Rewrite in Sine and Cosine

Simplify $\dfrac{\tan x}{\sec x}$.

Rewrite tangent and secant in sine and cosine: $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$ = $\dfrac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}}$

Divide by a fraction: multiply by its reciprocal: $\frac{\sin x}{\cos x} \cdot \frac{\cos x}{1}$ = $\frac{\sin x \cos x}{\cos x}$

Cancel the common factor $\cos x$: $\frac{\sin x \cancel{\cos x}}{\cancel{\cos x}}$ = $\sin x$

Answer: $\dfrac{\tan x}{\sec x} = \sin x$

Common Factor and a Pythagorean Identity

Simplify $\cos^2 x \tan^2 x + \cos^2 x$.

Both terms contain $\cos^2 x$: factor it out: $\cos^2 x \tan^2 x + \cos^2 x = \cos^2 x(\tan^2 x + 1)$ = $\cos^2 x(\tan^2 x + 1)$

Use the Pythagorean identity $1 + \tan^2 x = \sec^2 x$: $\cos^2 x \cdot \sec^2 x$ = $\cos^2 x \sec^2 x$

Use the reciprocal identity $\sec^2 x = \frac{1}{\cos^2 x}$: $\cos^2 x \cdot \frac{1}{\cos^2 x}$ = $1$

Answer: $\cos^2 x \tan^2 x + \cos^2 x = 1$ (for every $x$ where $\tan x$ is defined)

Difference of Squares

Simplify $\dfrac{\cos^2 x - \sin^2 x}{\cos x - \sin x}$.

Recognize the numerator as a difference of squares $a^2 - b^2$: $a = \cos x$, $b = \sin x$ = $\cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$

Write the factored numerator over the denominator: $\dfrac{(\cos x - \sin x)(\cos x + \sin x)}{\cos x - \sin x}$ = The factor $\cos x - \sin x$ appears above and below

Cancel the common factor: Allowed because it is a factor of the whole numerator = $\cos x + \sin x$

Answer: $\dfrac{\cos^2 x - \sin^2 x}{\cos x - \sin x} = \cos x + \sin x$ (where $\cos x \neq \sin x$)

Adding Fractions with a Common Denominator

Simplify $\dfrac{1}{1 + \cos x} + \dfrac{1}{1 - \cos x}$.

Use the common denominator $(1 + \cos x)(1 - \cos x)$: $\dfrac{(1 - \cos x) + (1 + \cos x)}{(1 + \cos x)(1 - \cos x)}$ = One fraction

Simplify the numerator: $1 - \cos x + 1 + \cos x = 2$ = $\dfrac{2}{(1 + \cos x)(1 - \cos x)}$

Multiply out the denominator (difference of squares): $(1 + \cos x)(1 - \cos x) = 1 - \cos^2 x$ = $\dfrac{2}{1 - \cos^2 x}$

Use the Pythagorean identity $1 - \cos^2 x = \sin^2 x$: $\dfrac{2}{\sin^2 x}$ = $\dfrac{2}{\sin^2 x}$

Use the reciprocal identity $\frac{1}{\sin^2 x} = \csc^2 x$: $2 \cdot \frac{1}{\sin^2 x}$ = $2\csc^2 x$

Answer: $\dfrac{1}{1 + \cos x} + \dfrac{1}{1 - \cos x} = 2\csc^2 x$

Checking a Simplification with an Angle

A classmate says $\dfrac{\sin x + \tan x}{\tan x} = \cos x + 1$. Check it at $x = 60°$, then confirm it with algebra.

Evaluate the left side at $x = 60°$: $\dfrac{\frac{\sqrt{3}}{2} + \sqrt{3}}{\sqrt{3}} = \frac{1}{2} + 1$ = $1.5$

Evaluate the right side at $x = 60°$: $\cos 60° + 1 = 0.5 + 1$ = $1.5$

Same value, so the claim passes the test. One angle cannot prove it, so do the algebra: split the fraction: $\dfrac{\sin x}{\tan x} + \dfrac{\tan x}{\tan x} = \sin x \cdot \frac{\cos x}{\sin x} + 1$ = $\cos x + 1$

Answer: The simplification is correct: $\dfrac{\sin x + \tan x}{\tan x} = \cos x + 1$. Had the two values at $60°$ been different, it would have been wrong for sure.

