What is the value of `[sin(90-A)+cos(180-2A)]//[cos(90-2A)+sin(180-A)]?`

Options

A. sin(A/2) cos A
B. cot (A/2)
C. tan (A/2)
D. sin A cos (A/2)

Answer

To solve the expression \(\frac{\sin(90^\circ - A) + \cos(180^\circ - 2A)}{\cos(90^\circ - 2A) + \sin(180^\circ - A)}\), we will simplify each part step by step.

### Step 1: Simplify \(\sin(90^\circ - A)\)
Using the co-function identity, we know that:
\[ \sin(90^\circ - A) = \cos(A) \]

### Step 2: Simplify \(\cos(180^\circ - 2A)\)
Using the identity for cosine, we have:
\[ \cos(180^\circ - \theta) = -\cos(\theta) \]
Thus,
\[ \cos(180^\circ - 2A) = -\cos(2A) \]

### Step 3: Combine the results from Steps 1 and 2
Now substituting back, we have:
\[ \sin(90^\circ - A) + \cos(180^\circ - 2A) = \cos(A) - \cos(2A) \]

### Step 4: Simplify \(\cos(90^\circ - 2A)\)
Using the co-function identity again:
\[ \cos(90^\circ - 2A) = \sin(2A) \]

### Step 5: Simplify \(\sin(180^\circ - A)\)
Using the identity for sine:
\[ \sin(180^\circ - \theta) = \sin(\theta) \]
Thus,
\[ \sin(180^\circ - A) = \sin(A) \]

### Step 6: Combine the results from Steps 4 and 5
Now substituting back, we have:
\[ \cos(90^\circ - 2A) + \sin(180^\circ - A) = \sin(2A) + \sin(A) \]

### Step 7: Substitute everything back into the original expression
Now we can rewrite the original expression:
\[ \frac{\cos(A) - \cos(2A)}{\sin(2A) + \sin(A)} \]

### Step 8: Use the sum-to-product identities
Using the identities:
- \(\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)\)
- \(\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\)

Let \(A = A\) and \(B = 2A\):
\[ \cos(A) - \cos(2A) = -2 \sin\left(\frac{A + 2A}{2}\right) \sin\left(\frac{A - 2A}{2}\right) = -2 \sin\left(\frac{3A}{2}\right) \sin\left(-\frac{A}{2}\right) \]
And,
\[ \sin(2A) + \sin(A) = 2 \sin\left(\frac{2A + A}{2}\right) \cos\left(\frac{2A - A}{2}\right) = 2 \sin\left(\frac{3A}{2}\right) \cos\left(\frac{A}{2}\right) \]

### Step 9: Substitute back into the expression
Now substituting these identities back into the expression gives:
\[ \frac{-2 \sin\left(\frac{3A}{2}\right) \sin\left(-\frac{A}{2}\right)}{2 \sin\left(\frac{3A}{2}\right) \cos\left(\frac{A}{2}\right)} \]

### Step 10: Simplify the expression
The \(2\) cancels out, and \(\sin\left(\frac{3A}{2}\right)\) cancels out as long as \(\sin\left(\frac{3A}{2}\right) \neq 0\):
\[ \frac{-\sin\left(-\frac{A}{2}\right)}{\cos\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{A}{2}\right)}{\cos\left(\frac{A}{2}\right)} = \tan\left(\frac{A}{2}\right) \]

### Final Result
Thus, the value of the expression is:
\[ \tan\left(\frac{A}{2}\right) \]