2.1 Prove that $$ \frac{\cos ^{4} x+\sin ^{2} x \cos ^{2} x}{1+\sin x}=1-\sin x $$

Question

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1. Start with the left-hand side (LHS):

$$
\frac{\cos^4 x + \sin^2 x \cos^2 x}{1+\sin x}
$$

2. Factor $$\cos^2 x$$ from the numerator:

$$
\cos^4 x + \sin^2 x \cos^2 x = \cos^2 x \left(\cos^2 x + \sin^2 x\right)
$$

3. Recall the Pythagorean identity:

$$
\cos^2 x + \sin^2 x = 1
$$

Thus, the numerator simplifies to:

$$
\cos^2 x \cdot 1 = \cos^2 x
$$

4. Substitute back into the expression:

$$
\frac{\cos^2 x}{1+\sin x}
$$

5. Multiply the numerator and the denominator by $$1-\sin x$$ (this is safe provided $$1+\sin x \neq 0$$):

$$
\frac{\cos^2 x (1-\sin x)}{(1+\sin x)(1-\sin x)}
$$

6. Recognize the difference of squares in the denominator:

$$
(1+\sin x)(1-\sin x) = 1 - \sin^2 x
$$

7. Use the Pythagorean identity once more:

$$
1-\sin^2 x = \cos^2 x
$$

8. Replace the denominator:

$$
\frac{\cos^2 x (1-\sin x)}{\cos^2 x}
$$

9. Cancel $$\cos^2 x$$ (assuming $$\cos x \neq 0$$, though the cancellation is valid in the regions where the original expression is defined):

$$
1-\sin x
$$

Thus, we have shown that:

$$
\frac{\cos^4 x+\sin^2 x \cos^2 x}{1+\sin x}=1-\sin x
$$
$$ \frac{\cos^4 x + \sin^2 x \cos^2 x}{1 + \sin x} = 1 - \sin x $$

2.1 Prove that \( \frac{\cos ^{4} x+\sin ^{2} x \cos ^{2} x}{1+\sin x}=1-\sin x \)