If `costheta=(5)/(13)`, then find the value of `tan^(2)theta+sec^(2)theta`.
`0^(@)ltthetalt90^(@)&tantheta=sqrt(6+sqrt(6+sqrt(6+....oo)))`, then find the value of `sec^(2)theta`.

Options

A. (a) `(39)/(23)`
B. (b) `(39)/(25)`
C. (c) `(313)/(25)`
D. (d) `(329)/(21)`

Answer

To solve the problem step by step, we start with the given information:

1. **Given**: \( \cos \theta = \frac{5}{13} \)

2. **Find**: \( \tan^2 \theta + \sec^2 \theta \)

### Step 1: Calculate \( \sec \theta \)
Using the identity:
\[ \sec \theta = \frac{1}{\cos \theta} \]
Substituting the value of \( \cos \theta \):
\[ \sec \theta = \frac{1}{\frac{5}{13}} = \frac{13}{5} \]

### Step 2: Calculate \( \sec^2 \theta \)
Using the identity:
\[ \sec^2 \theta = \left(\sec \theta\right)^2 \]
Substituting the value of \( \sec \theta \):
\[ \sec^2 \theta = \left(\frac{13}{5}\right)^2 = \frac{169}{25} \]

### Step 3: Calculate \( \tan \theta \)
Using the identity:
\[ \tan^2 \theta = \sec^2 \theta - 1 \]
Substituting the value of \( \sec^2 \theta \):
\[ \tan^2 \theta = \frac{169}{25} - 1 = \frac{169}{25} - \frac{25}{25} = \frac{144}{25} \]

### Step 4: Calculate \( \tan^2 \theta + \sec^2 \theta \)
Now we can find \( \tan^2 \theta + \sec^2 \theta \):
\[ \tan^2 \theta + \sec^2 \theta = \frac{144}{25} + \frac{169}{25} = \frac{144 + 169}{25} = \frac{313}{25} \]

### Final Result
Thus, the value of \( \tan^2 \theta + \sec^2 \theta \) is:
\[ \frac{313}{25} \]