7-1 Simplifying Trig Functions Worksheet | PDF - Free Printable

Let’s solve each problem one by one. We’ll use basic trig identities to simplify.

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1. $\frac{\tan^2 x + 1}{1 + \cot^2 x}$

We know:
- $\tan^2 x + 1 = \sec^2 x$
- $1 + \cot^2 x = \csc^2 x$

So expression becomes:
$\frac{\sec^2 x}{\csc^2 x} = \frac{1/\cos^2 x}{1/\sin^2 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x$

✔ Answer: $\tan^2 x$

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2. $\frac{1}{\sec x - \tan x} - \frac{1}{\sec x + \tan x}$

Common denominator: $(\sec x - \tan x)(\sec x + \tan x) = \sec^2 x - \tan^2 x = 1$ (identity!)

Numerator: $(\sec x + \tan x) - (\sec x - \tan x) = 2\tan x$

So overall: $\frac{2\tan x}{1} = 2\tan x$

✔ Answer: $2\tan x$

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3. $\sec x \tan x \cos x$

Rewrite: $\frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} \cdot \cos x = \frac{\sin x}{\cos x} = \tan x$

✔ Answer: $\tan x$

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4. $\sin^2 x \cot x \csc x$

Write in sin/cos:
$\sin^2 x \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = \sin^2 x \cdot \frac{\cos x}{\sin^2 x} = \cos x$

✔ Answer: $\cos x$

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5. $\frac{1 - \cos^2 t}{\sin^2 t}$

But $1 - \cos^2 t = \sin^2 t$, so:
$\frac{\sin^2 t}{\sin^2 t} = 1$

✔ Answer: $1$

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6. $\frac{\tan^2 x}{1 - \sec^2 x}$

Note: $1 - \sec^2 x = -(\sec^2 x - 1) = -\tan^2 x$

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

✔ Answer: $-1$

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7. $\tan^2 x (\csc^2 x - 1)$

We know: $\csc^2 x - 1 = \cot^2 x$

So: $\tan^2 x \cdot \cot^2 x = \tan^2 x \cdot \frac{1}{\tan^2 x} = 1$

✔ Answer: $1$

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8. $\frac{\cos^2 x}{1 - \cos^2 x}$

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

So: $\frac{\cos^2 x}{\sin^2 x} = \cot^2 x$

✔ Answer: $\cot^2 x$

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9. $\frac{\sec^2 x - 1}{\tan x}$

$\sec^2 x - 1 = \tan^2 x$, so:
$\frac{\tan^2 x}{\tan x} = \tan x$

✔ Answer: $\tan x$

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10. $\frac{\cos^2 x - 1}{\sin^2 x - 1}$

Numerator: $\cos^2 x - 1 = -\sin^2 x$
Denominator: $\sin^2 x - 1 = -\cos^2 x$

So: $\frac{-\sin^2 x}{-\cos^2 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x$

✔ Answer: $\tan^2 x$

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11. $\cos x (\sec x - \cos x)$

Distribute: $\cos x \cdot \sec x - \cos x \cdot \cos x = 1 - \cos^2 x = \sin^2 x$

✔ Answer: $\sin^2 x$

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12. $\cot x (\tan x + \cot x)$

Distribute: $\cot x \cdot \tan x + \cot x \cdot \cot x = 1 + \cot^2 x = \csc^2 x$

✔ Answer: $\csc^2 x$

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13. $\frac{\tan x + \cot x}{\cot x}$

Split: $\frac{\tan x}{\cot x} + \frac{\cot x}{\cot x} = \tan x \cdot \tan x + 1 = \tan^2 x + 1 = \sec^2 x$

Wait — let's do it carefully:

$\frac{\tan x}{\cot x} = \tan x \div \frac{1}{\tan x} = \tan^2 x$

And $\frac{\cot x}{\cot x} = 1$

So total: $\tan^2 x + 1 = \sec^2 x$

✔ Answer: $\sec^2 x$

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14. $\frac{\tan x}{\tan x + \cot x}$

Multiply numerator and denominator by $\tan x$:

Numerator: $\tan^2 x$

Denominator: $\tan^2 x + 1 = \sec^2 x$

So: $\frac{\tan^2 x}{\sec^2 x} = \frac{\sin^2 x / \cos^2 x}{1 / \cos^2 x} = \sin^2 x$

✔ Answer: $\sin^2 x$

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15. $\sec x \cot x - \cot x \cos x$

Factor out $\cot x$: $\cot x (\sec x - \cos x)$

Now, $\sec x - \cos x = \frac{1}{\cos x} - \cos x = \frac{1 - \cos^2 x}{\cos x} = \frac{\sin^2 x}{\cos x}$

So: $\cot x \cdot \frac{\sin^2 x}{\cos x} = \frac{\cos x}{\sin x} \cdot \frac{\sin^2 x}{\cos x} = \sin x$

