Tangent and Cotangent from Sine and Cosine

For an angle $\theta$ we know $\sin\theta = \frac{3}{5}$ and $\cos\theta = \frac{4}{5}$. Find $\tan\theta$ and $\cot\theta$.

Write the quotient identity for tangent: $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{4/5}$ = Divide the two fractions

Divide (the fifths cancel): $\frac{3/5}{4/5} = \frac{3}{5} \cdot \frac{5}{4} = \frac{3}{4}$ = $\tan\theta = \frac{3}{4}$

Use the quotient identity for cotangent: $\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{4/5}{3/5} = \frac{4}{3}$ = $\cot\theta = \frac{4}{3}$

Check with the reciprocal identity: $\tan\theta \cdot \cot\theta = \frac{3}{4} \cdot \frac{4}{3} = 1$ = Correct

Answer: $\tan\theta = \frac{3}{4}$ and $\cot\theta = \frac{4}{3}$

Exact Values at 60°

Use $\sin 60° = \frac{\sqrt{3}}{2}$ and $\cos 60° = \frac{1}{2}$ to find the exact values of $\tan 60°$ and $\cot 60°$.

Tangent is sine over cosine: $\tan 60° = \frac{\sqrt{3}/2}{1/2}$ = Same denominator 2 on top and bottom

Cancel the halves: $\frac{\sqrt{3}/2}{1/2} = \frac{\sqrt{3}}{2} \cdot \frac{2}{1} = \sqrt{3}$ = $\tan 60° = \sqrt{3}$

Cotangent is cosine over sine: $\cot 60° = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$ = $\cot 60° = \frac{1}{\sqrt{3}}$

Rationalize the denominator: $\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ = $\cot 60° = \frac{\sqrt{3}}{3} \approx 0.577$

Answer: $\tan 60° = \sqrt{3} \approx 1.732$ and $\cot 60° = \frac{\sqrt{3}}{3} \approx 0.577$

Rewrite in Sines and Cosines

Simplify $\cot\theta \cdot \sin\theta \cdot \sec\theta$.

Replace $\cot\theta$ with the quotient identity: $\frac{\cos\theta}{\sin\theta} \cdot \sin\theta \cdot \sec\theta$ = Only $\sec\theta$ is left to rewrite

Replace $\sec\theta$ with the reciprocal identity: $\frac{\cos\theta}{\sin\theta} \cdot \sin\theta \cdot \frac{1}{\cos\theta}$ = Everything is in $\sin$ and $\cos$

Write it as one fraction: $\frac{\cos\theta \cdot \sin\theta}{\sin\theta \cdot \cos\theta}$ = Numerator and denominator have the same factors

Cancel: $\frac{\cos\theta \sin\theta}{\sin\theta \cos\theta} = 1$ = $1$ (where $\sin\theta \neq 0$ and $\cos\theta \neq 0$)

Answer: $\cot\theta \cdot \sin\theta \cdot \sec\theta = 1$

A Short Simplification

Simplify $\frac{\tan\theta}{\sin\theta}$.

Replace $\tan\theta$: $\frac{\tan\theta}{\sin\theta} = \frac{\sin\theta / \cos\theta}{\sin\theta}$ = A fraction divided by $\sin\theta$

Dividing by $\sin\theta$ means multiplying by $\frac{1}{\sin\theta}$: $\frac{\sin\theta}{\cos\theta} \cdot \frac{1}{\sin\theta}$ = $\sin\theta$ appears on top and bottom

Cancel $\sin\theta$: $\frac{1}{\cos\theta}$ = $\frac{1}{\cos\theta}$

Name it with a reciprocal identity: $\frac{1}{\cos\theta} = \sec\theta$ = $\sec\theta$

Answer: $\frac{\tan\theta}{\sin\theta} = \sec\theta$ (for $\sin\theta \neq 0$, $\cos\theta \neq 0$)

An Angle in Quadrant III

The angle $\theta$ lies in quadrant III and $\sin\theta = -\frac{5}{13}$. Find $\tan\theta$ and $\cot\theta$.

