solve for x:\tan ^{-1}\left ( 2x \right )+\tan ^{-1}\left ( 3x \right )= \frac{\pi }{4}.

 

 

 

 
 
 
 
 

Answers (1)

 
given \ \ \tan ^{-1}2x+\tan ^{-1}3x= \frac{\pi }{4}
tan^{-1}\left [ \frac{2x+3x}{1-(2x)(3x)} \right ]= \frac{\pi }{4}
\therefore \tan ^{-1}x+\tan ^{-1}y= tan^{-1}\left ( \frac{x+y}{1-xy} \right ) , xy<1
\Rightarrow tan^{-1}\left ( \frac{5x}{1-6x^{2}} \right )= \frac{\pi }{4}
\Rightarrow \frac{5x}{1-6x^{2}}= \tan \frac{\pi }{4}
\Rightarrow \frac{5x}{1-6x^{2}}= 1
\Rightarrow 5x= 1-6x^{2}
\Rightarrow 6x^{2}+5x-1= 0
\Rightarrow \left ( 6x-1 \right )\left ( x+1 \right )= 0
\Rightarrow x= \frac{1}{6} \ \ or \ \ x= -1 \ \ \ \ \because \text{2x . 3x}<1 
\therefore x= \frac{1}{6}

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