Question

i) Solve for $x:{\tan }^{-1}(x-1)+{\tan }^{-1}x+{\tan }^{-1}(x+1)={\tan }^{-1}3x$.

ii) Prove that ${\tan }^{-1}\left(\frac{6x-8x^3}{1-12x^2}\right)-{\tan }^{-1}\left(\frac{4x}{1-4x^2}\right)={\tan }^{-1}2x;\left|2x\right|<\frac{1}{\sqrt{3}}$

Answer 1

(i) ${\tan }^{-1}(x-1)+{\tan }^{-1}x+{\tan }^{-1}(x+1)={\tan }^{-1}3x$

$\Rightarrow {\tan }^{-1}(x-1)+{\tan }^{-1}(x+1)={\tan }^{-1}3x-{\tan }^{-1}x$

$\Rightarrow {\tan }^{-1}\left(\frac{x-1+x+1}{1-(x-1)(x+1)}\right)={\tan }^{-1}\left(\frac{3x-x}{1+3x^2}\right)$

$\Rightarrow \frac{2x}{1-(x^2-1)}=\frac{2x}{1+3x^2}$

$\Rightarrow x(1+3x^2)=x(2-x^2)$

$\Rightarrow x(1+3x^2-2+x^2)=0$

$\Rightarrow x(4x^2-1)=0$

$\Rightarrow x=0;4x^2-1=0$

$\Rightarrow x^2=\frac{1}{4}$

$\Rightarrow x=\pm \frac{1}{2}$

$\Rightarrow x=0,\pm \frac{1}{2}$

(ii) Take LHS,

${\tan }^{-1}\left(\frac{6x-8x^3}{1-12x^2}\right)-{\tan }^{-1}\left(\frac{4x}{1-4x^2}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{\frac{6x-8x^3}{1-12x^2}-\frac{4x}{1-4x^2}}{1+\frac{6x-8x^3}{1-12x^2}\times \frac{4x}{1-4x^2}}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{(6x-8x^3)(1-4x^2)-4x(1-12x^2)}{(1-12x^2)(1-4x^2)+(6x-8x^3)4x}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{6x-24x^3-8x^3+32x^5-4x+48x^3}{1-4x^2-12x^2+48x^4+24x^2-32x^4}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{32x^5+16x^3+2x}{16x^4+8x^2+1}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{2x(16x^4+8x+1)}{16x^4+8x+1}\right)$

$\Rightarrow {\tan }^{-1}2x=$ RHS

Hence proved.