✔ Answer: $\sin x$

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16. $\sin x \tan x - \csc x \tan x$

Factor out $\tan x$: $\tan x (\sin x - \csc x)$

$\csc x = \frac{1}{\sin x}$, so:
$\tan x \left( \sin x - \frac{1}{\sin x} \right) = \tan x \cdot \frac{\sin^2 x - 1}{\sin x} = \tan x \cdot \frac{-\cos^2 x}{\sin x}$

Now $\tan x = \frac{\sin x}{\cos x}$, so:

$\frac{\sin x}{\cos x} \cdot \frac{-\cos^2 x}{\sin x} = -\cos x$

✔ Answer: $-\cos x$

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17. $\frac{\cot^2 x \cos^2 x}{\cot^2 x - \cos^2 x}$

Write everything in sin/cos:

Numerator: $\frac{\cos^2 x}{\sin^2 x} \cdot \cos^2 x = \frac{\cos^4 x}{\sin^2 x}$

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

So fraction: $\frac{\cos^4 x / \sin^2 x}{\cos^4 x / \sin^2 x} = 1$

✔ Answer: $1$

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18. $\frac{\sin^2 x - \tan^2 x}{\tan^2 x \sin^2 x}$

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

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

Denominator: $\frac{\sin^2 x}{\cos^2 x} \cdot \sin^2 x = \frac{\sin^4 x}{\cos^2 x}$

So overall: $\frac{ -\frac{\sin^4 x}{\cos^2 x} }{ \frac{\sin^4 x}{\cos^2 x} } = -1$

✔ Answer: $-1$

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19. $\frac{(\sin x + \tan x)^2 + \cos^2 x - \sec^2 x}{\tan x}$

First expand $(\sin x + \tan x)^2 = \sin^2 x + 2\sin x \tan x + \tan^2 x$

Add $\cos^2 x$: $\sin^2 x + \cos^2 x + 2\sin x \tan x + \tan^2 x = 1 + 2\sin x \tan x + \tan^2 x$

Subtract $\sec^2 x$: $1 + 2\sin x \tan x + \tan^2 x - \sec^2 x$

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

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

Now divide by $\tan x$: $\frac{2\sin x \tan x}{\tan x} = 2\sin x$

✔ Answer: $2\sin x$

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20. $\frac{2\sin x \cos x + (\sin x - \cos x)^2}{\sec x}$

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

Add to first term: $2\sin x \cos x + \sin^2 x - 2\sin x \cos x + \cos^2 x = \sin^2 x + \cos^2 x = 1$

Divide by $\sec x$: $\frac{1}{\sec x} = \cos x$

✔ Answer: $\cos x$

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

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

So: $\frac{\sin x (1 + \sin x)}{\sin^2 x} = \frac{1 + \sin x}{\sin x} = \frac{1}{\sin x} + 1$

Then subtract 1: $\frac{1}{\sin x} + 1 - 1 = \frac{1}{\sin x} = \csc x$

✔ Answer: $\csc x$

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22. $\frac{1 - \tan^2 x}{1 + \tan^2 x} + 1$

Note: $1 + \tan^2 x = \sec^2 x$

So: $\frac{1 - \tan^2 x}{\sec^2 x} + 1 = (1 - \tan^2 x)\cos^2 x + 1$

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

Because $\tan^2 x \cos^2 x = \sin^2 x$

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

Better: group as $(\cos^2 x - \sin^2 x) + 1 = \cos 2x + 1$? But maybe not simplest.

Wait — alternative approach:

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

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

Yes! Because $1 - \sin^2 x = \cos^2 x$, so:

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

✔ Answer: $2\cos^2 x$

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23. $\frac{\tan x - \tan x \sin^2 x}{2 \sin x \cos x}$

Factor numerator: $\tan x (1 - \sin^2 x) = \tan x \cos^2 x$

So: $\frac{\tan x \cos^2 x}{2 \sin x \cos x} = \frac{\frac{\sin x}{\cos x} \cdot \cos^2 x}{2 \sin x \cos x} = \frac{\sin x \cos x}{2 \sin x \cos x} = \frac{1}{2}$

✔ Answer: $\frac{1}{2}$

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Final Answer:
1. $\tan^2 x$
2. $2\tan x$
3. $\tan x$
4. $\cos x$
5. $1$
6. $-1$
7. $1$
8. $\cot^2 x$
9. $\tan x$
10. $\tan^2 x$
11. $\sin^2 x$
12. $\csc^2 x$
13. $\sec^2 x$
14. $\sin^2 x$
15. $\sin x$
16. $-\cos x$
17. $1$
18. $-1$
19. $2\sin x$
20. $\cos x$
21. $\csc x$
22. $2\cos^2 x$
23. $\frac{1}{2}$