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Here is the formatted question, ensuring it makes sense and is easier to read:

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Simplify the following expressions and verify the identity:

1. [tex](\cos \theta + \sin \theta)^2 + (\cos \theta - \cos \theta)^2 = 2[/tex]

2. [tex]\frac{\cos \theta}{1 + \sin \theta} = \sec \theta - \tan \theta[/tex]

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This version clearly states the task of simplifying and verifying the given mathematical identities.


Sagot :

Alright, let's delve into and prove the identities provided.

### Identity 1: [tex]\((\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta )^2 = 2\)[/tex]

First, let's expand both terms individually.

#### Expanding [tex]\((\cos \theta + \sin \theta)^2\)[/tex]

[tex]\[ (\cos \theta + \sin \theta)^2 = \cos^2 \theta + 2 \cos \theta \sin \theta + \sin^2 \theta \][/tex]

#### Expanding [tex]\((\cos \theta - \sin \theta)^2\)[/tex]

[tex]\[ (\cos \theta - \sin \theta)^2 = \cos^2 \theta - 2 \cos \theta \sin \theta + \sin^2 \theta \][/tex]

#### Adding the expanded forms together:

[tex]\[ (\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta)^2 = (\cos^2 \theta + 2 \cos \theta \sin \theta + \sin^2 \theta) + (\cos^2 \theta - 2 \cos \theta \sin \theta + \sin^2 \theta) \][/tex]

Combine the terms:

[tex]\[ = \cos^2 \theta + \sin^2 \theta + \cos^2 \theta + \sin^2 \theta \][/tex]

Simplify:

[tex]\[ = 2 \cos^2 \theta + 2 \sin^2 \theta \][/tex]

Using the Pythagorean identity:

[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]

So we have:

[tex]\[ 2(\cos^2 \theta + \sin^2 \theta) = 2 \cdot 1 = 2 \][/tex]

Thus, the first identity is proven:

[tex]\[ (\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta )^2 = 2 \][/tex]

### Identity 2: [tex]\(\frac{\cos \theta}{1+\sin \theta} = \sec \theta - \tan \theta\)[/tex]

First, let's express the right-hand side in terms of cosine and sine functions.

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

So,

[tex]\[ \sec \theta - \tan \theta = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} \][/tex]

Combine the fractions:

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

Now, let's focus on the left-hand side:

[tex]\[ \frac{\cos \theta}{1 + \sin \theta} \][/tex]

To simplify this to match the right-hand side, we can use an algebraic trick by multiplying the numerator and the denominator by the conjugate of the denominator:

[tex]\[ \frac{\cos \theta}{1 + \sin \theta} \cdot \frac{1 - \sin \theta}{1 - \sin \theta} \][/tex]

This simplifies to:

[tex]\[ = \frac{\cos \theta (1 - \sin \theta)}{(1 + \sin \theta) (1 - \sin \theta)} \][/tex]

Using the difference of squares:

[tex]\[ (1 + \sin \theta)(1 - \sin \theta) = 1 - \sin^2 \theta \][/tex]

Since [tex]\(1 - \sin^2 \theta = \cos^2 \theta\)[/tex], we get:

[tex]\[ = \frac{\cos \theta (1 - \sin \theta)}{\cos^2 \theta} \][/tex]

Simplify the fraction:

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

We can see this matches the right-hand side:

[tex]\[ = \sec \theta - \tan \theta \][/tex]

Thus, the second identity is proven:

[tex]\[ \frac{\cos \theta}{1 + \sin \theta} = \sec \theta - \tan \theta \][/tex]

In conclusion, we've proven both identities:

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