Question

If \( \frac{\pi}{2} \leq x \leq \frac{3\pi}{4} \), then \( \cos^{-1} \left( \frac{12}{13} \cos x + \frac{5}{13} \sin x \right) \) is equal to:

• A. \( x - \tan^{-1} \left(\frac{4}{3}\right) \)
• B. \( x - \tan^{-1} \left(\frac{5}{12}\right) \)
• C. \( x + \tan^{-1} \left(\frac{4}{5}\right) \)
• D. \( x + \tan^{-1} \left(\frac{5}{12}\right) \)

Hint:

To solve inverse trigonometric functions, use known identities such as \( \cos^{-1}(\cos \theta) = \theta \), ensuring that the angle is within the correct range.

Answer 1

The Correct Option is B

To solve this problem, we need to simplify the expression \( \cos^{-1} \left( \frac{12}{13} \cos x + \frac{5}{13} \sin x \right) \) given the range \( \frac{\pi}{2} \leq x \leq \frac{3\pi}{4} \). 

1. We recognize that the expression inside the inverse cosine function resembles the application of a known trigonometric identity:

Let \( a = \frac{12}{13} \) and \( b = \frac{5}{13} \).

1. Notice that \( \sqrt{a^2 + b^2} = \sqrt{\left(\frac{12}{13}\right)^2 + \left(\frac{5}{13}\right)^2} = 1 \). Therefore, the expression can be transformed into a standard cosine formula:

\(\frac{12}{13} \cos x + \frac{5}{13} \sin x = \cos(\theta) \cos x + \sin(\theta) \sin x = \cos(x - \theta)\).

1. We need to find the angle \( \theta \), where:
2. \(\theta = \tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(\frac{5}{12}\right)\).
3. Thus, we have \(\cos^{-1}(\cos(x - \theta))\).
4. Since the angles lie in the domain where cosine is non-increasing, \(\cos^{-1}(\cos(x - \theta)) = x - \theta\).
5. Subtracting the angle gives us: \(\cos^{-1} \left( \frac{12}{13} \cos x + \frac{5}{13} \sin x \right) = x - \tan^{-1}\left(\frac{5}{12}\right)\).

This matches with the given option:

\( x - \tan^{-1} \left(\frac{5}{12}\right) \).

Hence, the correct answer is 
Option: \( x - \tan^{-1} \left(\frac{5}{12}\right) \).

Answer 2

To solve the given problem, we need to evaluate \[\cos^{-1} \left( \frac{12}{13} \cos x + \frac{5}{13} \sin x \right)\] where \( \frac{\pi}{2} \leq x \leq \frac{3\pi}{4} \).

First, recognize that \[\frac{12}{13}\] and \[\frac{5}{13}\] are components of a vector and can be associated with the cosine and sine of an angle. Note that:

\({\cos \theta = \frac{12}{13}}\) and \({\sin \theta = \frac{5}{13}}\) for \({\theta}\) such that:

\({\cos^2 \theta + \sin^2 \theta = 1}\).

Therefore, the given expression inside the inverse cosine is:

\(\cos(x-\theta)\) where \(\theta = \tan^{-1}\left(\frac{5}{12}\right)\) because:

\(\tan \theta = \frac{5}{12}\).

Hence, we express the problem as:

\[\cos^{-1}(\cos(x - \theta))\].

Given \(x - \theta\) where \( \theta = \tan^{-1} \left(\frac{5}{12}\right)\) and considering the domain of \(x\), we need to evaluate this within principal values of cosine's inverse function, which means:

\(\cos^{-1}(\cos(x - \theta)) = x - \theta\).

So,

\(\cos^{-1} \left( \frac{12}{13} \cos x + \frac{5}{13} \sin x \right) = x - \tan^{-1} \left( \frac{5}{12} \right)\).

This confirms the correct answer: \( x - \tan^{-1} \left(\frac{5}{12}\right) \).