^2}={\tan }^{-1}\left(\frac{-11}{4}\right)$

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