If `theta ` and `phi` lie in the first quadrant such that sin `theta =(8)/(17) "and cos " phi =(12)/(13) ` find the values of
`(i) " sin"(theta- phi) " "(ii) " cos " (theta + phi) " " (iii) " tan " (theta - phi)`

Answer

To solve the problem, we need to find the values of \( \sin(\theta - \phi) \), \( \cos(\theta + \phi) \), and \( \tan(\theta - \phi) \) given that \( \sin \theta = \frac{8}{17} \) and \( \cos \phi = \frac{12}{13} \).

### Step 1: Find \( \cos \theta \)

We know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting the value of \( \sin \theta \):
\[ \left(\frac{8}{17}\right)^2 + \cos^2 \theta = 1 \]
\[ \frac{64}{289} + \cos^2 \theta = 1 \]
\[ \cos^2 \theta = 1 - \frac{64}{289} = \frac{225}{289} \]
Taking the square root (since \( \theta \) is in the first quadrant):
\[ \cos \theta = \frac{15}{17} \]

### Step 2: Find \( \sin \phi \)

We also know:
\[ \sin^2 \phi + \cos^2 \phi = 1 \]
Substituting the value of \( \cos \phi \):
\[ \sin^2 \phi + \left(\frac{12}{13}\right)^2 = 1 \]
\[ \sin^2 \phi + \frac{144}{169} = 1 \]
\[ \sin^2 \phi = 1 - \frac{144}{169} = \frac{25}{169} \]
Taking the square root (since \( \phi \) is in the first quadrant):
\[ \sin \phi = \frac{5}{13} \]

### Step 3: Calculate \( \sin(\theta - \phi) \)

Using the sine subtraction formula:
\[ \sin(\theta - \phi) = \sin \theta \cos \phi - \cos \theta \sin \phi \]
Substituting the values:
\[ \sin(\theta - \phi) = \left(\frac{8}{17}\right) \left(\frac{12}{13}\right) - \left(\frac{15}{17}\right) \left(\frac{5}{13}\right) \]
Calculating each term:
\[ = \frac{96}{221} - \frac{75}{221} = \frac{21}{221} \]

### Step 4: Calculate \( \cos(\theta + \phi) \)

Using the cosine addition formula:
\[ \cos(\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi \]
Substituting the values:
\[ \cos(\theta + \phi) = \left(\frac{15}{17}\right) \left(\frac{12}{13}\right) - \left(\frac{8}{17}\right) \left(\frac{5}{13}\right) \]
Calculating each term:
\[ = \frac{180}{221} - \frac{40}{221} = \frac{140}{221} \]

### Step 5: Calculate \( \tan(\theta - \phi) \)

Using the tangent subtraction formula:
\[ \tan(\theta - \phi) = \frac{\sin(\theta - \phi)}{\cos(\theta - \phi)} \]
We already have \( \sin(\theta - \phi) = \frac{21}{221} \). Now we need \( \cos(\theta - \phi) \):
\[ \cos(\theta - \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi \]
Substituting the values:
\[ \cos(\theta - \phi) = \left(\frac{15}{17}\right) \left(\frac{12}{13}\right) + \left(\frac{8}{17}\right) \left(\frac{5}{13}\right) \]
Calculating each term:
\[ = \frac{180}{221} + \frac{40}{221} = \frac{220}{221} \]
Now substituting into the tangent formula:
\[ \tan(\theta - \phi) = \frac{\frac{21}{221}}{\frac{220}{221}} = \frac{21}{220} \]

### Final Answers:
(i) \( \sin(\theta - \phi) = \frac{21}{221} \)
(ii) \( \cos(\theta + \phi) = \frac{140}{221} \)
(iii) \( \tan(\theta - \phi) = \frac{21}{220} \)