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# If $\tan^{-1}\large\frac{1}{a-1}$$=\tan^{-1}x+\tan^{-1}(x+1)=\tan^{-1}3x is \begin{array}{1 1}(a)\;\large\frac{a}{2}&(b)\;a^3\\(c)\;a^2-a+1&(d)\;a^2+a-1\end{array} Can you answer this question? ## 1 Answer 0 votes We have \tan^{-1}\large\frac{1}{a-1}$$=\tan^{-1}\large\frac{1}{x}$$+\tan^{-1}\large\frac{1}{a^2-x+1} \Rightarrow \tan^{-1}\large\frac{1}{a-1}$$=\tan^{-1}\large\frac{1/x+1/a^2-x+1}{1-1/x(a^2-x+1)}$
$\Rightarrow \large\frac{1}{a-1}=\frac{a^2+1}{(a^2+1)(x-x^2-1)}$
$x^2-x(a^2+1)+1+(a-1)(a^2+1)=0$
$x^2-a^2-(a^2+1)(x-a)=0$
$(x-a)(x+a-a^2-1)=0$
$x=a$
Or $x=a^2-a+1$
Hence (c) is the correct option.