Question

If \( \sin\left(\frac{\pi}{4}\cot\theta\right) = \cos\left(\frac{\pi}{4}\tan\theta\right) \), then the general solution of \( \theta \) is: 

• A. \( n\pi + \frac{\pi}{4}, n \in \mathbb{Z} \)
• B. \( n\pi + (-1)^n \frac{\pi}{4}, n \in \mathbb{Z} \)
• C. \( 2n\pi \pm \frac{\pi}{4}, n \in \mathbb{Z} \)
• D. \( 2n\pi + \frac{\pi}{4}, n \in \mathbb{Z} \)

Hint:

When solving trigonometric equations, use standard trigonometric identities and consider the periodicity of trigonometric functions to equate angles and solve for the variable.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Convert the cosine on the right side to sine using the property \( \cos \text{A} = \sin(\pi/2 - \text{A}) \).
Step 2: Detailed Explanation:
\( \sin \left( \frac{\pi}{4}\cot \theta \right) = \sin \left( \frac{\pi}{2} - \frac{\pi}{4}\tan \theta \right) \).
Equating the arguments:
\( \frac{\pi}{4}\cot \theta = \frac{\pi}{2} - \frac{\pi}{4}\tan \theta \)
\( \cot \theta = 2 - \tan \theta \)
\( \tan \theta + \cot \theta = 2 \).
Using \( \cot \theta = 1 / \tan \theta \):
\( \tan \theta + \frac{1}{\tan \theta} = 2 \Rightarrow \tan^2 \theta - 2\tan \theta + 1 = 0 \).
\( (\tan \theta - 1)^2 = 0 \Rightarrow \tan \theta = 1 \).
The general solution for \( \tan \theta = 1 \) is \( \theta = n\pi + \pi/4 \).
Step 3: Final Answer:
The solution is \( n\pi + \frac{\pi}{4} \).