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 will follow these steps:

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

### Step 2: Set the arguments equal
Since \( \tan A = \tan B \) implies \( A = B + n\pi \) (where \( n \) is any integer), we can write:
\[ \frac{\pi}{2} \sin \theta = \frac{\pi}{2}(1 - \cos \theta) + n\pi \]

### Step 3: Simplify the equation
Dividing through by \( \frac{\pi}{2} \):
\[ \sin \theta = 1 - \cos \theta + 2n \]
Rearranging gives:
\[ \sin \theta + \cos \theta = 1 + 2n \]

### Step 4: Use the identity for sine and cosine
We can use the identity \( \sin \theta + \cos \theta = \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) \). Thus, we have:
\[ \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) = 1 + 2n \]

### Step 5: Solve for the sine function
Dividing both sides by \( \sqrt{2} \):
\[ \sin\left(\theta + \frac{\pi}{4}\right) = \frac{1 + 2n}{\sqrt{2}} \]

### Step 6: Determine the general solution
The sine function has a range of \([-1, 1]\), so \( \frac{1 + 2n}{\sqrt{2}} \) must also be in this range. We can express the general solution as:
\[ \theta + \frac{\pi}{4} = \arcsin\left(\frac{1 + 2n}{\sqrt{2}}\right) + k\pi \]
where \( k \) is any integer.

### Step 7: Solve for \( \theta \)
Thus, we have:
\[ \theta = \arcsin\left(\frac{1 + 2n}{\sqrt{2}}\right) - \frac{\pi}{4} + k\pi \]

### Final General Solution
The general solution for the original equation is:
\[ \theta = \arcsin\left(\frac{1 + 2n}{\sqrt{2}}\right) - \frac{\pi}{4} + k\pi \]