1/sec thetha + Tan thetha = sec thetha - Tan thetha

We are given the equation:

\[
\frac{1}{\sec \theta} + \tan \theta = \sec \theta - \tan \theta
\]

Let's try to solve it step-by-step:

### Step 1: Simplify \( \frac{1}{\sec \theta} \)

We know that \( \sec \theta = \frac{1}{\cos \theta} \), so \( \frac{1}{\sec \theta} = \cos \theta \). 

Now the equation becomes:

\[
\cos \theta + \tan \theta = \sec \theta - \tan \theta
\]

### Step 2: Express \( \tan \theta \)

We can express \( \tan \theta \) in terms of sine and cosine:

\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\]

Substituting this into the equation:

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

### Step 3: Combine terms

Both sides of the equation have a denominator of \( \cos \theta \), so we can simplify:

Left-hand side (LHS):
\[
\frac{\cos^2 \theta + \sin \theta}{\cos \theta}
\]

Right-hand side (RHS):
\[
\frac{1 - \sin \theta}{\cos \theta}
\]

So the equation becomes:

\[
\frac{\cos^2 \theta + \sin \theta}{\cos \theta} = \frac{1 - \sin \theta}{\cos \theta}
\]

### Step 4: Multiply both sides by \( \cos \theta \)

This eliminates the denominators:

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

### Step 5: Move terms around

Move all terms to one side of the equation:

\[
\cos^2 \theta + \sin \theta + \sin \theta - 1 = 0
\]

This simplifies to:

\[
\cos^2 \theta + 2\sin \theta - 1 = 0
\]

### Step 6: Use the identity \( \cos^2 \theta = 1 - \sin^2 \theta \)

Substitute \( \cos^2 \theta \) with \( 1 - \sin^2 \theta \):

\[
1 - \sin^2 \theta + 2\sin \theta - 1 = 0
\]

This simplifies to:

\[
-\sin^2 \theta + 2\sin \theta = 0
\]

### Step 7: Factor the equation

\[
-\sin \theta (\sin \theta - 2) = 0
\]

This gives two possible solutions:

1. \( \sin \theta = 0 \)
2. \( \sin \theta = 2 \) (which is not possible since the sine function has values between -1 and 1)

### Step 8: Solve for \( \theta \)

For \( \sin \theta = 0 \), \( \theta = n\pi \), where \( n \) is an integer.

### Final Solution:

\[
\theta = n\pi, \quad n \in \mathbb{Z}
\]

Would you like more details or have any questions?

Here are 5 related questions:

1. What are the properties of the sine and cosine functions?
2. How can we use trigonometric identities to simplify expressions?
3. What are the general solutions for \( \sin \theta = 0 \)?
4. How does the secant function relate to cosine?
5. Why is it important to express tangent in terms of sine and cosine?

**Tip:** Always check the domain of trigonometric functions to avoid invalid solutions.

1/sec theta + Tan theta = sec theta - Tan theta

Explore the step-by-step solution of the trigonometric equation 1/sec(theta) + tan(theta) = sec(theta) - tan(theta). This problem involves trigonometric identities such as secant, tangent, and Pythagorean identity, and is suitable for students in Grades 10-12.

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