SAT Prep trigonometry practice sheet with 18 identities to prove.

Let's solve and prove each of the given trigonometric identities step by step. We'll use fundamental trigonometric identities such as:

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

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1. $\tan x \sin x + \cos x = \sec x$

Left-hand side (LHS):

$$
\tan x \sin x + \cos x = \left(\frac{\sin x}{\cos x}\right)\sin x + \cos x = \frac{\sin^2 x}{\cos x} + \cos x
$$

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

✔ Proven.

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

LHS:

$$
\frac{1}{\tan x} + \tan x = \cot x + \tan x = \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}
$$

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

✔ Proven.

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3. $\sin x - \sin x \cos^2 x = \sin^3 x$

Factor LHS:

$$
\sin x(1 - \cos^2 x) = \sin x(\sin^2 x) = \sin^3 x
$$

✔ Proven.

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4. $\frac{\cos \alpha}{1 + \sin \alpha} + \frac{1 + \sin \alpha}{\cos \alpha} = 2 \sec \alpha$

Let’s combine the two terms:

$$
\frac{\cos \alpha}{1 + \sin \alpha} + \frac{1 + \sin \alpha}{\cos \alpha} = \frac{\cos^2 \alpha + (1 + \sin \alpha)^2}{(1 + \sin \alpha)\cos \alpha}
$$

Numerator:

$$
\cos^2 \alpha + 1 + 2\sin \alpha + \sin^2 \alpha = (\cos^2 \alpha + \sin^2 \alpha) + 1 + 2\sin \alpha = 1 + 1 + 2\sin \alpha = 2(1 + \sin \alpha)
$$

So,

$$
\frac{2(1 + \sin \alpha)}{(1 + \sin \alpha)\cos \alpha} = \frac{2}{\cos \alpha} = 2 \sec \alpha
$$

✔ Proven.

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5. $\frac{\cos x}{1 - \sin x} - \frac{\cos x}{1 + \sin x} = 2 \tan x$

Common denominator: $(1 - \sin x)(1 + \sin x) = 1 - \sin^2 x = \cos^2 x$

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

$$
= \frac{\cos x(2\sin x)}{\cos^2 x} = \frac{2\sin x}{\cos x} = 2 \tan x
$$

✔ Proven.

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6. $\cos^2 x = \frac{\csc x \cos x}{\tan x + \cot x}$

First, simplify RHS.

We know:
- $\csc x = \frac{1}{\sin x}$
- $\tan x = \frac{\sin x}{\cos x}, \quad \cot x = \frac{\cos x}{\sin x}$

So numerator: $\csc x \cos x = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x}$

Denominator: $\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}$

Now:

$$
\text{RHS} = \frac{\frac{\cos x}{\sin x}}{\frac{1}{\sin x \cos x}} = \frac{\cos x}{\sin x} \cdot \sin x \cos x = \cos x \cdot \cos x = \cos^2 x
$$

✔ Proven.

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

Note: $a^4 - b^4 = (a^2 - b^2)(a^2 + b^2)$

So:

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

✔ Proven.

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8. $\frac{\tan^2 x}{\tan^2 x + 1} = \sin^2 x$

Recall: $1 + \tan^2 x = \sec^2 x$, so:

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

✔ Proven.

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9. $\frac{1 - \sin x}{\cos x} = \frac{\cos x}{1 + \sin x}$

Cross-multiply to verify:

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

Right: $\cos x \cdot \cos x = \cos^2 x$

So both sides equal when cross-multiplied → identity holds.

Alternatively, multiply LHS by $\frac{1 + \sin x}{1 + \sin x}$:

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

✔ Proven.

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10. $1 - 2\cos^2 x = \frac{\tan^2 x - 1}{\tan^2 x + 1}$

Start with RHS:

Let $t = \tan^2 x$. Then:

$$
\frac{t - 1}{t + 1}
$$

But we want to express in terms of $\cos x$.

