If $ an A=\frac{1}{\sqrt{x\left(x^{2}+x+1 ight)}}, an B=\frac{\sqrt{x}}{\sqrt{x^{2}+x+1}}$ and $

Question

If $	an A=\frac{1}{\sqrt{x\left(x^{2}+x+1ight)}}, 	an B=\frac{\sqrt{x}}{\sqrt{x^{2}+x+1}}$ and $	an C=\left(x^{-3}+x^{-2}+x^{-1}ight)^{1 / 2}, 0<A, B, C<\pi / 2$, then $A+B$ is equal to :\

Professor PC Thomas Classes · Answer

(c) : As $	an (A+B)=\frac{	an A+	an B}{1-	an A 	an B}$ $$ \begin{aligned} & =\frac{\frac{1}{\sqrt{x\left(x^{2}+x+1ight)}}+\frac{\sqrt{x}}{\sqrt{x^{2}+x+1}}}{1-\frac{1}{x^{2}+x+1}} \ & =\frac{1+x}{\sqrt{x\left(x^{2}+x+1ight)}} 	imes \frac{x^{2}+x+1}{x^{2}+x} \ & =\frac{1+x}{\sqrt{x}} 	imes \frac{\sqrt{x^{2}+x+1}}{x(1+x)} \ & =\sqrt{\frac{x^{2}+x+1}{x^{3}}}=\sqrt{x^{-1}+x^{-2}+x^{-3}}=	an C \ & \Rightarrow A+B=C \end{aligned} $$