Prove that `(1+sin theta)/(1-sin theta)=(sec theta+tantheta)^(2)`

Answer

To prove that

\[ \frac{1 + \sin \theta}{1 - \sin \theta} = (\sec \theta + \tan \theta)^2, \]

we will start with the left-hand side (LHS) and manipulate it to show that it equals the right-hand side (RHS).

### Step 1: Start with the LHS

\[ \text{LHS} = \frac{1 + \sin \theta}{1 - \sin \theta} \]

### Step 2: Rationalize the denominator

To simplify the expression, we can multiply the numerator and denominator by the conjugate of the denominator, which is \(1 + \sin \theta\):

\[ \text{LHS} = \frac{(1 + \sin \theta)(1 + \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)} \]

### Step 3: Simplify the denominator

Using the difference of squares formula, we have:

\[ 1 - \sin^2 \theta = \cos^2 \theta \]

Thus, the denominator simplifies to:

\[ \text{LHS} = \frac{(1 + \sin \theta)^2}{\cos^2 \theta} \]

### Step 4: Rewrite the expression

Now, we can express the LHS as:

\[ \text{LHS} = \frac{(1 + \sin \theta)^2}{\cos^2 \theta} = \left(\frac{1 + \sin \theta}{\cos \theta}\right)^2 \]

### Step 5: Separate the terms

We can separate the terms in the numerator:

\[ \frac{1 + \sin \theta}{\cos \theta} = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} \]

### Step 6: Identify secant and tangent

Recognizing the trigonometric identities, we have:

\[ \frac{1}{\cos \theta} = \sec \theta \quad \text{and} \quad \frac{\sin \theta}{\cos \theta} = \tan \theta \]

Thus, we can rewrite the expression as:

\[ \text{LHS} = (\sec \theta + \tan \theta)^2 \]

### Step 7: Conclusion

We have shown that:

\[ \frac{1 + \sin \theta}{1 - \sin \theta} = (\sec \theta + \tan \theta)^2 \]

Therefore, we have proved the equation.