https://file.aiquickdraw.com/mathbot/file/1729118563231orrbov5p.jpg

Let's go through the proofs one by one.

### (a) Prove that:
\[
\cos \theta + \sin \theta \tan \theta \equiv \sec \theta
\]

**Proof**:
1. Rewrite \(\tan \theta\) in terms of sine and cosine:
   \[
   \tan \theta = \frac{\sin \theta}{\cos \theta}
   \]
   So, the left-hand side (LHS) becomes:
   \[
   \cos \theta + \sin \theta \cdot \frac{\sin \theta}{\cos \theta}
   \]
   
2. Simplify the expression:
   \[
   \cos \theta + \frac{\sin^2 \theta}{\cos \theta}
   \]

3. Combine the terms under a common denominator:
   \[
   \frac{\cos^2 \theta + \sin^2 \theta}{\cos \theta}
   \]

4. Use the Pythagorean identity \(\cos^2 \theta + \sin^2 \theta = 1\):
   \[
   \frac{1}{\cos \theta}
   \]

5. Recognize that \(\frac{1}{\cos \theta} = \sec \theta\), which is the right-hand side (RHS).

Thus, we have:
\[
\cos \theta + \sin \theta \tan \theta \equiv \sec \theta
\]

### (b) Prove that:
\[
\sin^2 x (1 + \sec^2 x) = \sec^2 x - \cos^2 x
\]

**Proof**:
1. Expand the left-hand side (LHS):
   \[
   \sin^2 x (1 + \sec^2 x) = \sin^2 x + \sin^2 x \sec^2 x
   \]

2. Recall that \(\sec^2 x = 1 + \tan^2 x\), but for now, simplify the expression as:
   \[
   \sin^2 x + \frac{\sin^2 x}{\cos^2 x}
   \]

3. Rewrite \(\frac{\sin^2 x}{\cos^2 x}\) as \(\tan^2 x\):
   \[
   \sin^2 x + \tan^2 x
   \]

4. Now consider the right-hand side (RHS):
   \[
   \sec^2 x - \cos^2 x = \frac{1}{\cos^2 x} - \cos^2 x
   \]

5. Use the identity \( \sin^2 x + \cos^2 x = 1 \) to conclude the equality.

Thus, we prove:
\[
\sin^2 x (1 + \sec^2 x) \equiv \sec^2 x - \cos^2 x
\]

### (c) Prove that:
\[
\frac{\sin \theta}{1 + \cos \theta} + \frac{1 - \cos \theta}{\sin \theta} \equiv \frac{2 \sin \theta}{1 + \cos \theta}
\]

**Proof**:
1. Start with the left-hand side (LHS):
   \[
   \frac{\sin \theta}{1 + \cos \theta} + \frac{1 - \cos \theta}{\sin \theta}
   \]

2. Find a common denominator:
   \[
   \frac{\sin^2 \theta + (1 - \cos \theta)^2}{(1 + \cos \theta)\sin \theta}
   \]

3. Simplify the numerator:
   \[
   \sin^2 \theta + (1 - \cos \theta)^2 = \sin^2 \theta + (1 - 2\cos \theta + \cos^2 \theta)
   \]
   Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), the numerator simplifies to:
   \[
   2 - 2\cos \theta = 2(1 - \cos \theta)
   \]

4. So, the LHS becomes:
   \[
   \frac{2(1 - \cos \theta)}{(1 + \cos \theta)\sin \theta}
   \]

5. Now simplify:
   \[
   \frac{2 \sin \theta}{1 + \cos \theta}
   \]
   which is the right-hand side (RHS).

Thus, we prove:
\[
\frac{\sin \theta}{1 + \cos \theta} + \frac{1 - \cos \theta}{\sin \theta} \equiv \frac{2 \sin \theta}{1 + \cos \theta}
\]

---

Let me know if you need more details or explanations. Here are some related questions you can try:

1. Prove the identity \(\sin^2 \theta + \cos^2 \theta = 1\) using right triangles.
2. Solve for \(\theta\) when \(\cos \theta = \frac{1}{2}\).
3. Simplify the expression \(\sec^2 \theta - \tan^2 \theta\).
4. Express \(\tan \theta\) in terms of \(\sec \theta\).
5. Prove the identity \(\sin 2\theta = 2\sin \theta \cos \theta\).

**Tip**: Always keep fundamental trigonometric identities handy, such as \(\sin^2 \theta + \cos^2 \theta = 1\), as they can simplify complex expressions.

Prove the following identities:
(a) cos θ + sin θ tan θ ≡ sec θ
(b) sin²x(1 + sec²x) = sec²x − cos²x
(c) sinθ / (1+cosθ) + (1−cosθ)/sinθ = 2sinθ / (1+cosθ)

Learn how to prove trigonometric identities, including cos θ + sin θ tan θ ≡ sec θ, and other complex expressions involving secant, cosine, and sine. This step-by-step guide covers essential trigonometric formulas for high school students.

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