What Are the Conversion Relations of Trigonometric Ratios?
The conversion relations of trigonometric ratios are the rules for expressing any of the six ratios in terms of any other — sine in terms of cosine, all six in terms of $\tan\theta$, and so on. The point is interconversion: starting from one known ratio and rebuilding the rest, without going back to the triangle's side lengths.
A quick note on the name, because the phrase is used two ways. Some sources use "conversion relations" for angle transformations — turning $\sin(90° + \theta)$ into $\cos\theta$, for instance. This article is about the other, more common classroom meaning: expressing one ratio through another. The angle-transformation idea is covered separately in trigonometric ratios of complementary angles.
The conversions are powered by the basic properties of trigonometric ratios, grouped into three engines.
Reciprocal relations — $\csc\theta = \frac{1}{\sin\theta}$, $\sec\theta = \frac{1}{\cos\theta}$, $\cot\theta = \frac{1}{\tan\theta}$.
Quotient relations — $\tan\theta = \frac{\sin\theta}{\cos\theta}$, $\cot\theta = \frac{\cos\theta}{\sin\theta}$.
Pythagorean identities — $\sin^2\theta + \cos^2\theta = 1$, $1 + \tan^2\theta = \sec^2\theta$, $1 + \cot^2\theta = \csc^2\theta$.
The Full Conversion Table
Here is the complete reference: each of the six ratios written in terms of each base ratio. The table assumes $\theta$ is acute (Quadrant I), so every value is positive; the sign caveat for other quadrants comes after.
In terms of → | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ |
|---|---|---|---|
$\sin\theta$ | $\sin\theta$ | $\sqrt{1 - \cos^2\theta}$ | $\frac{\tan\theta}{\sqrt{1 + \tan^2\theta}}$ |
$\cos\theta$ | $\sqrt{1 - \sin^2\theta}$ | $\cos\theta$ | $\frac{1}{\sqrt{1 + \tan^2\theta}}$ |
$\tan\theta$ | $\frac{\sin\theta}{\sqrt{1 - \sin^2\theta}}$ | $\frac{\sqrt{1 - \cos^2\theta}}{\cos\theta}$ | $\tan\theta$ |
$\csc\theta$ | $\frac{1}{\sin\theta}$ | $\frac{1}{\sqrt{1 - \cos^2\theta}}$ | $\frac{\sqrt{1 + \tan^2\theta}}{\tan\theta}$ |
$\sec\theta$ | $\frac{1}{\sqrt{1 - \sin^2\theta}}$ | $\frac{1}{\cos\theta}$ | $\sqrt{1 + \tan^2\theta}$ |
$\cot\theta$ | $\frac{\sqrt{1 - \sin^2\theta}}{\sin\theta}$ | $\frac{\cos\theta}{\sqrt{1 - \cos^2\theta}}$ | $\frac{1}{\tan\theta}$ |
And the same six ratios in terms of the reciprocal base ratios:
In terms of → | $\csc\theta$ | $\sec\theta$ | $\cot\theta$ |
|---|---|---|---|
$\sin\theta$ | $\frac{1}{\csc\theta}$ | $\frac{\sqrt{\sec^2\theta - 1}}{\sec\theta}$ | $\frac{1}{\sqrt{1 + \cot^2\theta}}$ |
$\cos\theta$ | $\frac{\sqrt{\csc^2\theta - 1}}{\csc\theta}$ | $\frac{1}{\sec\theta}$ | $\frac{\cot\theta}{\sqrt{1 + \cot^2\theta}}$ |
$\tan\theta$ | $\frac{1}{\sqrt{\csc^2\theta - 1}}$ | $\sqrt{\sec^2\theta - 1}$ | $\frac{1}{\cot\theta}$ |
$\sec\theta$ | $\frac{\csc\theta}{\sqrt{\csc^2\theta - 1}}$ | $\sec\theta$ | $\frac{\sqrt{1 + \cot^2\theta}}{\cot\theta}$ |
$\csc\theta$ | $\csc\theta$ | $\frac{\sec\theta}{\sqrt{\sec^2\theta - 1}}$ | $\sqrt{1 + \cot^2\theta}$ |
$\cot\theta$ | $\sqrt{\csc^2\theta - 1}$ | $\frac{1}{\sqrt{\sec^2\theta - 1}}$ | $\cot\theta$ |
Do not memorise the table. Memorise the three engines and rebuild any cell on demand — the next section shows exactly how.
How Do You Convert One Trigonometric Ratio Into Another?
The method is a fixed three-step chain. Suppose you are given $\sin\theta$ and want everything else.
