Question

If ${\tan }^{-1}\frac{1}{a-1}={\tan }^{-1}\frac{1}{x}+{\tan }^{-1}\frac{1}{a^2-x+1},thenxis$

• A. $a$
• B. $a^3$
• C. $a^2-a+1$
• D. $a^2+a-1$

Answer 1

The correct option is C $a^2-a+1$

Given,

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

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

or, ${\tan }^{-1}\frac{1}{a-1}={\tan }^{-1}\frac{a^2+1}{a^{2}x-x^2+x-1}$

or, $(a-1)(a^2+1)=a^{2}x-x^2+x-1$

or, $a^3+a-a^2-1=a^{2}x-x^2+x-1$

or, $a^2(a-x)-(a^2-x^2)+(a-x)=0$

or, $(a-x)(a^2-a-x+1)=0$

or, $x=a,a^2-a+1$.