Question

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

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

Answer 1

Answer: (A)

$a$

$ta{n}^{-}\left(\frac{1}{a-1}\right)-ta{n}^{-}\left(\frac{1}{x}\right)=ta{n}^{-}\left(\frac{1}{a^2-x+1}\right)ta{n}^{-}\left(\frac{\left(\frac{1}{a-1}\right)-\left(\frac{1}{x}\right)}{1+\left(\frac{1}{a-1}\right)\cdot \left(\frac{1}{x}\right)}\right)=ta{n}^{-}\left(\right(\frac{1}{a^2-x+1}\left)\right)\left(\frac{x-(a-1)}{(a-1)x}\right)\cdot \left(\frac{(a-1)x}{(a-1)x+1}\right)=\left(\frac{1}{a^2-x+1}\right)\left(\frac{x-a+1}{(a-1)x+1}\right)=\left(\frac{1}{a^2-x+1}\right)(x-a+1)(a^2-x+1)=(a-1)x+1(a^2+1)x-x^2+(1-a)(a^2+1)-(1-a)x=(a-1)x+1(a^2+1)x-x^2+(a^2+1)-a^3-a=1x^2-(1+a^2)x+(a^3-a^2+a)=0clearlyx=asatisfiesthegivenequation$