prove cosec (a+b)= secA secB CosecA cosecB /secA cosecB + cosecA secB

To prove the identity \( \csc(a + b) = \frac{\sec A \sec B \csc A \csc B}{\sec A \csc B + \csc A \sec B} \), we can start by recalling some fundamental trigonometric identities and relationships. Let’s break this down step by step.

Understanding the Left Side

The left-hand side is \( \csc(a + b) \), which can be expressed in terms of sine:

\[ \csc(a + b) = \frac{1}{\sin(a + b)} \]

Using the sine addition formula, we know that:

\[ \sin(a + b) = \sin a \cos b + \cos a \sin b \]

Thus, we have:

\[ \csc(a + b) = \frac{1}{\sin a \cos b + \cos a \sin b} \]

Examining the Right Side

Now let’s analyze the right-hand side:

\[ \frac{\sec A \sec B \csc A \csc B}{\sec A \csc B + \csc A \sec B} \]

First, rewrite the secant and cosecant functions in terms of sine and cosine:

• \( \sec A = \frac{1}{\cos A} \)
• \( \sec B = \frac{1}{\cos B} \)
• \( \csc A = \frac{1}{\sin A} \)
• \( \csc B = \frac{1}{\sin B} \)

Substituting into the Right Side

Substituting these definitions into the right-hand side gives us:

\[ \frac{\left(\frac{1}{\cos A}\right)\left(\frac{1}{\cos B}\right)\left(\frac{1}{\sin A}\right)\left(\frac{1}{\sin B}\right)}{\left(\frac{1}{\cos A}\right)\left(\frac{1}{\sin B}\right) + \left(\frac{1}{\sin A}\right)\left(\frac{1}{\cos B}\right)} \]

This simplifies to:

\[ \frac{\frac{1}{\cos A \cos B \sin A \sin B}}{\frac{1}{\cos A \sin B} + \frac{1}{\sin A \cos B}} \]

Finding a Common Denominator

Next, let’s find a common denominator for the expression in the denominator:

\[ \frac{1}{\cos A \sin B} + \frac{1}{\sin A \cos B} = \frac{\sin A \cos B + \cos A \sin B}{\sin A \sin B \cos A \cos B} \]

So the right-hand side becomes:

\[ \frac{\frac{1}{\cos A \cos B \sin A \sin B}}{\frac{\sin A \cos B + \cos A \sin B}{\sin A \sin B \cos A \cos B}} \]

Simplifying Further

When we divide by a fraction, it’s equivalent to multiplying by its reciprocal. Thus, we have:

\[ \frac{1}{\cos A \cos B \sin A \sin B} \cdot \frac{\sin A \sin B \cos A \cos B}{\sin A \cos B + \cos A \sin B} \]

This simplifies to:

\[ \frac{1}{\sin A \cos B + \cos A \sin B} \]

Final Comparison

Notice that this expression matches our earlier expression for \( \csc(a + b) \):

\[ \csc(a + b) = \frac{1}{\sin(a + b)} = \frac{1}{\sin A \cos B + \cos A \sin B} \]

Since both sides are equal, we have successfully proven the identity:

\[ \csc(a + b) = \frac{\sec A \sec B \csc A \csc B}{\sec A \csc B + \csc A \sec B} \]

Wrap-Up

This proof illustrates the power of trigonometric identities and how they can be manipulated to show equalities. Understanding these identities not only helps in proofs but also enhances problem-solving skills in trigonometry. Keep practicing, and these concepts will become second nature!