(Sin A - cos A)/(sin A + cos A ) +(sinaA + cosA) / (sinA-cosA) =2/(sin square S -1)

Let's solve the given equation step by step.

We are given the equation:

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

### Step 1: Simplify the left-hand side (LHS)
We have two terms in the LHS, and we can rewrite this as:

\[
\frac{\sin A - \cos A}{\sin A + \cos A} + \frac{\sin A + \cos A}{\sin A - \cos A}
\]

Now, we can take a common denominator for the two fractions. The common denominator would be:

\[
(\sin A + \cos A)(\sin A - \cos A)
\]

This is a difference of squares, so we can simplify the denominator:

\[
(\sin A + \cos A)(\sin A - \cos A) = \sin^2 A - \cos^2 A
\]

So, the LHS becomes:

\[
\frac{(\sin A - \cos A)^2 + (\sin A + \cos A)^2}{\sin^2 A - \cos^2 A}
\]

### Step 2: Expand the numerators
We will expand both \((\sin A - \cos A)^2\) and \((\sin A + \cos A)^2\):

\[
(\sin A - \cos A)^2 = \sin^2 A - 2\sin A \cos A + \cos^2 A
\]

\[
(\sin A + \cos A)^2 = \sin^2 A + 2\sin A \cos A + \cos^2 A
\]

Now, adding these two expressions:

\[
(\sin^2 A - 2\sin A \cos A + \cos^2 A) + (\sin^2 A + 2\sin A \cos A + \cos^2 A) = 2\sin^2 A + 2\cos^2 A
\]

Since \(\sin^2 A + \cos^2 A = 1\), we have:

\[
2\sin^2 A + 2\cos^2 A = 2
\]

Thus, the LHS becomes:

\[
\frac{2}{\sin^2 A - \cos^2 A}
\]

### Step 3: Simplify the right-hand side (RHS)
The RHS of the equation is:

\[
\frac{2}{\sin^2 A - 1}
\]

### Step 4: Compare both sides
We now compare the LHS and RHS:

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

This implies:

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

### Step 5: Solve the equation
Canceling \(\sin^2 A\) from both sides, we are left with:

\[
-\cos^2 A = -1
\]

Thus:

\[
\cos^2 A = 1
\]

This gives us:

\[
\cos A = \pm 1
\]

### Step 6: Conclusion
The solution to the equation is when \(\cos A = 1\) or \(\cos A = -1\). This corresponds to specific values of \(A\), such as \(A = 0\) or \(A = \pi\).

Would you like more details or clarifications on any step?

Here are 5 follow-up questions:
1. What are the specific angles where \(\cos A = 1\) or \(\cos A = -1\)?
2. Can we derive other trigonometric identities from \(\sin^2 A + \cos^2 A = 1\)?
3. What is the significance of the difference of squares in trigonometry?
4. How do you solve trigonometric equations for general values of \(A\)?
5. How does the sign of \(\cos A\) affect the behavior of the equation?

**Tip:** Always remember that \(\sin^2 A + \cos^2 A = 1\) is a fundamental identity and can help simplify many trigonometric expressions.

(Sin A - cos A)/(sin A + cos A ) +(sin A + cos A) / (sin A - cos A) = 2/(sin^2 A - 1)

Explore how to simplify and solve the trigonometric equation (Sin A - Cos A)/(Sin A + Cos A) + (Sin A + Cos A)/(Sin A - Cos A) = 2/(Sin^2 A - 1). This step-by-step solution covers key trigonometric identities, including the Pythagorean identity and the difference of squares, making it suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation