Solving Inverse Trigonometric Equations A Comprehensive Guide

Inverse trigonometric functions, often shrouded in complexity, play a pivotal role in various branches of mathematics, physics, and engineering. Mastering these functions is crucial for solving a wide array of problems, from calculating angles in geometric figures to analyzing alternating current circuits. In this comprehensive guide, we will delve into the intricacies of two intriguing problems involving inverse trigonometric functions, providing step-by-step solutions and insightful explanations. By carefully dissecting these problems, we aim to demystify the concepts and empower you with the skills to confidently tackle similar challenges. This exploration is designed to enhance your understanding of inverse trigonometric functions, their properties, and their applications in real-world scenarios. We will focus on clarity, accuracy, and a pedagogical approach that ensures a firm grasp of the underlying principles.

13. Solving ${\sec^{-1}x = \csc^{-1}y}$: A Step-by-Step Approach

When faced with the equation ${\sec^{-1}x = \csc^{-1}y}$, our primary goal is to determine the value of ${\cos^{-1}\frac{1}{x} + \cos^{-1}\frac{1}{y}}$. This problem elegantly combines the concepts of inverse secant and cosecant functions, demanding a keen understanding of their properties and relationships. To embark on this journey, let's first establish a solid foundation by revisiting the definitions of these functions and their connections to other trigonometric counterparts. The inverse secant function, denoted as ${\sec^{-1}x}$, yields the angle whose secant is x. Similarly, the inverse cosecant function, represented as ${\csc^{-1}y}$, gives the angle whose cosecant is y. These definitions are pivotal in transforming the given equation into a more manageable form. We will explore how these functions relate to the more familiar inverse cosine function, which will serve as a bridge in our solution. Understanding these relationships is not just about memorizing formulas; it's about grasping the inherent connections within the trigonometric world. This approach will enable us to not only solve this specific problem but also to tackle a broader range of challenges involving inverse trigonometric functions. By the end of this section, you will have a clear, step-by-step understanding of how to navigate such equations and arrive at the correct solution.

Transforming the Equation

To effectively tackle this problem involving inverse trigonometric functions, we begin by introducing a crucial substitution. Let's assume that ${\sec^{-1}x = \csc^{-1}y = \theta}$. This substitution is the cornerstone of our approach, allowing us to express both inverse secant and inverse cosecant in terms of a single variable, ${\theta}$. This seemingly simple step is a powerful technique that simplifies the equation and paves the way for further manipulation. With this substitution in place, we can now express x and y in terms of ${\theta}$ using the definitions of secant and cosecant. Since ${\sec^{-1}x = \theta}$, it follows that ${x = \sec \theta}$. Similarly, from ${\csc^{-1}y = \theta}$, we deduce that ${y = \csc \theta}$. These transformations are crucial because they allow us to move from the realm of inverse trigonometric functions to the more familiar territory of standard trigonometric functions. This transition is a key strategy in solving many trigonometric problems. Next, we will leverage the fundamental relationships between secant, cosecant, cosine, and sine to further simplify our expressions. This involves recognizing that ${\sec \theta}$ is the reciprocal of ${\cos \theta}$ and ${\csc \theta}$ is the reciprocal of ${\sin \theta}$. By making these substitutions, we will be able to express ${\cos^{-1}\frac{1}{x}}$ and ${\cos^{-1}\frac{1}{y}}$ in terms of ${\theta}$, bringing us closer to our ultimate goal of finding the value of ${\cos^{-1}\frac{1}{x} + \cos^{-1}\frac{1}{y}}$.

Utilizing Trigonometric Identities

With ${x = \sec \theta}$ and ${y = \csc \theta}$ established, we can now express ${\frac{1}{x}}$ and ${\frac{1}{y}}$ in terms of cosine and sine, respectively. Since ${x = \sec \theta}$, it follows that ${\frac{1}{x} = \frac{1}{\sec \theta} = \cos \theta}$. Similarly, since ${y = \csc \theta}$, we have ${\frac{1}{y} = \frac{1}{\csc \theta} = \sin \theta}$. These transformations are pivotal as they directly connect the given expressions to the standard trigonometric functions, cosine and sine, which we are more adept at handling. Now, let's consider the expressions ${\cos^{-1}\frac{1}{x}}$ and ${\cos^{-1}\frac{1}{y}}$. Substituting the values we just derived, we get ${\cos^{-1}(\cos \theta)}$ and ${\cos^{-1}(\sin \theta)}$. The first expression, ${\cos^{-1}(\cos \theta)}$, simplifies directly to ${\theta}$, thanks to the fundamental property of inverse trigonometric functions. However, the second expression, ${\cos^{-1}(\sin \theta)}$, requires a bit more finesse. To simplify it, we recall the cofunction identity: ${\sin \theta = \cos(\frac{\pi}{2} - \theta)}$. This identity is a cornerstone of trigonometry, linking sine and cosine through a complementary angle relationship. Applying this identity, we can rewrite ${\cos^{-1}(\sin \theta)}$ as ${\cos^{-1}(\cos(\frac{\pi}{2} - \theta))}$, which simplifies to ${\frac{\pi}{2} - \theta}$. This step is crucial as it allows us to express both terms in our target expression in terms of ${\theta}$, setting the stage for the final calculation.

Calculating the Final Value

Now that we have simplified ${\cos^{-1}\frac{1}{x}}$ to ${\theta}$ and ${\cos^{-1}\frac{1}{y}}$ to ${\frac{\pi}{2} - \theta}$, we are in a prime position to determine the value of ${\cos^{-1}\frac{1}{x} + \cos^{-1}\frac{1}{y}}$. This final step involves a straightforward addition of the simplified expressions. Substituting the values, we get: ${\cos^{-1}\frac{1}{x} + \cos^{-1}\frac{1}{y} = \theta + (\frac{\pi}{2} - \theta)}$. This equation beautifully encapsulates the essence of our step-by-step simplification process. The ${\theta}$ and ${-\theta}$ terms elegantly cancel each other out, leaving us with a remarkably simple result. After the cancellation, we are left with: ${\cos^{-1}\frac{1}{x} + \cos^{-1}\frac{1}{y} = \frac{\pi}{2}}$. This result is not only the solution to the problem but also a testament to the power of strategic substitution and the application of trigonometric identities. Therefore, the value of ${\cos^{-1}\frac{1}{x} + \cos^{-1}\frac{1}{y}}$ is ${\frac{\pi}{2}}$, which corresponds to option (D). This conclusion highlights the importance of a methodical approach to problem-solving, where each step builds upon the previous one to unveil the final answer. The journey from the initial equation to this solution has been a demonstration of how inverse trigonometric functions can be manipulated and simplified using fundamental trigonometric principles.

14. Solving ${\sin^{-1}x - \cos^{-1}x = \frac{\pi}{6}}$: A Comprehensive Solution

The equation ${\sin^{-1}x - \cos^{-1}x = \frac{\pi}{6}}$ presents a fascinating challenge that requires a blend of trigonometric identities and algebraic manipulation. Our primary objective is to find the value of x that satisfies this equation. This problem is particularly intriguing because it involves the interplay between inverse sine and inverse cosine functions, demanding a deep understanding of their properties and relationships. To embark on this solution, we will first strategically transform the equation into a more manageable form. This involves leveraging the fundamental identity that connects ${\sin^{-1}x}$ and ${\cos^{-1}x}$. By skillfully applying this identity, we can rewrite the equation in terms of a single inverse trigonometric function, simplifying the problem considerably. This step is crucial as it streamlines the subsequent algebraic manipulations. We will then employ algebraic techniques to isolate x and determine its value. This process will involve careful consideration of the domains and ranges of the inverse trigonometric functions to ensure the validity of our solution. Furthermore, we will provide a detailed explanation of each step, ensuring that you not only understand the solution but also the underlying principles and techniques. This approach will empower you to tackle similar problems with confidence and proficiency. By the end of this section, you will have a clear understanding of how to solve equations involving inverse trigonometric functions and how to apply these techniques to a broader range of mathematical challenges.

