Question:
If \(\tan^{-1} x = 3y\), then
If \(\tan^{-1} x = 3y\), then
Show Hint
The principal value of tan inverse x lies strictly between minus pi by 2 and pi by 2. Divide that range by 3.
Updated On: Oct 22, 2026
- \(-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}\)
- \(-\dfrac{3\pi}{2} < y < \dfrac{3\pi}{2}\)
- \(-\dfrac{\pi}{6} < y < \dfrac{\pi}{6}\)
- \(-\dfrac{\pi}{6} \leq y \leq \dfrac{\pi}{6}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
The principal value of \(\tan^{-1} x\) lies strictly between \(-\dfrac{\pi}{2}\) and \(\dfrac{\pi}{2}\), for every real \(x\).
The end values are never reached, because \(\tan \theta\) is undefined at \(\pm \dfrac{\pi}{2}\).
Step 2: Write the range:
Since \(\tan^{-1} x = 3y\), we get:
\[ -\frac{\pi}{2} < 3y < \frac{\pi}{2} \]
Step 3: Divide by 3:
3 is positive, so the inequality signs do not change:
\[ -\frac{\pi}{6} < y < \frac{\pi}{6} \]
Step 4: Why the other options are wrong:
Option (A) forgets the factor 3 and uses the range of \(3y\) as the range of \(y\).
Option (B) multiplies the range by 3 instead of dividing by 3.
Option (D) uses closed end values. \(y = \pm \dfrac{\pi}{6}\) would need \(\tan^{-1} x = \pm \dfrac{\pi}{2}\), which no real \(x\) gives.
Final Answer:
The correct range is option (C). \[ \boxed{-\frac{\pi}{6} < y < \frac{\pi}{6}} \]
The principal value of \(\tan^{-1} x\) lies strictly between \(-\dfrac{\pi}{2}\) and \(\dfrac{\pi}{2}\), for every real \(x\).
The end values are never reached, because \(\tan \theta\) is undefined at \(\pm \dfrac{\pi}{2}\).
Step 2: Write the range:
Since \(\tan^{-1} x = 3y\), we get:
\[ -\frac{\pi}{2} < 3y < \frac{\pi}{2} \]
Step 3: Divide by 3:
3 is positive, so the inequality signs do not change:
\[ -\frac{\pi}{6} < y < \frac{\pi}{6} \]
Step 4: Why the other options are wrong:
Option (A) forgets the factor 3 and uses the range of \(3y\) as the range of \(y\).
Option (B) multiplies the range by 3 instead of dividing by 3.
Option (D) uses closed end values. \(y = \pm \dfrac{\pi}{6}\) would need \(\tan^{-1} x = \pm \dfrac{\pi}{2}\), which no real \(x\) gives.
Final Answer:
The correct range is option (C). \[ \boxed{-\frac{\pi}{6} < y < \frac{\pi}{6}} \]
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