Trigonometric identities worksheet featuring eight problems to prove using trigonometric functions and identities.

Let's go through each trigonometric identity one by one and prove them step-by-step using fundamental trigonometric identities.

---

1. $ \sec x - \tan x \sin x = \frac{1}{\sec x} $

#### Step 1: Recall identities:
- $ \sec x = \frac{1}{\cos x} $
- $ \tan x = \frac{\sin x}{\cos x} $

#### Left-hand side (LHS):
$$
\sec x - \tan x \sin x = \frac{1}{\cos x} - \left( \frac{\sin x}{\cos x} \right) \cdot \sin x
= \frac{1}{\cos x} - \frac{\sin^2 x}{\cos x}
= \frac{1 - \sin^2 x}{\cos x}
$$

Now use identity: $ 1 - \sin^2 x = \cos^2 x $
$$
= \frac{\cos^2 x}{\cos x} = \cos x
$$

But right-hand side (RHS): $ \frac{1}{\sec x} = \cos x $

✔ So, LHS = RHS → Identity is proven.

---

2. $ \frac{1 + \cos x}{\sin x} = \csc x + \cot x $

#### Right-hand side (RHS):
Recall:
- $ \csc x = \frac{1}{\sin x} $
- $ \cot x = \frac{\cos x}{\sin x} $

So,
$$
\csc x + \cot x = \frac{1}{\sin x} + \frac{\cos x}{\sin x} = \frac{1 + \cos x}{\sin x}
$$

Which is exactly the left-hand side.

✔ Proven.

---

3. $ \frac{\sec \theta \sin \theta}{\tan \theta + \cot \theta} = \sin^2 \theta $

#### Start with LHS:

Express everything in terms of $ \sin $ and $ \cos $:

- $ \sec \theta = \frac{1}{\cos \theta} $
- $ \tan \theta = \frac{\sin \theta}{\cos \theta} $
- $ \cot \theta = \frac{\cos \theta}{\sin \theta} $

So numerator:
$$
\sec \theta \sin \theta = \frac{1}{\cos \theta} \cdot \sin \theta = \frac{\sin \theta}{\cos \theta}
$$

Denominator:
$$
\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}
= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}
= \frac{1}{\sin \theta \cos \theta} \quad (\text{since } \sin^2 + \cos^2 = 1)
$$

Now put together:
$$
\frac{\frac{\sin \theta}{\cos \theta}}{\frac{1}{\sin \theta \cos \theta}} = \frac{\sin \theta}{\cos \theta} \cdot \sin \theta \cos \theta = \sin \theta \cdot \sin \theta = \sin^2 \theta
$$

✔ Proven.

---

4. $ \frac{\sec \theta}{\cos \theta} - \frac{\tan \theta}{\cot \theta} = 1 $

#### Simplify each term:

- $ \frac{\sec \theta}{\cos \theta} = \frac{1/\cos \theta}{\cos \theta} = \frac{1}{\cos^2 \theta} $
- $ \frac{\tan \theta}{\cot \theta} = \frac{\tan \theta}{1/\tan \theta} = \tan^2 \theta $

So expression becomes:
$$
\frac{1}{\cos^2 \theta} - \tan^2 \theta
$$

But $ \frac{1}{\cos^2 \theta} = \sec^2 \theta $, and $ \tan^2 \theta = \sec^2 \theta - 1 $

So:
$$
\sec^2 \theta - (\sec^2 \theta - 1) = \sec^2 \theta - \sec^2 \theta + 1 = 1
$$

✔ Proven.

---

5. $ \cos^2 y - \sin^2 y = 1 - 2\sin^2 y $

#### Start from left-hand side:

Use identity: $ \cos^2 y = 1 - \sin^2 y $

So:
$$
\cos^2 y - \sin^2 y = (1 - \sin^2 y) - \sin^2 y = 1 - 2\sin^2 y
$$

✔ Proven.

---

6. $ \csc^2 \theta \tan^2 \theta - 1 = \tan^2 \theta $

#### Work on LHS:

Recall:
- $ \csc^2 \theta = \frac{1}{\sin^2 \theta} $
- $ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} $

So:
$$
\csc^2 \theta \tan^2 \theta = \left( \frac{1}{\sin^2 \theta} \right) \left( \frac{\sin^2 \theta}{\cos^2 \theta} \right) = \frac{1}{\cos^2 \theta} = \sec^2 \theta
$$

So:
$$
\csc^2 \theta \tan^2 \theta - 1 = \sec^2 \theta - 1 = \tan^2 \theta
$$

✔ Proven.

---

7. $ \frac{\sec^2 \theta}{\sec^2 \theta - 1} = \csc^2 \theta $

#### Simplify denominator:

We know:
- $ \sec^2 \theta - 1 = \tan^2 \theta $

So:
$$
\frac{\sec^2 \theta}{\sec^2 \theta - 1} = \frac{\sec^2 \theta}{\tan^2 \theta}
$$

Now write in terms of sin and cos:
- $ \sec^2 \theta = \frac{1}{\cos^2 \theta} $
- $ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} $

So:
$$
\frac{1/\cos^2 \theta}{\sin^2 \theta / \cos^2 \theta} = \frac{1}{\sin^2 \theta} = \csc^2 \theta
$$

✔ Proven.

---

8. $ \tan^2 x \sin^2 x = \tan^2 x - \sin^2 x $

#### Start with LHS:
$$
\tan^2 x \sin^2 x = \left( \frac{\sin^2 x}{\cos^2 x} \right) \cdot \sin^2 x = \frac{\sin^4 x}{\cos^2 x}
$$

Now look at RHS: $ \tan^2 x - \sin^2 x = \frac{\sin^2 x}{\cos^2 x} - \sin^2 x $

Factor $ \sin^2 x $:
$$
= \sin^2 x \left( \frac{1}{\cos^2 x} - 1 \right) = \sin^2 x \left( \frac{1 - \cos^2 x}{\cos^2 x} \right)
= \sin^2 x \left( \frac{\sin^2 x}{\cos^2 x} \right) = \frac{\sin^4 x}{\cos^2 x}
$$

So both sides equal $ \frac{\sin^4 x}{\cos^2 x} $

✔ Proven.

---

✔ Final Answer:

All eight identities are proven using standard trigonometric identities and algebraic manipulation.

Let me know if you'd like a written version for your worksheet!