• # question_answer The number of real solutions of the equation, ${{\cot }^{-1}}(x-1)+{{\cot }^{-1}}(6-x)={{\cot }^{-1}}(x-2)$ is. A) 0 B) 1 C) 2 D) infinite

 $\because$${{\cot }^{-1}}(x-1)+co{{t}^{-1}}(6-x)=co{{t}^{-1}}(x-2)$ take cot of both sides, we get $\Rightarrow$$\left[ \frac{(x-1)\,\,(6-x)-1}{6-x+x-1} \right]=(x-2)$$\Rightarrow$$-\frac{{{x}^{2}}+7x-7}{5}=x-2$$\Rightarrow$            ${{x}^{2}}-2x-3=0$   $\Rightarrow$$(x-3)\,\,(x+1)=0$$\Rightarrow$   $x=3$;$x=-1.$ both satisfy the given equation.