Question

solve for x : cot-1x - cot-1(x+2) = pi/12 .

Answer 1

The value of x by solving this equation $\cot ^{-1 x}-\cot ^{-1(x+2)}=\frac{\pi}{12}$ is $\bold{\sqrt{3}}$.

As we know that,

$\cot ^{-1 x}=\tan x$.

$\cot ^{-1(x+2)}=\tan x+2$.

As we know that the value of pi ($\pi=180^{\circ}$),

Then,

$\frac{\pi}{12}=15^{\circ}$

Substitute the values in given equation,

$\tan ^{-1} \frac{1}{x}-\tan ^{-1} \frac{1}{x+2}=15$

$\frac{\tan ^{-1}\left(\left(\frac{1}{x}-\frac{1}{x+2}\right)\right)}{1+\frac{1}{x(x+2)}}=15$

$\tan ^{-1}\left(\frac{2}{x^{2}+2 x+1}\right)=15$

$\frac{2}{x^{2}+2 x+1}=\tan 15$

$\frac{2}{x^{2}+2 x+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

Take conjugate for ($\sqrt{3}+1$),

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

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

$(x+1)^{2}=(\sqrt{3}+1)^{2}$

Therefore, $\bold{x=\sqrt{3}}$