Use identity: $\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{1 - \cos^2 x}{\cos^2 x}$

So:

$$
\tan^2 x = \frac{1 - \cos^2 x}{\cos^2 x}
$$

Plug into RHS:

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

✔ Proven.

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

Start from RHS:

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

But $\sec^2 \theta - 1 = \tan^2 \theta$

So RHS = $\tan^2 \theta$

✔ Proven.

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

Multiply RHS by $\frac{1 + \sin x}{1 + \sin x}$:

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

Now:

$$
\frac{1 + \sin x}{\cos x} = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \sec x + \tan x
$$

✔ Proven.

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13. $\frac{\csc \beta}{\sin \beta} - \frac{\cot \beta}{\tan \beta} = 1$

Simplify each term:

- $\frac{\csc \beta}{\sin \beta} = \frac{1/\sin \beta}{\sin \beta} = \frac{1}{\sin^2 \beta}$
- $\frac{\cot \beta}{\tan \beta} = \frac{\frac{\cos \beta}{\sin \beta}}{\frac{\sin \beta}{\cos \beta}} = \frac{\cos^2 \beta}{\sin^2 \beta}$

So:

$$
\frac{1}{\sin^2 \beta} - \frac{\cos^2 \beta}{\sin^2 \beta} = \frac{1 - \cos^2 \beta}{\sin^2 \beta} = \frac{\sin^2 \beta}{\sin^2 \beta} = 1
$$

✔ Proven.

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14. $\sin^4 x - \cos^4 x = 1 - 2\cos^2 x$

As in #7:

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

But $\sin^2 x = 1 - \cos^2 x$, so:

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

✔ Proven.

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15. $(\sin x - \cos x)^2 + (\sin x + \cos x)^2 = 2$

Expand both squares:

First: $(\sin x - \cos x)^2 = \sin^2 x - 2\sin x \cos x + \cos^2 x$

Second: $(\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x$

Add:

$$
(\sin^2 x + \cos^2 x) + (-2\sin x \cos x + 2\sin x \cos x) + (\sin^2 x + \cos^2 x) = 1 + 0 + 1 = 2
$$

✔ Proven.

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16. $\frac{\sin^2 x + 4\sin x + 3}{\cos^2 x} = \frac{3 + \sin x}{1 - \sin x}$

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

Numerator: $\sin^2 x + 4\sin x + 3 = (\sin x + 1)(\sin x + 3)$

Wait — factor:

$$
\sin^2 x + 4\sin x + 3 = (\sin x + 1)(\sin x + 3)
$$

But let’s try another approach.

Let $s = \sin x$, then:

LHS: $\frac{s^2 + 4s + 3}{1 - s^2} = \frac{(s+1)(s+3)}{(1 - s)(1 + s)} = \frac{s+3}{1 - s}$

Because $(s+1)/(1+s) = 1$, so cancels.

So:

$$
\frac{s+3}{1 - s} = \frac{3 + s}{1 - s} = \frac{3 + \sin x}{1 - \sin x}
$$

✔ Proven.

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

We already did a similar one earlier.

Recall from #9: $\frac{\cos x}{1 - \sin x} = \frac{1 + \sin x}{\cos x}$

So:

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

✔ Proven.

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18. $\tan^2 x + 1 + \tan x \sec x = \frac{1 + \sin x}{\cos^2 x}$

Left side: $\tan^2 x + 1 = \sec^2 x$, so:

$$
\sec^2 x + \tan x \sec x
$$

Factor: $\sec x (\sec x + \tan x)$

Now recall from #12: $\sec x + \tan x = \frac{1 + \sin x}{\cos x}$

So:

$$
\sec x (\sec x + \tan x) = \frac{1}{\cos x} \cdot \frac{1 + \sin x}{\cos x} = \frac{1 + \sin x}{\cos^2 x}
$$

✔ Proven.

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## ✔ All 18 identities are proven!

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