If `tan A = (3)/(4), " then " 1- {(1+cos A) (1- cosA)//(1+sin A) (1-sin A)} + (7)/(16)`= ?

Options

A. `(5)/(8)`
B. 1
C. 0
D. `(7)/(8)`

Answer

To solve the equation given in the question, we will follow these steps:

### Step 1: Understand the given information
We know that \( \tan A = \frac{3}{4} \). From this, we can derive the values of \( \sin A \) and \( \cos A \) using the identity \( \tan A = \frac{\sin A}{\cos A} \).

### Step 2: Find \( \sin A \) and \( \cos A \)
Using the Pythagorean identity:
- Let \( \sin A = 3k \) and \( \cos A = 4k \) for some \( k \).
- Then, \( \sin^2 A + \cos^2 A = 1 \) implies:
\[ (3k)^2 + (4k)^2 = 1 \\ 9k^2 + 16k^2 = 1 \\ 25k^2 = 1 \\ k^2 = \frac{1}{25} \\ k = \frac{1}{5} \]
Thus, \( \sin A = \frac{3}{5} \) and \( \cos A = \frac{4}{5} \).

### Step 3: Substitute \( \sin A \) and \( \cos A \) into the expression
We need to evaluate:
\[ 1 - \frac{(1 + \cos A)(1 - \cos A)}{(1 + \sin A)(1 - \sin A)} + \frac{7}{16} \]
Substituting \( \sin A \) and \( \cos A \):
- \( 1 + \cos A = 1 + \frac{4}{5} = \frac{9}{5} \)
- \( 1 - \cos A = 1 - \frac{4}{5} = \frac{1}{5} \)
- \( 1 + \sin A = 1 + \frac{3}{5} = \frac{8}{5} \)
- \( 1 - \sin A = 1 - \frac{3}{5} = \frac{2}{5} \)

### Step 4: Calculate the numerator and denominator
Now calculate:
\[ (1 + \cos A)(1 - \cos A) = \left(\frac{9}{5}\right)\left(\frac{1}{5}\right) = \frac{9}{25} \]
\[ (1 + \sin A)(1 - \sin A) = \left(\frac{8}{5}\right)\left(\frac{2}{5}\right) = \frac{16}{25} \]

### Step 5: Substitute back into the expression
Now we substitute back into the expression:
\[ 1 - \frac{\frac{9}{25}}{\frac{16}{25}} + \frac{7}{16} = 1 - \frac{9}{16} + \frac{7}{16} \]
Calculating:
\[ 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16} \]
Now add \( \frac{7}{16} \):
\[ \frac{7}{16} + \frac{7}{16} = \frac{14}{16} = \frac{7}{8} \]

### Final Answer
Thus, the final answer is:
\[ \frac{7}{8} \]

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