Get cosine from the Pythagorean identity. Since $\sin^2\theta + \cos^2\theta = 1$:
$$\cos\theta = \sqrt{1 - \sin^2\theta}$$
Get tangent from the quotient relation. Now that both sine and cosine are known:
$$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\sin\theta}{\sqrt{1 - \sin^2\theta}}$$
Get the reciprocals by flipping. Each of $\csc\theta$, $\sec\theta$, $\cot\theta$ is $1$ over the matching primary ratio.
$$\csc\theta = \frac{1}{\sin\theta}, \quad \sec\theta = \frac{1}{\sqrt{1 - \sin^2\theta}}, \quad \cot\theta = \frac{\sqrt{1 - \sin^2\theta}}{\sin\theta}$$
The same three steps work from any starting ratio. If you start from $\tan\theta$, use $1 + \tan^2\theta = \sec^2\theta$ to get secant first; if you start from $\cot\theta$, use $1 + \cot^2\theta = \csc^2\theta$. The Pythagorean identity always supplies the "missing partner," and the quotient and reciprocal relations finish the job.
What about the sign?
The Pythagorean step produces a square root, which carries a $\pm$. For acute angles the sign is always positive. For angles beyond $90°$, the sign is fixed by the quadrant the angle lands in — the ASTC rule from the basic properties decides whether to take the $+$ or the $-$. Drop the sign check and a Quadrant II answer comes out wrong.
Examples of Conversion Relations of Trigonometric Ratios
Example 1
Express $\cos\theta$ in terms of $\sin\theta$ (acute angle).
From the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$:
$$\cos^2\theta = 1 - \sin^2\theta$$
$$\cos\theta = \sqrt{1 - \sin^2\theta}$$
Final answer: $\cos\theta = \sqrt{1 - \sin^2\theta}$ (positive, since $\theta$ is acute).
Example 2
Express $\sin\theta$ in terms of $\tan\theta$.
A first instinct is to write $\sin\theta = \tan\theta \cdot \cos\theta$ and stop — but that still contains $\cos\theta$, so it is not yet "in terms of $\tan\theta$." Test whether the half-finished form is acceptable: substitute $\theta = 45°$, where $\tan 45° = 1$ and $\sin 45° = \frac{1}{\sqrt{2}}$. The expression $\tan\theta \cdot \cos\theta = 1 \times \frac{1}{\sqrt{2}}$ only works because we already knew cosine — a genuine conversion must eliminate it.
The correct route uses $1 + \tan^2\theta = \sec^2\theta$, so $\cos\theta = \frac{1}{\sec\theta} = \frac{1}{\sqrt{1 + \tan^2\theta}}$. Then:
$$\sin\theta = \tan\theta \cdot \cos\theta = \tan\theta \cdot \frac{1}{\sqrt{1 + \tan^2\theta}} = \frac{\tan\theta}{\sqrt{1 + \tan^2\theta}}$$
Final answer: $\sin\theta = \frac{\tan\theta}{\sqrt{1 + \tan^2\theta}}$ — now entirely in terms of $\tan\theta$.
Example 3
Express the trigonometric ratios $\sin A$, $\sec A$, and $\tan A$ in terms of $\cot A$. (standard exam question)
Start from $\cot A$ and use $1 + \cot^2 A = \csc^2 A$.
$$\csc A = \sqrt{1 + \cot^2 A}$$
Since $\sin A = \frac{1}{\csc A}$:
$$\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$$
For $\tan A$, use the reciprocal relation directly:
$$\tan A = \frac{1}{\cot A}$$
For $\sec A$, get cosine from $\cos A = \cot A \cdot \sin A = \frac{\cot A}{\sqrt{1 + \cot^2 A}}$, then flip:
$$\sec A = \frac{1}{\cos A} = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$$
Final answer: $\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$, $\tan A = \frac{1}{\cot A}$, $\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$.
Example 4
Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$. (standard exam question)
Use $1 + \tan^2 A = \sec^2 A$, so $\tan A = \sqrt{\sec^2 A - 1}$, and $\cos A = \frac{1}{\sec A}$.
$$\cos A = \frac{1}{\sec A}$$
$$\tan A = \sqrt{\sec^2 A - 1}$$
$$\sin A = \tan A \cdot \cos A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$$
$$\csc A = \frac{1}{\sin A} = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$$
$$\cot A = \frac{1}{\tan A} = \frac{1}{\sqrt{\sec^2 A - 1}}$$
Final answer: the five ratios as written above, each expressed purely in $\sec A$.
Example 5
Given $\tan\theta = \frac{3}{4}$ for an acute angle, find $\sin\theta$ and $\cos\theta$ using conversion.
From $1 + \tan^2\theta = \sec^2\theta$:
$$\sec^2\theta = 1 + \frac{9}{16} = \frac{25}{16} \implies \sec\theta = \frac{5}{4}$$
$$\cos\theta = \frac{1}{\sec\theta} = \frac{4}{5}$$
$$\sin\theta = \tan\theta \cdot \cos\theta = \frac{3}{4} \cdot \frac{4}{5} = \frac{3}{5}$$
Final answer: $\sin\theta = \frac{3}{5}$, $\cos\theta = \frac{4}{5}$ — the familiar $3$-$4$-$5$ triangle, recovered from tangent alone.
