QJEEM24Ma076d: JEE Main 2024 Mathematics - Class 11 JEEM24Ma076d Solution: If and , then is equal to :

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

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

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

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

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

Now, we can apply the addition formula:

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

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

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

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

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

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

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

We see that:

$\tan (A + B) = \tan C$

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 : $C$