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Home  >>  CBSE XII  >>  Math  >>  Model Papers

Solve for x : \( \tan^{-1}3x+\tan^{-1}2x=\large\frac{\pi}{4} \)

1 Answer

Toolbox:
  • $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\bigg[\large\frac{x+y}{1-xy}\bigg]$
Step 1:
$\tan^{-1}3x+\tan^{-1}2x=\large\frac{\pi}{4}$
$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\bigg[\large\frac{x+y}{1-xy}\bigg]$
$\Rightarrow \tan^{-1}\bigg[\large\frac{3x+2x}{1-3x\times 2x}\bigg]=\large\frac{\pi}{4}$
$\Rightarrow \bigg[\large\frac{3x+2x}{1-3x\times 2x}\bigg]$$=\tan(\large\frac{\pi}{4})$
But $\tan(\large\frac{\pi}{4})$$=1$
$\large\frac{5x}{1-6x^2}$$=1$
$\Rightarrow 5x=1-6x^2$
$\therefore 6x^2+5x-1=0$
Step 2:
On factorizing we get,
$6x^2+6x-x-1=0$
$6x(x+1)-1(x+1)=0$
$\therefore (6x-1)(x+1)=0$
$6x-1=0$
$6x=1$
$x=\large\frac{1}{6}$
$x+1=0$
$x=-1$
answered Sep 23, 2013 by sreemathi.v
 
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