JEE Mathematics 2024 Question with Solution

Given:

• $\tan A = \frac{1}{\sqrt{x(x^2+x+1)}}$
• $\tan B = \frac{\sqrt{x}}{\sqrt{x^2+x+1}}$
• $0 < A, B, C < \frac{\pi}{2}$

Find: $A+B$

Use the tangent addition formula:

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

First simplify the numerator:

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

Now simplify the denominator:

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

So,

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

Therefore,

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

Cancelling $x+1$,

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

That is,

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

From the solution working, this is compared with $\tan C$ to get

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

Using

$$\cot C=\tan\left(\frac{\pi}{2}-C\right)$$

and since the angles are acute, we obtain

$$A+B=\frac{\pi}{2}-C$$

Therefore, the correct option is D.

Note: The answer key marks option A, but the extracted solution concludes $A+B=\frac{\pi}{2}-C$. Hence the solution has been followed.

The extracted solution itself indicates a mismatch in the printed expression for $\tan C$ and discusses a likely typo. However, the algebra shown gives

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

If

$$\tan C=(x^{-1}+x^{-2}+x^{-3})^{-1/2}$$

then

$$\tan(A+B)=\frac{1}{\tan C}=\cot C$$

Hence,

$$A+B=\frac{\pi}{2}-C$$

for acute angles. So the defensible answer from the solution working is D.