Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Ask your questions and receive precise answers from experienced professionals across different disciplines. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

(C) [tex]\frac{\cos A - \sin A}{\cos A + \sin A} = \sec 2A - \tan 2A[/tex]

Sagot :

Let's solve the equation \(\frac{\cos A - \sin A}{\cos A + \sin A} = \sec(2A) - \tan(2A)\) step by step.

First, consider the left-hand side of the equation:

[tex]\[ \text{Left-hand side: } \frac{\cos A - \sin A}{\cos A + \sin A} \][/tex]

Next, let's consider the right-hand side of the equation:

[tex]\[ \text{Right-hand side: } \sec(2A) - \tan(2A) \][/tex]

Now let's break it down.

### Simplifying the Left-Hand Side
The left-hand side is already in a simplified form involving trigonometric functions:

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

### Simplifying the Right-Hand Side
Evaluate the right-hand side using known trigonometric identities:

[tex]\[ \sec(2A) = \frac{1}{\cos(2A)}, \quad \tan(2A) = \frac{\sin(2A)}{\cos(2A)} \][/tex]

Thus, the right-hand side can be written as:

[tex]\[ \sec(2A) - \tan(2A) = \frac{1}{\cos(2A)} - \frac{\sin(2A)}{\cos(2A)} = \frac{1 - \sin(2A)}{\cos(2A)} \][/tex]

### Comparing Both Sides
To check the equality, let's compare:

[tex]\[ \frac{\cos A - \sin A}{\cos A + \sin A} \quad \text{and} \quad \frac{1 - \sin(2A)}{\cos(2A)} \][/tex]

Upon simplifying expressions and calculations, we find that:

[tex]\[ \frac{\cos A - \sin A}{\cos A + \sin A} = \frac{1}{\tan(A + \frac{\pi}{4})} \][/tex]

Whereas:

[tex]\[ \sec(2A) - \tan(2A) = -\tan(2A) + \sec(2A) \][/tex]

### Conclusion
Upon simplification, it is evident that the two expressions:

[tex]\[ \frac{\cos A - \sin A}{\cos A + \sin A} \quad \text{and} \quad \sec(2A) - \tan(2A) \][/tex]

do not simplify to the same value. Hence, the given equation:

[tex]\[ \frac{\cos A - \sin A}{\cos A + \sin A} = \sec(2A) - \tan(2A) \][/tex]

is not true.