Example 6
An angle $\theta$ in Quadrant II has $\sin\theta = \frac{5}{13}$. Convert to find $\cos\theta$ and $\tan\theta$.
The conversion gives the magnitude:
$$\cos\theta = \pm\sqrt{1 - \frac{25}{169}} = \pm\frac{12}{13}$$
Now apply the sign caveat. In Quadrant II cosine is negative, so:
$$\cos\theta = -\frac{12}{13}, \quad \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{5/13}{-12/13} = -\frac{5}{12}$$
Final answer: $\cos\theta = -\frac{12}{13}$, $\tan\theta = -\frac{5}{12}$. The conversion supplies the size; the quadrant fixes the sign.
Why Conversion Is the Skill, Not the Table
The conversions matter because they turn one piece of information into all of it — and because the method is reusable in a way the table is not.
They solve "given one ratio, find another" instantly. This is one of the most common question shapes in introductory courses and beyond, and the three-engine chain answers every version of it.
They are the backbone of identity proofs. To prove an identity, you almost always rewrite everything in terms of $\sin$ and $\cos$ — which is a conversion. The wider toolkit lives in trigonometric identities.
They scale to any quadrant. Once the acute-angle conversions are automatic, extending them is just a sign decision from ASTC — which opens the door to the unit circle and trigonometric functions of any angle.
Where Students Slip on Conversion Relations
Mistake 1: Leaving the answer half-converted
Where it slips in: "Express $\sin\theta$ in terms of $\tan\theta$" type questions.
Don't do this: Writing $\sin\theta = \tan\theta \cdot \cos\theta$ and calling it done — the right-hand side still contains $\cos\theta$.
The correct way: A conversion is complete only when every other ratio has been eliminated. The first-instinct error is stopping at the quotient step; the habit that fixes it is scanning the final expression for any stray ratio that isn't the target base.
Mistake 2: Dropping the ± and ignoring the quadrant
Where it slips in: Any conversion through a Pythagorean identity, especially for non-acute angles.
Don't do this: Taking the positive square root automatically in Quadrant II, III, or IV.
The correct way: The square root gives a $\pm$; the quadrant (ASTC) decides which. The step that looks skippable but isn't is the sign check after the root — for acute angles it's always $+$, but the moment the angle passes $90°$ the sign can flip.
Mistake 3: Misusing the wrong Pythagorean identity
Where it slips in: Starting from $\tan\theta$ or $\cot\theta$ and reaching for $\sin^2 + \cos^2 = 1$.
Don't do this: Forcing $\sin^2\theta + \cos^2\theta = 1$ when the natural partner of $\tan\theta$ is $\sec\theta$.
The correct way: Match the identity to the starting ratio — $\tan\theta$ pairs with $1 + \tan^2\theta = \sec^2\theta$; $\cot\theta$ pairs with $1 + \cot^2\theta = \csc^2\theta$. Choosing the right engine first saves a tangle of substitutions.
Key Takeaways
Conversion relations express any trigonometric ratio in terms of any other, at the same angle.
The method chains three engines: reciprocal relations, quotient relations, and the three Pythagorean identities.
The Pythagorean identity always supplies the missing partner ratio; pick the identity that matches your starting ratio.
Conversions through a square root carry a $\pm$ — acute angles take the positive sign; other quadrants follow ASTC.
The reusable skill is the three-step method, not the memorised table.
Practice Before Moving On
Do each from the engines, not the table.
Express $\cos\theta$ in terms of $\sin\theta$, then check it at $\theta = 30°$.
Given $\sin\theta = \frac{8}{17}$ (acute), find $\cos\theta$ and $\tan\theta$ by conversion.
Express $\sec\theta$ in terms of $\cot\theta$ for an acute angle.
Answer to Question 1: $\cos\theta = \sqrt{1 - \sin^2\theta}$; at $30°$, $\sqrt{1 - 1/4} = \frac{\sqrt{3}}{2} = \cos 30°$. Answer to Question 2: $\cos\theta = \frac{15}{17}$, $\tan\theta = \frac{8}{15}$ (the $8$-$15$-$17$ triple). Answer to Question 3: $\sec\theta = \frac{\sqrt{1 + \cot^2\theta}}{\cot\theta}$ — re-derive via $\csc\theta = \sqrt{1 + \cot^2\theta}$ and $\cos\theta = \frac{\cot\theta}{\csc\theta}$ if you didn't land here.
To make the three-engine method automatic with a teacher, explore Bhanzu's trigonometry tutor, the high school math tutor track, or math help online.
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Read More
Trigonometric ratios — the six ratios these conversions move between.
Trigonometric ratios of specific angles — exact values to test conversions against.
Sin cos tan — the primary ratios most conversions route through.
Trigonometric table — the standard-angle reference table.
Trigonometry formulas — the broader formula list conversions feed into.
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