Solve: \tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{4}

 

 

 

 
 
 
 
 

Answers (1)

\tan^{-1} 4x + \tan^{-1}6x = \frac{\pi}{4}

\tan^{-1}\left(\frac{4x + 6x}{1 - 4x\cdot6x} \right ) = \frac{\pi}{4}

\frac{10x}{1 - 24x^2} = \tan\frac{\pi}{4}

\frac{10x}{1 - 24x^2} = 1

{10x} = {1 - 24x^2}

24x^2 + 10x -1 = 0

x = \frac{-10\pm \sqrt{100 + 96}}{48}

x = \frac{-10\pm 14}{48}

\therefore x = \frac{-1}{2}, \frac{1}{12}

Range of  \tan^{-1} \theta is (-\frac{\pi}{2},\frac{\pi}{2}) 

So if we take x=\frac{-1}{2} then 4x={-2} \ \& \ 6x=-3 won't be in the range  (-\frac{\pi}{2},\frac{\pi}{2})

 \therefore x = \frac{1}{12} is only answer

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