If `sin theta = (a^2 - 1)/(a^2 + 1)`, then the value of `sectheta + tantheta` will be

Options

A. `a/sqrt2`
B. `a/(a^2 +1)`
C. `sqrt2a`
D. a

Answer

To solve the problem, we need to find the value of \( \sec \theta + \tan \theta \) given that \( \sin \theta = \frac{a^2 - 1}{a^2 + 1} \).

### Step 1: Identify the values of sine, cosine, and tangent
We start with the given value of \( \sin \theta \):
\[ \sin \theta = \frac{a^2 - 1}{a^2 + 1} \]

### Step 2: Use the Pythagorean identity to find \( \cos \theta \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can find \( \cos \theta \):
\[ \sin^2 \theta = \left(\frac{a^2 - 1}{a^2 + 1}\right)^2 = \frac{(a^2 - 1)^2}{(a^2 + 1)^2} \]
Now, substituting into the identity:
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{(a^2 - 1)^2}{(a^2 + 1)^2} \]
\[ = \frac{(a^2 + 1)^2 - (a^2 - 1)^2}{(a^2 + 1)^2} \]

### Step 3: Simplify the expression for \( \cos^2 \theta \)
Now, we simplify the numerator:
\[ (a^2 + 1)^2 - (a^2 - 1)^2 = (a^2 + 1 + a^2 - 1)(a^2 + 1 - (a^2 - 1)) = (2a^2)(2) = 4a^2 \]
Thus, we have:
\[ \cos^2 \theta = \frac{4a^2}{(a^2 + 1)^2} \]
Taking the square root gives:
\[ \cos \theta = \frac{2a}{a^2 + 1} \]

### Step 4: Find \( \sec \theta \) and \( \tan \theta \)
Now we can find \( \sec \theta \) and \( \tan \theta \):
\[ \sec \theta = \frac{1}{\cos \theta} = \frac{a^2 + 1}{2a} \]
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{a^2 - 1}{a^2 + 1}}{\frac{2a}{a^2 + 1}} = \frac{a^2 - 1}{2a} \]

### Step 5: Calculate \( \sec \theta + \tan \theta \)
Now we can add \( \sec \theta \) and \( \tan \theta \):
\[ \sec \theta + \tan \theta = \frac{a^2 + 1}{2a} + \frac{a^2 - 1}{2a} \]
Combining the fractions:
\[ = \frac{(a^2 + 1) + (a^2 - 1)}{2a} = \frac{2a^2}{2a} = a \]

### Final Answer
Thus, the value of \( \sec \theta + \tan \theta \) is:
\[ \boxed{a} \]