If A=1x(x^2+x+1), B=xx^2+x+1 and C=(x^-3+x^-2+x^-1)^1/2,0

Question

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

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

Accepted Answer

To find the sum of two angles in terms of tangent, we can use the tangent addition formula:

$	an (A+B)=\frac{	an A+	an B}{1-	an A\cdot 	an B}$

Let's compute $	an (A+B)$ using the given $	an A$ and $	an B$:

$	an A=\frac{1}{\sqrt{x({x}^{2}+x+1)}}$

$	an B=\frac{\sqrt{x}}{\sqrt{{x}^{2}+x+1}}$

Now, we can apply the addition formula:

$	an (A+B)=\frac{\frac{1}{\sqrt{x({x}^{2}+x+1)}}+\frac{\sqrt{x}}{\sqrt{{x}^{2}+x+1}}}{1-\frac{1}{\sqrt{x({x}^{2}+x+1)}}\cdot \frac{\sqrt{x}}{\sqrt{{x}^{2}+x+1}}}$

$	an (A+B)=\frac{\frac{1+\sqrt{{x}^{2}}}{\sqrt{x({x}^{2}+x+1)}}}{1-\frac{1}{({x}^{2}+x+1)}}$

$	an (A+B)=\frac{\frac{\sqrt{{x}^{2}}+1}{\sqrt{x({x}^{2}+x+1)}}}{\frac{({x}^{2}+x+1)-1}{({x}^{2}+x+1)}}$

$	an (A+B)=\frac{(x+1)\sqrt{({x}^{2}+x+1)}}{({x}^{2}+x)\sqrt{x}}$

$	an (A+B)=\frac{(x+1)\sqrt{({x}^{2}+x+1)}}{x(x+1)\sqrt{x}}$

$	an (A+B)=\frac{\sqrt{{x}^{2}+x+1}}{{x}^{3/2}}$

Let us first simplify $	an C$:

$	an C=\sqrt{{x}^{-3}+{x}^{-2}+{x}^{-1}}=\sqrt{\frac{1}{{x}^{3}}+\frac{1}{{x}^{2}}+\frac{1}{x}}=\sqrt{\frac{1+x+{x}^{2}}{{x}^{3}}}=\frac{\sqrt{{x}^{2}+x+1}}{{x}^{3/2}}.$

We see that:

$	an (A+B)=	an C$

Since tangent is positive and all the angles $A$, $B$, and $C$ are in the first quadrant, we can say that:

$A+B=C$

Hence, the answer is:

Option A : $C$

Solutions · 1