If sin `x=(12)/(13) "and sin " y= (4)/(5) " where " (pi)/(2) lt x lt pi " and " 0 lt y lt .(pi)/(2) ` find the values of
`(i) " sin " (x+y) " "(ii) " cos " (x+y) " "(iii) " tan " (x-y)`

Explanation

To solve the problem, we need to find the values of \( \sin(x+y) \), \( \cos(x+y) \), and \( \tan(x-y) \) given that \( \sin x = \frac{12}{13} \) and \( \sin y = \frac{4}{5} \). We also know the ranges for \( x \) and \( y \).

### Step 1: Find \( \cos x \) and \( \cos y \)

**For \( \cos x \):**
Since \( x \) is in the second quadrant, \( \cos x \) will be negative. We use the identity:
\[ \cos x = -\sqrt{1 - \sin^2 x} \]
Calculating:
\[ \sin^2 x = \left(\frac{12}{13}\right)^2 = \frac{144}{169} \]
So,
\[ \cos x = -\sqrt{1 - \frac{144}{169}} = -\sqrt{\frac{25}{169}} = -\frac{5}{13} \]

**For \( \cos y \):**
Since \( y \) is in the first quadrant, \( \cos y \) will be positive. We use the same identity:
\[ \cos y = \sqrt{1 - \sin^2 y} \]
Calculating:
\[ \sin^2 y = \left(\frac{4}{5}\right)^2 = \frac{16}{25} \]
So,
\[ \cos y = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 2: Calculate \( \sin(x+y) \)

Using the sine addition formula:
\[ \sin(x+y) = \sin x \cos y + \cos x \sin y \]
Substituting the values:
\[ \sin(x+y) = \left(\frac{12}{13}\right) \left(\frac{3}{5}\right) + \left(-\frac{5}{13}\right) \left(\frac{4}{5}\right) \]
Calculating:
\[ = \frac{36}{65} - \frac{20}{65} = \frac{16}{65} \]

### Step 3: Calculate \( \cos(x+y) \)

Using the cosine addition formula:
\[ \cos(x+y) = \cos x \cos y - \sin x \sin y \]
Substituting the values:
\[ \cos(x+y) = \left(-\frac{5}{13}\right) \left(\frac{3}{5}\right) - \left(\frac{12}{13}\right) \left(\frac{4}{5}\right) \]
Calculating:
\[ = -\frac{15}{65} - \frac{48}{65} = -\frac{63}{65} \]

### Step 4: Calculate \( \tan(x-y) \)

Using the tangent subtraction formula:
\[ \tan(x-y) = \frac{\tan x - \tan y}{1 + \tan x \tan y} \]
First, we need to find \( \tan x \) and \( \tan y \):
\[ \tan x = \frac{\sin x}{\cos x} = \frac{\frac{12}{13}}{-\frac{5}{13}} = -\frac{12}{5} \]
\[ \tan y = \frac{\sin y}{\cos y} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \]
Now substituting into the formula:
\[ \tan(x-y) = \frac{-\frac{12}{5} - \frac{4}{3}}{1 + \left(-\frac{12}{5}\right) \left(\frac{4}{3}\right)} \]
Calculating the numerator:
\[ = -\frac{12}{5} - \frac{4}{3} = -\frac{36}{15} - \frac{20}{15} = -\frac{56}{15} \]
Calculating the denominator:
\[ 1 - \frac{48}{15} = \frac{15 - 48}{15} = -\frac{33}{15} \]
Now substituting back:
\[ \tan(x-y) = \frac{-\frac{56}{15}}{-\frac{33}{15}} = \frac{56}{33} \]

### Final Answers:
(i) \( \sin(x+y) = \frac{16}{65} \)
(ii) \( \cos(x+y) = -\frac{63}{65} \)
(iii) \( \tan(x-y) = \frac{56}{33} \)