Understanding the Problem: Trigonometric Value Calculation

The question asks us to find the value of a trigonometric expression involving inverse trigonometric functions. Specifically, we need to calculate \(\cot \left( {cose{c^{ - 1}}\frac{5}{3} + {{\tan }^{ - 1}}\frac{2}{3}\;} \right)\). To solve this, we will convert the inverse trigonometric functions into a more convenient form, usually \(\tan^{-1}\), and then use standard trigonometric identities.

Converting Inverse Cosec to Inverse Tan

We have the term \(cose{c^{ - 1}}\frac{5}{3}\). Let \({\theta = cose{c^{ - 1}}\frac{5}{3}}\). This means \({\text{cosec }\theta = \frac{5}{3}}\). In a right-angled triangle, cosec is defined as \(\frac{\text{hypotenuse}}{\text{opposite side}}\). So, we have a hypotenuse of 5 and an opposite side of 3. Using the Pythagorean theorem, the adjacent side is \(\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\). The tangent of this angle \(\theta\) is \(\frac{\text{opposite side}}{\text{adjacent side}} = \frac{3}{4}\). Therefore, \({\theta = \tan^{ - 1}}\frac{3}{4}\). So, \(cose{c^{ - 1}}\frac{5}{3} = \tan^{ - 1}\frac{3}{4}\). This conversion is a key step in this type of inverse trigonometry problem.

Applying the Tan Inverse Addition Formula

Now the expression becomes \(\cot \left( {\tan^{ - 1}\frac{3}{4} + {{\tan }^{ - 1}}\frac{2}{3}\;} \right)\). We can use the formula for the sum of two inverse tangents: \({\tan^{ - 1}x + \tan^{ - 1}y = \tan^{ - 1}\left(\frac{x+y}{1-xy}\right)}\), provided \(xy < 1\). Here, \(x = \frac{3}{4}\) and \(y = \frac{2}{3}\).

Let's calculate \(xy\): \({\frac{3}{4} \times \frac{2}{3} = \frac{6}{12} = \frac{1}{2}}\). Since \({\frac{1}{2} < 1}\), we can use the formula.

Now calculate the sum inside the \(\tan^{-1}\):

\[\frac{x+y}{1-xy} = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}} = \frac{\frac{3 \times 3 + 2 \times 4}{4 \times 3}}{1 - \frac{6}{12}} = \frac{\frac{9+8}{12}}{1 - \frac{1}{2}} = \frac{\frac{17}{12}}{\frac{1}{2}}\]

Dividing the fractions:

\[\frac{\frac{17}{12}}{\frac{1}{2}} = \frac{17}{12} \times \frac{2}{1} = \frac{17 \times 2}{12} = \frac{34}{12} = \frac{17}{6}\]

So, \({\tan^{ - 1}\frac{3}{4} + {{\tan }^{ - 1}}\frac{2}{3} = \tan^{ - 1}\frac{17}{6}}\). This step demonstrates the application of trigonometric identities.

Final Cot Value Calculation

The original expression is now reduced to \({\cot \left( \tan^{ - 1}\frac{17}{6} \right)}\). We use the identity \({\cot(\tan^{ - 1}z) = \frac{1}{z}}\). Here \(z = \frac{17}{6}\).

Therefore, \({\cot \left( \tan^{ - 1}\frac{17}{6} \right) = \frac{1}{\frac{17}{6}}}\).

Calculating the final value:

\[\frac{1}{\frac{17}{6}} = 1 \times \frac{6}{17} = \frac{6}{17}\]

Thus, the value of the expression is \(\frac{6}{17}\). This completes the trigonometric value calculation.

Summary of Steps

Let's recap the steps for this trigonometric value calculation:

• Convert \(cose{c^{ - 1}}\frac{5}{3}\) to an equivalent \(\tan^{ - 1}\) form using a right-angled triangle approach, finding \({\tan^{ - 1}\frac{3}{4}}\).
• Apply the sum of \(\tan^{ - 1}\) formula: \({\tan^{ - 1}x + \tan^{ - 1}y = \tan^{ - 1}\left(\frac{x+y}{1-xy}\right)}\) with \(x = \frac{3}{4}\) and \(y = \frac{2}{3}\) to get \({\tan^{ - 1}\frac{17}{6}}\). This uses trigonometric identities effectively.
• Finally, calculate the cotangent of the resulting inverse tangent using \({\cot(\tan^{ - 1}z) = \frac{1}{z}}\) to find the cot value.

The final answer obtained is \(\frac{6}{17}\).