Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Let $x=\tan \theta$, where $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$.

The equation transforms to ${\tan }^{-1}(\tan 2\theta )+{\tan }^{-1}(\tan 3\theta )=\frac{\pi }{2}$.

We need to analyze different ranges for $\theta$:

1. For $\theta \in (0,\frac{\pi }{6})$, we have $2\theta \in (0,\frac{\pi }{3})$ and $3\theta \in (0,\frac{\pi }{2})$.

So, ${\tan }^{-1}(\tan 2\theta )=2\theta$ and ${\tan }^{-1}(\tan 3\theta )=3\theta$.

The equation becomes $2\theta +3\theta =\frac{\pi }{2}⟹5\theta =\frac{\pi }{2}⟹\theta =\frac{\pi }{10}$.

Since $\frac{\pi }{10}\in (0,\frac{\pi }{6})$, $x=\tan (\frac{\pi }{10})$ is a valid solution.

2. For $\theta \in (\frac{\pi }{6},\frac{\pi }{4})$, we have $2\theta \in (\frac{\pi }{3},\frac{\pi }{2})$ and $3\theta \in (\frac{\pi }{2},\frac{3\pi }{4})$.

So, ${\tan }^{-1}(\tan 2\theta )=2\theta$ and ${\tan }^{-1}(\tan 3\theta )=3\theta -\pi$.

The equation becomes $2\theta +(3\theta -\pi )=\frac{\pi }{2}⟹5\theta =\frac{3\pi }{2}⟹\theta =\frac{3\pi }{10}$.

This value is not in $(\frac{\pi }{6},\frac{\pi }{4})$ (as $\frac{3\pi }{10}=54^{\circ }$, while the range is $30^{\circ }$ to $45^{\circ }$). No solution here.

3. For $\theta \in (\frac{\pi }{4},\frac{\pi }{2})$, we have $2\theta \in (\frac{\pi }{2},\pi )$ and $3\theta \in (\frac{3\pi }{4},\frac{3\pi }{2})$.

So, ${\tan }^{-1}(\tan 2\theta )=2\theta -\pi$ and ${\tan }^{-1}(\tan 3\theta )=3\theta -\pi$.

The equation becomes $(2\theta -\pi )+(3\theta -\pi )=\frac{\pi }{2}⟹5\theta =\frac{5\pi }{2}⟹\theta =\frac{\pi }{2}$.

This implies $x=\tan (\frac{\pi }{2})$, which is undefined. No solution here.

4. For $\theta \in (-\frac{\pi }{6},0)$, we have $2\theta \in (-\frac{\pi }{3},0)$ and $3\theta \in (-\frac{\pi }{2},0)$.

So, ${\tan }^{-1}(\tan 2\theta )=2\theta$ and ${\tan }^{-1}(\tan 3\theta )=3\theta$.

The equation becomes $2\theta +3\theta =\frac{\pi }{2}⟹5\theta =\frac{\pi }{2}⟹\theta =\frac{\pi }{10}$.

This value is not in $(-\frac{\pi }{6},0)$. No solution here.

5. For $\theta \in (-\frac{\pi }{4},--\frac{\pi }{6})$, we have $2\theta \in (-\frac{\pi }{2},-\frac{\pi }{3})$ and $3\theta \in (-\frac{3\pi }{4},-\frac{\pi }{2})$.

So, ${\tan }^{-1}(\tan 2\theta )=2\theta$ and ${\tan }^{-1}(\tan 3\theta )=3\theta +\pi$.

The equation becomes $2\theta +(3\theta +\pi )=\frac{\pi }{2}⟹5\theta =-\frac{\pi }{2}⟹\theta =-\frac{\pi }{10}$.

This value is not in $(-\frac{\pi }{4},-\frac{\pi }{6})$. No solution here.

6. For $\theta \in (-\frac{\pi }{3},-\frac{\pi }{4})$, we have $2\theta \in (-\frac{2\pi }{3},-\frac{\pi }{2})$ and $3\theta \in (-\pi ,-\frac{3\pi }{4})$.

So, ${\tan }^{-1}(\tan 2\theta )=2\theta +\pi$ and ${\tan }^{-1}(\tan 3\theta )=3\theta +\pi$.

The equation becomes $(2\theta +\pi )+(3\theta +\pi )=\frac{\pi }{2}⟹5\theta =-\frac{3\pi }{2}⟹\theta =-\frac{3\pi }{10}$.

Since $-\frac{3\pi }{10}=-54^{\circ }$, which is in $(-\frac{\pi }{3},-\frac{\pi }{4})=(-60^{\circ },-45^{\circ })$, $x=\tan (-\frac{3\pi }{10})=-\tan (\frac{3\pi }{10})$ is a valid solution.

7. For $\theta \in (-\frac{\pi }{2},-\frac{\pi }{3})$, we have $2\theta \in (-\pi ,-\frac{2\pi }{3})$ and $3\theta \in (-\frac{3\pi }{2},-\pi )$.

So, ${\tan }^{-1}(\tan 2\theta )=2\theta +\pi$ and ${\tan }^{-1}(\tan 3\theta )=3\theta +\pi$.

The equation becomes $(2\theta +\pi )+(3\theta +\pi )=\frac{\pi }{2}⟹5\theta =-\frac{3\pi }{2}⟹\theta =-\frac{3\pi }{10}$.

This value is not in $(-\frac{\pi }{2},-\frac{\pi }{3})$. No solution here.

Thus, the equation has exactly two real solutions: $x_1=\tan (\frac{\pi }{10})$ (positive) and $x_2=-\tan (\frac{3\pi }{10})$ (negative).

All options A, B, C, D are correct.