Trig Identities worksheet 3.3 featuring eight problems to prove using trigonometric identities.

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

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1. $\csc\theta = \frac{\cot\theta}{\cos\theta}$

Proof:

Start with the right-hand side (RHS):

$$
\frac{\cot\theta}{\cos\theta} = \frac{\frac{\cos\theta}{\sin\theta}}{\cos\theta} = \frac{\cos\theta}{\sin\theta} \cdot \frac{1}{\cos\theta} = \frac{1}{\sin\theta} = \csc\theta
$$

✔ Proven.

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2. $\frac{1}{\sec^2 x} + \frac{1}{\csc^2 x} = 1$

Recall:
- $\sec x = \frac{1}{\cos x} \Rightarrow \sec^2 x = \frac{1}{\cos^2 x} \Rightarrow \frac{1}{\sec^2 x} = \cos^2 x$
- $\csc x = \frac{1}{\sin x} \Rightarrow \csc^2 x = \frac{1}{\sin^2 x} \Rightarrow \frac{1}{\csc^2 x} = \sin^2 x$

So:
$$
\frac{1}{\sec^2 x} + \frac{1}{\csc^2 x} = \cos^2 x + \sin^2 x = 1
$$

✔ Proven.

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3. $\csc^2 y \tan^2 y - 1 = \tan^2 y$

Start with LHS:

$$
\csc^2 y \tan^2 y - 1
$$

We know:
- $\csc^2 y = 1 + \cot^2 y$, but better to use definitions.
- $\csc y = \frac{1}{\sin y} \Rightarrow \csc^2 y = \frac{1}{\sin^2 y}$
- $\tan y = \frac{\sin y}{\cos y} \Rightarrow \tan^2 y = \frac{\sin^2 y}{\cos^2 y}$

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

Then:
$$
\csc^2 y \tan^2 y - 1 = \sec^2 y - 1 = \tan^2 y
$$

✔ Proven.

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

Simplify each term:

- $\sec\theta = \frac{1}{\cos\theta}$, so $\frac{\sec\theta}{\cos\theta} = \frac{1/\cos\theta}{\cos\theta} = \frac{1}{\cos^2\theta} = \sec^2\theta$

- $\tan\theta = \frac{\sin\theta}{\cos\theta}$, $\cot\theta = \frac{\cos\theta}{\sin\theta}$, so:
$$
\frac{\tan\theta}{\cot\theta} = \frac{\frac{\sin\theta}{\cos\theta}}{\frac{\cos\theta}{\sin\theta}} = \frac{\sin\theta}{\cos\theta} \cdot \frac{\sin\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta
$$

Now:
$$
\frac{\sec\theta}{\cos\theta} - \frac{\tan\theta}{\cot\theta} = \sec^2\theta - \tan^2\theta
$$

But we know: $\sec^2\theta = 1 + \tan^2\theta$, so:
$$
\sec^2\theta - \tan^2\theta = (1 + \tan^2\theta) - \tan^2\theta = 1
$$

✔ Proven.

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5. $\csc^4 x - \cot^4 x = \csc^2 x + \cot^2 x$

Left-hand side (LHS): difference of squares

$$
\csc^4 x - \cot^4 x = (\csc^2 x)^2 - (\cot^2 x)^2 = (\csc^2 x - \cot^2 x)(\csc^2 x + \cot^2 x)
$$

Recall identity: $\csc^2 x = 1 + \cot^2 x$, so:
$$
\csc^2 x - \cot^2 x = (1 + \cot^2 x) - \cot^2 x = 1
$$

Therefore:
$$
(\csc^2 x - \cot^2 x)(\csc^2 x + \cot^2 x) = (1)(\csc^2 x + \cot^2 x) = \csc^2 x + \cot^2 x
$$

✔ Proven.

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6. $\sec^4 y - \tan^4 y = \tan^2 y + \sec^2 y$

Again, factor LHS as difference of squares:

$$
\sec^4 y - \tan^4 y = (\sec^2 y)^2 - (\tan^2 y)^2 = (\sec^2 y - \tan^2 y)(\sec^2 y + \tan^2 y)
$$

We know: $\sec^2 y = 1 + \tan^2 y$, so:
$$
\sec^2 y - \tan^2 y = (1 + \tan^2 y) - \tan^2 y = 1
$$

Thus:
$$
(\sec^2 y - \tan^2 y)(\sec^2 y + \tan^2 y) = 1 \cdot (\sec^2 y + \tan^2 y) = \sec^2 y + \tan^2 y
$$

But the RHS is $\tan^2 y + \sec^2 y$, which is the same.

✔ Proven.

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7. $(1 - \tan\theta)^2 = \sec^2\theta - 2\tan\theta$

Expand the left-hand side:

$$
(1 - \tan\theta)^2 = 1 - 2\tan\theta + \tan^2\theta
$$

Now recall: $\sec^2\theta = 1 + \tan^2\theta$, so:
$$
\sec^2\theta - 2\tan\theta = (1 + \tan^2\theta) - 2\tan\theta = 1 - 2\tan\theta + \tan^2\theta
$$

Which matches the expanded LHS.

✔ Proven.

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8. $(1 - \sin^2 x)(1 + \tan^2 x) = 1$

Use identities:

- $1 - \sin^2 x = \cos^2 x$
- $1 + \tan^2 x = \sec^2 x$

So:
$$
(1 - \sin^2 x)(1 + \tan^2 x) = \cos^2 x \cdot \sec^2 x
$$

But $\sec x = \frac{1}{\cos x} \Rightarrow \sec^2 x = \frac{1}{\cos^2 x}$, so:
$$
\cos^2 x \cdot \frac{1}{\cos^2 x} = 1
$$

✔ Proven.

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✔ Final Answer:

All eight identities are proven using basic trigonometric identities such as:

- $\sin^2 x + \cos^2 x = 1$
- $\tan x = \frac{\sin x}{\cos x}$
- $\cot x = \frac{1}{\tan x}$
- $\sec x = \frac{1}{\cos x}$
- $\csc x = \frac{1}{\sin x}$
- $\sec^2 x = 1 + \tan^2 x$
- $\csc^2 x = 1 + \cot^2 x$

Each identity has been verified algebraically.

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