Question

tan^-1(2x) + tan^-1(3x) = π/4
find x...
pls help

Answer 1

\[tan^{-1} \frac{5x}{1 - 6x^2} = \frac{\pi}{4}\]

Answer 2

should we use tan(A+B) ?

Answer 3

1 - 6x^2 = 5x

6x^2 + 5x -1 =0

(6x - 1)(x + 1) =0

x = -1 , 1/6

Answer 4

yea we should

Answer 5

thanks a lot!! :)
can u pls tell how u got 1-6x^2 =5x

Answer 6

\[tan^{-1} a + tan^{-1} b = tan^{-1} \frac{ a + b}{1 - ab}\]

the real thing is down here ...i wrote the above by mistake

\[tan^{-1} x = y\]

\[x = tany\]

Answer 7

okie...tnx.. :)