The general solution of `"tan" ((pi)/(2)"sin" theta) ="cot"((pi)/(2)"cos"theta)`, is

Options

A. `theta= 2r pi + (pi)/(2)`
B. `theta =2r pi`
C. `theta =2r pi + (pi)/(2) " and " theta = 2r pi`
D. none of these

Answer

To solve the equation \( \tan\left(\frac{\pi}{2} \sin \theta\right) = \cot\left(\frac{\pi}{2} \cos \theta\right) \), we can follow these steps:

### Step 1: Rewrite the cotangent
We know that \( \cot x = \frac{1}{\tan x} \). Therefore, we can rewrite the equation as:
\[ \tan\left(\frac{\pi}{2} \sin \theta\right) = \frac{1}{\tan\left(\frac{\pi}{2} \cos \theta\right)} \]

### Step 2: Cross multiply
Cross multiplying gives us:
\[ \tan\left(\frac{\pi}{2} \sin \theta\right) \tan\left(\frac{\pi}{2} \cos \theta\right) = 1 \]

### Step 3: Use the identity
We know that \( \tan A \tan B = 1 \) implies \( A + B = \frac{\pi}{2} + n\pi \) for some integer \( n \). Thus, we have:
\[ \frac{\pi}{2} \sin \theta + \frac{\pi}{2} \cos \theta = \frac{\pi}{2} + n\pi \]

### Step 4: Simplify the equation
Dividing the entire equation by \( \frac{\pi}{2} \):
\[ \sin \theta + \cos \theta = 1 + 2n \]

### Step 5: Rearranging
Rearranging gives us:
\[ \sin \theta + \cos \theta - 1 = 2n \]

### Step 6: Use the identity for sine and cosine
Using the identity \( \sin \theta + \cos \theta = \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) \):
\[ \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) - 1 = 2n \]

### Step 7: Solve for \( \sin\left(\theta + \frac{\pi}{4}\right) \)
Rearranging gives:
\[ \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) = 2n + 1 \]
Thus,
\[ \sin\left(\theta + \frac{\pi}{4}\right) = \frac{2n + 1}{\sqrt{2}} \]

### Step 8: General solution
The general solution for \( \sin x = k \) is given by:
\[ x = \arcsin(k) + 2m\pi \quad \text{or} \quad x = \pi - \arcsin(k) + 2m\pi \]
Applying this to our equation, we have:
\[ \theta + \frac{\pi}{4} = \arcsin\left(\frac{2n + 1}{\sqrt{2}}\right) + 2m\pi \quad \text{or} \quad \theta + \frac{\pi}{4} = \pi - \arcsin\left(\frac{2n + 1}{\sqrt{2}}\right) + 2m\pi \]

### Step 9: Isolate \( \theta \)
From these, we can isolate \( \theta \):
1. \( \theta = \arcsin\left(\frac{2n + 1}{\sqrt{2}}\right) - \frac{\pi}{4} + 2m\pi \)
2. \( \theta = \pi - \arcsin\left(\frac{2n + 1}{\sqrt{2}}\right) - \frac{\pi}{4} + 2m\pi \)

### Final General Solution
The final general solution can be expressed as:
\[ \theta = 2n\pi + \frac{\pi}{4} \quad \text{or} \quad \theta = 2n\pi + \frac{3\pi}{4} \]