Mistake: Cancelling a term instead of a factor: $\dfrac{\sin x + \cos x}{\sin x} = \cos x$

Why: Only factors of the WHOLE numerator and denominator cancel. Here $\sin x$ is just one term of the sum $\sin x + \cos x$. At $x = 45°$ the left side is $2$, while $\cos 45° \approx 0.71$.

Correct: Split the fraction: $\dfrac{\sin x}{\sin x} + \dfrac{\cos x}{\sin x} = 1 + \cot x$.

Mistake: Dropping the squares: writing $1 - \sin x = \cos x$

Why: The Pythagorean identity is about squares: $\sin^2 x + \cos^2 x = 1$. Without the squares it is false: at $x = 30°$, $1 - \sin 30° = 0.5$ but $\cos 30° \approx 0.87$.

Correct: Only $1 - \sin^2 x = \cos^2 x$ is true. If you see $1 - \sin x$, think of the difference of squares or a conjugate instead.

Mistake: Adding fractions by adding denominators: $\dfrac{1}{\sin x} + \dfrac{1}{\cos x} = \dfrac{2}{\sin x + \cos x}$

Why: Trig fractions follow the same rules as number fractions, and $\frac{1}{2} + \frac{1}{3}$ is not $\frac{2}{5}$.

Correct: Use the common denominator $\sin x \cos x$: $\dfrac{\cos x + \sin x}{\sin x \cos x}$.

Mistake: Trusting a check at one special angle such as $45°$

Why: At $x = 45°$ we have $\sin x = \cos x$, so many wrong simplifications (like $\sin x\tan x = \cos x$) give the same number there by accident.

Correct: Check with one or two "ordinary" angles, for example $20°$ and $50°$. A difference proves a mistake; equal values only make the result likely, the algebra proves it.

Engineering Formulas

Engineers often get force and stress formulas as long trigonometric expressions after combining several parts. Simplifying first makes the numbers easy and reduces errors.

A force on a slanted beam comes out as $F = 50(1 - \cos^2\theta)\csc\theta$ newtons. Since $(1 - \cos^2\theta)\csc\theta = \sin\theta$, this is just $F = 50\sin\theta$.

Faster Code in Games and Graphics

Game engines compute trig expressions millions of times per second. Every function call saved by simplifying makes the program faster.

A rotation routine computes $\dfrac{\sin x}{\tan x}$ in every frame. Since $\dfrac{\sin x}{\tan x} = \sin x \cdot \dfrac{\cos x}{\sin x} = \cos x$, two function calls and a division become one call.

Ramps and Slopes in Construction

Builders and surveyors work with the rise of a ramp over a horizontal distance. Spreadsheet formulas built from several steps often hide a simple tangent.

A spreadsheet computes the rise of a ramp as $h = d\sin x\sec x$, where $d$ is the horizontal run. Because $\sin x \sec x = \frac{\sin x}{\cos x} = \tan x$, the formula is really $h = d\tan x$.

Q: How do I know when an expression is fully simplified?

A: Usually when it has as few terms and as few different functions as possible, with no fraction inside a fraction. If the task names a target ("in terms of $\sin x$", "as one function"), stop when you reach that form.

Q: Should I always rewrite everything in sine and cosine?

A: It is the safest general strategy, because then only two functions are left and the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ applies. Sometimes a shortcut is faster, for example $\sec^2 x - 1 = \tan^2 x$ directly.

Q: Is checking with a calculator enough?

A: It is a great way to catch mistakes: if the two forms give different values at the same angle, something is wrong. But equal values at one or two angles do not prove the simplification, the algebra steps do.

Q: Why do some simplified results come with conditions like "where $\cos x \neq \sin x$"?

A: Cancelling a factor is only allowed where that factor is not zero. The original expression is undefined there, while the simplified one may not be, so strictly the two agree everywhere except at those angles.