Question

The value of \(\tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3})\) is:

• A. \(\pi\)
• B. \(\dfrac{-\pi}{2}\)
• C. \(0\)
• D. \(2\sqrt{3}\)

Hint:

Use tan⁻¹(√3)=π/3 and cot⁻¹(-x)=π-cot⁻¹(x) with cot⁻¹(√3)=π/6.

Answer 1

The Correct Option is B

Step 1: Convert everything to sine/cosine mentally:
$\tan^{-1}(\sqrt3)$ asks: at what angle in $(-\pi/2,\pi/2)$ does tangent equal $\sqrt3$? That is $60^\circ = \pi/3$.

Step 2: Handle the negative argument of cot inverse separately:
$\cot^{-1}$ is defined on $(0,\pi)$, where cotangent is positive on $(0,\pi/2)$ and negative on $(\pi/2,\pi)$. Since we want cot to equal $-\sqrt3$, the angle must lie in the second quadrant part, at $180^\circ - 30^\circ = 150^\circ = 5\pi/6$ (because $\cot(30^\circ)=\sqrt3$).

Step 3: Subtract in degrees first, then convert:
$60^\circ - 150^\circ = -90^\circ = -\pi/2$.

Final Answer:
The value equals $-\pi/2$. \[ \boxed{-\dfrac{\pi}{2}} \]