ext{If } \sec \alpha + an \alpha = p ext{, then the value of } an \alpha ext{ is :}

Q: \text{If } \sec \alpha + \tan \alpha = p \text{, then the value of } \tan \alpha \text{ is :}

\text{If } \sec \alpha + \tan \alpha = p \text{, then the value of } \tan \alpha \text{ is :}

Question

ext{If } \sec \alpha + 	an \alpha = p 	ext{, then the value of } 	an \alpha 	ext{ is :}

Accepted Answer

To solve the problem and justify why Option 1 is correct, we need to find the value of $\tan \alpha$ given that $\sec \alpha + \tan \alpha = p$.

### Step-by-step Explanation:

1. **Expression for $\sec \alpha$ and $\tan \alpha$:**
   - We know that $\sec \alpha = \frac{1}{\cos \alpha}$ and $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$.

2. **Given Equation:**
   - $\sec \alpha + \tan \alpha = p$
   - Substitute the expressions: $\frac{1}{\cos \alpha} + \frac{\sin \alpha}{\cos \alpha} = p$
   - Combine the terms: $\frac{1 + \sin \alpha}{\cos \alpha} = p$

3. **Rearrange to find $\sin \alpha$:**
   - Multiply both sides by $\cos \alpha$: $1 + \sin \alpha = p \cos \alpha$
   - Rearrange: $\sin \alpha = p \cos \alpha - 1$

4. **Use Pythagorean Identity:**
   - $\sin^2 \alpha + \cos^2 \alpha = 1$
   - Substitute $\sin \alpha = p \cos \alpha - 1$ into the identity:
     $(p \cos \alpha - 1)^2 + \cos^2 \alpha = 1$

5. **Simplify the Equation:**
   - Expand: $p^2 \cos^2 \alpha - 2p \cos \alpha + 1 + \cos^2 \alpha = 1$
   - Combine terms: $(p^2 + 1) \cos^2 \alpha - 2p \cos \alpha = 0$

6. **Solve for $\cos \alpha$:**
   - Factor out $\cos \alpha$: $\cos \alpha ((p^2 + 1) \cos \alpha - 2p) = 0$
   - Since $\cos \alpha \neq 0$, solve $(p^2 + 1) \cos \alpha = 2p$
   - $\cos \alpha = \frac{2p}{p^2 + 1}$

7. **Find $\tan \alpha$:**
   - $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$
   - Substitute $\sin \alpha = p \cos \alpha - 1$ and $\cos \alpha = \frac{2p}{p^2 + 1}$:
   - $\tan \alpha = \frac{p \left(\frac{2p}{p^2 + 1}\right) - 1}{\frac{2p}{p^2 + 1}}$
   - Simplify: $\tan \alpha = \frac{\frac{2p^2}{p^2 + 1} - 1}{\frac{2p}{p^2 + 1}}$
   - $\tan \alpha = \frac{2p^2 - (p^2 + 1)}{2p}$
   - $\tan \alpha = \frac{p^2 - 1}{2p}$

### Conclusion:

- **Option 1**: $\frac{p^2 - 1}{2p}$ is indeed the correct value for $\tan \alpha$.

### Explanation of Incorrect Options:

- **Option 2**: $\frac{p^2 - 1}{p^2 + 1}$ does not match the derived expression for $\tan \alpha$.
- **Option 3**: $\frac{2p}{p^2 - 1}$ is incorrect as it is the reciprocal of the correct expression.
- **Option 4**: $\frac{p^2 - 1}{p}$ is incorrect because it lacks the factor of 2 in the denominator, which is necessary for the correct expression.

Thus, Option 1 is the correct answer, as it accurately represents the value of $\tan \alpha$ derived from the given equation.