Question

Solve the equation: $cot^{-1}(\frac{1 + x}{1 - x}) = \frac{1}{2}cot^{-1}(\frac{1}{x}),\ x \ \textgreater \ 0\ and \ x\neq1$.

Answer 1

Answer:

x = 1/√3

x = -1/√3

Step-by-step explanation:

Solve the equation:  

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

here as we know that

$cot^{-1}x=tan^{-1}(\frac{1}{x})$

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

as

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

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

$=tan^{-1}\frac{\frac{2(1-x)}{(1+x)} }{\frac{(1+x^{2}+2x-1-x^{2}+2x)}{(1+x)^{2}} } \\\\\\=tan^{-1}\frac{2(1-x)(1+x)}{4x} \\\\=tan^{-1}\frac{2(1-x^{2})}{2x}$

taking cot both sides

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

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

x = 1/√3

x = -1/√3