Find $\cos^2\theta$ with the Pythagorean identity: $\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$ = $\cos\theta = \pm\frac{12}{13}$

Choose the sign: in quadrant III, $x < 0$: $\cos\theta < 0$ = $\cos\theta = -\frac{12}{13}$

Quotient identity for tangent: $\tan\theta = \frac{-5/13}{-12/13} = \frac{5}{12}$ = $\tan\theta = \frac{5}{12}$ (positive, as in quadrant III)

Quotient identity for cotangent: $\cot\theta = \frac{-12/13}{-5/13} = \frac{12}{5}$ = $\cot\theta = \frac{12}{5}$

Answer: $\tan\theta = \frac{5}{12}$ and $\cot\theta = \frac{12}{5}$

Mistake: Writing $\tan\theta = \frac{\cos\theta}{\sin\theta}$

Why: The fraction is upside down: that is cotangent. On the unit circle, tangent is the slope $\frac{y}{x}$, and $y = \sin\theta$.

Correct: $\tan\theta = \frac{\sin\theta}{\cos\theta}$ (sine on top, like $y$ over $x$) and $\cot\theta = \frac{\cos\theta}{\sin\theta}$

Mistake: Saying $\tan 90° = 0$ or $\cot 0° = 0$

Why: At $90°$ the cosine is $0$, and you cannot divide by $0$. At $0°$ the sine is $0$, so cotangent has a zero denominator.

Correct: $\tan 90° = \frac{1}{0}$ is undefined, and $\cot 0° = \frac{1}{0}$ is undefined. The graph of $\tan$ has vertical asymptotes there.

Mistake: Mixing two angles: $\frac{\sin 40°}{\cos 20°} = \tan 20°$ or $\tan 40°$

Why: The identity only works when sine and cosine have the **same** angle.

Correct: $\frac{\sin 40°}{\cos 40°} = \tan 40°$, but $\frac{\sin 40°}{\cos 20°} \approx 0.684$ is neither $\tan 20° \approx 0.364$ nor $\tan 40° \approx 0.839$.

Mistake: Losing the sign in quadrants II, III and IV

Why: Students often use the positive square root for $\cos\theta$ or $\sin\theta$ and get $\tan\theta > 0$ everywhere.

Correct: Take the signs from the quadrant first: in quadrant II $\sin > 0$, $\cos < 0$, so $\tan\theta < 0$; in quadrant III both are negative, so $\tan\theta > 0$.

Ski Slopes and Road Gradients

The slope of a hill (rise over horizontal run) is the tangent of its angle of inclination. A sensor in a phone measures sine and cosine of the tilt, and the app divides them.

A slope with $\sin\theta = 0.6$ and $\cos\theta = 0.8$ has $\tan\theta = 0.75$: it drops $75$ m for every $100$ m of horizontal distance.

Friction on a Ramp

A box on a ramp stays at rest as long as the friction coefficient $\mu$ is at least $\tan\theta$. Engineers use this to decide how steep conveyor belts and loading ramps may be.

On a ramp with $\tan\theta = 0.5$, a box with $\mu = 0.6$ stays put, but a box with $\mu = 0.4$ slides.

Shadow Lengths

When the sun stands at an elevation angle $\theta$, a pole of height $h$ casts a shadow of length $h \cot\theta = h \cdot \frac{\cos\theta}{\sin\theta}$. Architects use this to plan shade and sunlight.

With the sun at $45°$, $\cot 45° = 1$, so every object casts a shadow as long as it is tall.

Q: Why are they called quotient identities?

A: A quotient is the result of a division. These identities express $\tan$ and $\cot$ as one function divided by another: $\sin\theta \div \cos\theta$ and $\cos\theta \div \sin\theta$.

Q: Is $\cot\theta = \frac{\cos\theta}{\sin\theta}$ a quotient identity or a reciprocal identity?

A: Both views lead to the same place. The reciprocal identity says $\cot\theta = \frac{1}{\tan\theta}$; replacing $\tan\theta$ with $\frac{\sin\theta}{\cos\theta}$ and flipping gives the quotient form $\frac{\cos\theta}{\sin\theta}$. The quotient form also works at $90°$, where $\tan$ is undefined but $\cot 90° = \frac{0}{1} = 0$.

Q: Why do $y = \tan x$ and $y = \frac{\sin x}{\cos x}$ have the same graph?

A: Because they are the same function. Both have vertical asymptotes where $\cos x = 0$ ($x = \pm 90°, \pm 270°, ...$) and zeros where $\sin x = 0$. Toggle the two curves in the graph above: they lie exactly on top of each other.

Q: Can I check a quotient identity with a calculator?

A: Yes. Type $\sin(40) / \cos(40)$ and $\tan(40)$ in degree mode: both give about $0.8391$. A numeric check at a few angles is a good test, but only an algebraic argument proves an identity for all angles.