Question

\(\cos \left(\tan ^{-1}\left(\sin \left(\cot ^{-1} x\right)\right)\right)\) is (a) \(\sqrt{\frac{x^{2}+1}{x^{2}+2}}\) (b) \(\sqrt{\frac{x^{2}+2}{x^{2}+3}}\) (c) \(\sqrt{\frac{x^{2}}{x^{2}+1}}\) (d) \(\sqrt{\frac{x^{2}-1}{x^{2}+2}}\)

Short answer

Answer: \(\sqrt{\frac{x^{2}+1}{x^{2}+2}}\)

Step-by-step solution

Find the equivalent tangent value

Since \(\cot(\theta) = x\), we can write it as \(\cot(\theta) = \frac{1}{\tan(\theta)}\). So \(\tan(\theta) = \frac{1}{x}\).

Determine \(\sin{\theta}\) by solving the right triangle

Since \(\tan(\theta) = \frac{1}{x}\), we can create a right triangle with the opposite side as 1 and the adjacent side as x. Using the Pythagorean theorem, the hypotenuse will be \(\sqrt{1^2 + x^2} = \sqrt{x^2+1}\). Therefore, \(\sin(\theta) = \frac{1}{\sqrt{x^2+1}}\).

Calculate \(\tan^{-1}(\sin(\cot^{-1}(x)))\)

According to step 2, we have \(\sin \left(\cot ^{-1} x\right) = \frac{1}{\sqrt{x^2+1}}\). Thus, \(\tan^{-1}\left(\sin(\cot^{-1}(x))\right) = \tan^{-1}\left(\frac{1}{\sqrt{x^2+1}}\right)\).

Calculate the cosine of the result of Step 3

Let \(\alpha = \tan^{-1}\left(\frac{1}{\sqrt{x^2+1}}\right)\). Then we have \(\tan(\alpha) = \frac{1}{\sqrt{x^2+1}}\). We can create a right triangle with the opposite side as 1 and the adjacent side as \(\sqrt{x^2+1}\). The hypotenuse can be calculated using the Pythagorean theorem: Hypotenuse\( = \sqrt{(1)^2 + (\sqrt{x^2+1})^2} = \sqrt{1 + x^2+1} = \sqrt{x^2+2}\). Finally, we find the cosine of angle \(\alpha\): \(\cos(\alpha)=\cos\left(\tan^{-1}\left(\frac{1}{\sqrt{x^2+1}}\right)\right) = \frac{\sqrt{x^2+1}}{\sqrt{x^2+2}}\). Hence, the correct option is (a) \(\sqrt{\frac{x^{2}+1}{x^{2}+2}}\).

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the arcsine, arccosine, arctangent, etc. They allow us to find angles when given a trigonometric ratio.

These functions are crucial for solving problems where you need to determine the angle based on a known function value. For example, if we know that \( \sin(\theta) = \frac{1}{2} \), we can find \( \theta \) by calculating \( \sin^{-1}(\frac{1}{2}) \).

In this exercise, we're finding \( \tan^{-1} \left( \sin(\cot^{-1}(x)) \right) \). This process involves reversing the trigonometric functions to find angles, which helps us compute more complex expressions.

• The key here is to understand that an inverse function "undoes" the original function, converting a ratio back into an angle.
• They help in real-world applications like solving navigation and engineering problems.

Pythagorean Theorem

The Pythagorean theorem is fundamental in connecting the sides of a right triangle. It states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: \( a^2 + b^2 = c^2 \).

This theorem is not only a cornerstone of geometry but essential for trigonometry. In our solved problem, it helps construct a right triangle from the given tangent value, \( \tan(\theta) = \frac{1}{x} \).

By recognizing this relationship, we can calculate side lengths needed to determine values for sine and cosine:

• Opposite side: 1
• Adjacent side: \( x \)
• Hypotenuse: \( \sqrt{x^2+1} \)

The hypotenuse is essential for calculating sine and cosine functions in subsequent steps. This process is vital because it transforms trigonometric problems into geometric ones, making it easier to visualize and solve.

Trigonometric Ratios

Trigonometric ratios deal with the relationships between the angles and sides of a right triangle. The basic ratios are sine, cosine, and tangent.

These functions serve as the foundation for calculating angles and sides within triangles and have numerous applications in sciences and engineering.

In context, knowing \( \tan(\theta) = \frac{1}{x} \) lets us find \( \sin(\theta) \) and \( \cos(\theta) \) using the right triangle derived from the tangent value.

• \( \sin(\theta) = \frac{1}{\sqrt{x^2+1}} \) from the constructed triangle.
• \( \cos(\theta) \) is then found using \( \cos(\alpha)=\frac{\sqrt{x^2+1}}{\sqrt{x^2+2}} \).

Understanding these ratios helps simplify complex trigonometric expressions by turning them into basic functions that are easier to interpret and solve. By working through the ratios systematically, we can find the solution to even the most convoluted expressions, illuminating the way trigonometric concepts interlock.