Transforming the Equation Using Identities

The cornerstone of solving this equation, ${\sin^{-1}x - \cos^{-1}x = \frac{\pi}{6}}$, lies in strategically employing the fundamental identity that relates inverse sine and inverse cosine functions. This identity, a bedrock of inverse trigonometry, states that ${\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}}$ for all x in the domain [-1, 1]. This identity is not merely a formula to be memorized; it embodies a deep connection between the two inverse trigonometric functions, stemming from the complementary relationship between sine and cosine in the unit circle. To effectively utilize this identity, we need to manipulate our original equation to create an opportunity for substitution. We begin by isolating ${\cos^{-1}x}$ in the identity, which gives us ${\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x}$. This rearrangement is a pivotal step, as it expresses ${\cos^{-1}x}$ in terms of ${\sin^{-1}x}$, allowing us to substitute this expression into our original equation. By making this substitution, we effectively transform the original equation into one involving only ${\sin^{-1}x}$, significantly simplifying the problem. The substitution process is not just about replacing one term with another; it's about strategically transforming the equation into a form that is easier to solve. This technique is a common thread in mathematical problem-solving, highlighting the importance of recognizing and leveraging fundamental identities to simplify complex expressions. Now, with the equation expressed solely in terms of ${\sin^{-1}x}$, we can proceed with algebraic manipulations to isolate ${\sin^{-1}x}$ and ultimately determine the value of x.

Isolating and Solving for ${x}$

Having strategically transformed the original equation, ${\sin^{-1}x - \cos^{-1}x = \frac{\pi}{6}}$, using the identity ${\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x}$, we now have a simplified equation in terms of ${\sin^{-1}x}$. Substituting the identity into the original equation, we get: ${\sin^{-1}x - (\frac{\pi}{2} - \sin^{-1}x) = \frac{\pi}{6}}$. This substitution is a crucial step, as it consolidates the equation into a single inverse trigonometric function, making it much easier to manipulate algebraically. Next, we simplify the equation by distributing the negative sign and combining like terms. This gives us: ${\sin^{-1}x - \frac{\pi}{2} + \sin^{-1}x = \frac{\pi}{6}}$, which further simplifies to ${2\sin^{-1}x = \frac{\pi}{6} + \frac{\pi}{2}}$. This algebraic manipulation is a straightforward application of basic principles, but it is essential in isolating the term containing ${\sin^{-1}x}$. Now, we need to find a common denominator and add the fractions on the right side of the equation. The common denominator for 6 and 2 is 6, so we rewrite ${\frac{\pi}{2}}$ as ${\frac{3\pi}{6}}$. This gives us: ${2\sin^{-1}x = \frac{\pi}{6} + \frac{3\pi}{6}}$, which simplifies to ${2\sin^{-1}x = \frac{4\pi}{6}}$. We can further simplify the fraction on the right side by dividing both the numerator and the denominator by 2, resulting in ${2\sin^{-1}x = \frac{2\pi}{3}}$. To isolate ${\sin^{-1}x}$, we divide both sides of the equation by 2, yielding: ${\sin^{-1}x = \frac{\pi}{3}}$. This step is a critical juncture in our solution, as we have now isolated the inverse sine function. To finally solve for x, we take the sine of both sides of the equation. This gives us: ${x = \sin(\frac{\pi}{3})}$. We know that ${\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}}$, so the solution for x is ${x = \frac{\sqrt{3}}{2}}$. This final step beautifully illustrates the inverse relationship between the sine and inverse sine functions. By taking the sine of both sides, we effectively