Understanding the Inverse Trigonometric Problem

The question asks us to find the value of a composite function involving inverse trigonometric functions. Specifically, we need to evaluate \(\tan^{-1}(\cot(\operatorname{cosec}^{-1} 2))\). To solve this, we will work from the innermost function outwards.

Step 1: Evaluate the Innermost Function - \(\operatorname{cosec}^{-1} 2\)

Let \(\theta = \operatorname{cosec}^{-1} 2\). By definition, this means \(\operatorname{cosec} \theta = 2\).

We know that \(\operatorname{cosec} \theta = \frac{1}{\sin \theta}\). So, \(\frac{1}{\sin \theta} = 2\), which implies \(\sin \theta = \frac{1}{2}\).

For the principal value branch of \(\operatorname{cosec}^{-1} x\), the angle \(\theta\) lies in the interval \([-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]\).

We need to find an angle \(\theta\) in this interval such that \(\sin \theta = \frac{1}{2}\). The angle that satisfies this is \(\theta = \frac{\pi}{6}\).

Therefore, \(\operatorname{cosec}^{-1} 2 = \frac{\pi}{6}\).

Step 2: Evaluate the Next Function - \(\cot(\operatorname{cosec}^{-1} 2)\)

Now we need to find the value of \(\cot\) of the angle we found in Step 1.

We need to calculate \(\cot(\frac{\pi}{6})\).

Recall the definition of \(\cot \theta = \frac{\cos \theta}{\sin \theta}\).

We know that \(\sin(\frac{\pi}{6}) = \frac{1}{2}\) and \(\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\).

So, \(\cot(\frac{\pi}{6}) = \frac{\cos(\frac{\pi}{6})}{\sin(\frac{\pi}{6})} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}\).

Thus, \(\cot(\operatorname{cosec}^{-1} 2) = \sqrt{3}\).

Step 3: Evaluate the Outermost Function - \(\tan^{-1}(\sqrt{3})\)

Finally, we need to find the value of \(\tan^{-1}\) of the result from Step 2.

We need to calculate \(\tan^{-1}(\sqrt{3})\).

Let \(\phi = \tan^{-1}(\sqrt{3})\). By definition, this means \(\tan \phi = \sqrt{3}\).

For the principal value branch of \(\tan^{-1} x\), the angle \(\phi\) lies in the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\).

We need to find an angle \(\phi\) in this interval such that \(\tan \phi = \sqrt{3}\). The angle that satisfies this is \(\phi = \frac{\pi}{3}\).

Therefore, \(\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}\).

Combining the Steps

Putting it all together:
\(\tan^{-1}(\cot(\operatorname{cosec}^{-1} 2))\)
First, \(\operatorname{cosec}^{-1} 2 = \frac{\pi}{6}\).
So, we have \(\tan^{-1}(\cot(\frac{\pi}{6}))\).
Next, \(\cot(\frac{\pi}{6}) = \sqrt{3}\).
So, we have \(\tan^{-1}(\sqrt{3})\).
Finally, \(\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}\).

The value of \(\tan^{-1}(\cot(\operatorname{cosec}^{-1} 2))\) is \(\frac{\pi}{3}\).

Comparison with Options

Let's look at the given options:

• Option 1: \(\frac{\pi}{8}\)
• Option 2: \(\frac{\pi}{6}\)
• Option 3: \(\frac{\pi}{4}\)
• Option 4: \(\frac{\pi}{3}\)

Our calculated value is \(\frac{\pi}{3}\), which matches Option 4.

Revision Table: Key Trigonometric Values

Additional Information on Inverse Trigonometric Functions

Inverse trigonometric functions, also known as arc functions, are the inverse functions of the basic trigonometric functions. They are used to find the angle when the value of the trigonometric ratio is given.

For example, if \(\sin \theta = x\), then \(\theta = \sin^{-1} x\) (read as "theta is the angle whose sine is x").

Since trigonometric functions are periodic, they are not one-to-one over their entire domain. To define inverse functions, we restrict the domain of each trigonometric function to an interval where it is one-to-one. This restricted domain is called the principal value branch.

• \(\sin^{-1} x\): Domain [-1, 1], Range \([-\frac{\pi}{2}, \frac{\pi}{2}]\)
• \(\cos^{-1} x\): Domain [-1, 1], Range \([0, \pi]\)
• \(\tan^{-1} x\): Domain \((-\infty, \infty)\), Range \((-\frac{\pi}{2}, \frac{\pi}{2})\)
• \(\operatorname{cosec}^{-1} x\): Domain \((-\infty, -1] \cup [1, \infty)\), Range \([-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]\)
• \(\sec^{-1} x\): Domain \((-\infty, -1] \cup [1, \infty)\), Range \([0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]\)
• \(\cot^{-1} x\): Domain \((-\infty, \infty)\), Range \((0, \pi)\)

Understanding the principal value branches is crucial for correctly evaluating inverse trigonometric expressions. In this problem, we used the principal values for \(\operatorname{cosec}^{-1} 2\) and \(\tan^{-1} \sqrt{3}\).