tan⁻¹3–√+sec⁻¹2–cos⁻¹1 is equal to

The correct option is 2π/3.

To find the value of the expression ${\tan }^{\mathrm{-1}}\sqrt{3}+{\sec }^{\mathrm{-1}}2-{\cos }^{\mathrm{-1}}1$, we can evaluate each term individually using the principal value branches of the inverse trigonometric functions.

First, let us find the value of the first term: ${\tan }^{\mathrm{-1}}\sqrt{3}$.
We know that the principal value branch of ${\tan }^{\mathrm{-1}}$ is $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.
Since $\tan \left(\frac{\pi }{3}\right)=\sqrt{3}$, we have:
${\tan }^{\mathrm{-1}}\sqrt{3}=\frac{\pi }{3}$

Second, let us find the value of the second term: ${\sec }^{\mathrm{-1}}2$.
We know that the principal value branch of ${\sec }^{\mathrm{-1}}$ is $\left[0,\pi \right]\backslash \left\{\frac{\pi }{2}\right\}$.
Since $\sec \left(\frac{\pi }{3}\right)=2$, we have:
${\sec }^{\mathrm{-1}}2=\frac{\pi }{3}$

Third, let us find the value of the third term: ${\cos }^{\mathrm{-1}}1$.
We know that the principal value branch of ${\cos }^{\mathrm{-1}}$ is $\left[0,\pi \right]$.
Since $\cos \left(0\right)=1$, we have:
${\cos }^{\mathrm{-1}}1=0$

Now, substitute these principal values back into the original expression:
${\tan }^{\mathrm{-1}}\sqrt{3}+{\sec }^{\mathrm{-1}}2-{\cos }^{\mathrm{-1}}1=\frac{\pi }{3}+\frac{\pi }{3}-0$

Adding the fractions, we get:
$\frac{\pi }{3}+\frac{\pi }{3}=\frac{2\pi }{3}$

Therefore, the expression is equal to 2π/3.