Reciprocal Functions and the Pythagorean Identities

Give one trigonometric function's value and a quadrant, and the page finds the angle, the partner value through the identity that ties the two functions together, and the other four from the point P. Or drag the angle and watch sec θ, csc θ and cot θ change as lengths in the picture.

Reciprocal functions	$sec θ = \frac{1}{cos θ}, csc θ = \frac{1}{sin θ} cot θ = \frac{cos θ}{sin θ}$
Pythagorean identity	$sin^{2} θ + cos^{2} θ = 1$
Divided by cos²θ	$1 + tan^{2} θ = sec^{2} θ$
Divided by sin²θ	$1 + cot^{2} θ = csc^{2} θ$

Worked example — Reciprocal Functions and the Pythagorean Identities

Angle	143.1301°
From the identity, with quadrant II's sign	−0.8
Sine	0.6
Cosine	−0.8
Tangent	−0.75
Secant	−1.25
Cosecant	1.6667
Cotangent	−1.3333

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Reciprocal Functions and the Pythagorean Identities

The reciprocal functions are sec θ = 1/cos θ, csc θ = 1/sin θ and cot θ = cos θ/sin θ, each defined wherever its denominator is not zero. The Pythagorean identities sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ and 1 + cot²θ = csc²θ hold at every angle where both sides are defined.

Where the formula comes from

Start from the circle. P = (cos θ, sin θ) is at distance 1 from O at every angle, so cos²θ + sin²θ = 1 in every quadrant. The squares do not see the signs, which is why the triangle with legs |cos θ| and |sin θ| gives the same equation.
$x_{P}^{2} + y_{P}^{2} = 1 \Rightarrow cos^{2} θ + sin^{2} θ = 1$
Divide by cos²θ, or by sin²θ. Each division is allowed exactly where the new functions are defined.
$\div cos^{2} θ : \div sin^{2} θ : 1 + tan^{2} θ = sec^{2} θ cot^{2} θ + 1 = csc^{2} θ$
Take the root, then choose the sign. An identity gives the square of the unknown, so it fixes the size; the quadrant fixes the sign.
$cos θ = \pm 1 - sin^{2} θ, sec θ = \pm 1 + tan^{2} θ$

Assumes: Each identity holds only where its functions are defined: the second fails at odd multiples of 90°, the third at multiples of 180°.

Problems solved in full

Given the sine of an obtuse angle 2 steps
sin θ = 3/5 and θ is obtuse. Find the other five functions exactly.

Open this problem in the tool
1. An obtuse angle lies in quadrant II, where cosine is negative. From sin²θ + cos²θ = 1:
  
  $cos θ = - 1 - (\frac{3}{5})^{2} = - \frac{16}{25} = - \frac{4}{5}$
2. The other four are quotients and reciprocals:
  
  $tan θ csc θ = \frac{3/5}{- 4/5} = - \frac{3}{4}, = \frac{1}{sin θ} = \frac{5}{3}, sec θ cot θ = \frac{1}{cos θ} = - \frac{5}{4} = \frac{1}{tan θ} = - \frac{4}{3}$
Answer

cos θ = −4/5, tan θ = −3/4, sec θ = −5/4, csc θ = 5/3, cot θ = −4/3.

$cos θ = - \frac{4}{5}, tan θ = - \frac{3}{4}, sec θ = - \frac{5}{4}, csc θ = \frac{5}{3}, cot θ = - \frac{4}{3}$
Given the tangent, find the secant 2 steps
tan θ = −12/5 and 270° < θ < 360°. Find sec θ, cos θ and sin θ.

Open this problem in the tool
Show solution
1. In quadrant IV cosine is positive, and so is secant. From 1 + tan²θ = sec²θ:
  
  $sec θ = + 1 + (- \frac{12}{5})^{2} = \frac{169}{25} = \frac{13}{5}$
2. Then cos θ is the reciprocal, and sin θ = tan θ · cos θ:
  
  $cos θ = \frac{5}{13}, sin θ = - \frac{12}{5} \cdot \frac{5}{13} = - \frac{12}{13}$
Answer

sec θ = 13/5 = 2.6, cos θ = 5/13 ≈ 0.3846, sin θ = −12/13 ≈ −0.9231.

$sec θ = \frac{13}{5}, cos θ = \frac{5}{13} \approx 0.3846, sin θ = - \frac{12}{13} \approx - 0.9231$
Given the cotangent, find the cosecant 2 steps
cot θ = 3/4 and 180° < θ < 270°. Find csc θ, sin θ and cos θ.

Open this problem in the tool
Show solution
1. In quadrant III sine is negative, and so is cosecant. From 1 + cot²θ = csc²θ:
  
  $csc θ = - 1 + (\frac{3}{4})^{2} = - \frac{25}{16} = - \frac{5}{4}$
2. Then sin θ is the reciprocal, and cos θ = cot θ · sin θ:
  
  $sin θ = - \frac{4}{5}, cos θ = \frac{3}{4} \cdot (- \frac{4}{5}) = - \frac{3}{5}$
Answer

csc θ = −5/4, sin θ = −4/5, cos θ = −3/5.

$csc θ = - \frac{5}{4}, sin θ = - \frac{4}{5}, cos θ = - \frac{3}{5}$
Exact values at 30° 3 steps
Find sec 30°, csc 30° and cot 30° exactly.

Open this problem in the tool
Show solution
1. Half of an equilateral triangle with side 2 has sides 1, √3 and 2, so
  
  $sin 3 0^{\circ} = \frac{1}{2}, cos 3 0^{\circ} = \frac{3}{2}$
2. Invert, and rationalise the denominator:
  
  $sec 3 0^{\circ} = \frac{2}{3} = \frac{2 3}{3}, csc 3 0^{\circ} = 2, cot 3 0^{\circ} = \frac{3 /2}{1/2} = 3$
3. Check with the second identity:
  
  $1 + tan^{2} 3 0^{\circ} = 1 + \frac{1}{3} = \frac{4}{3} = (\frac{2}{3})^{2}$
Answer

sec 30° = 2√3/3 ≈ 1.1547, csc 30° = 2, cot 30° = √3 ≈ 1.7321.

$sec 3 0^{\circ} = \frac{2 3}{3} \approx 1.1547, csc 3 0^{\circ} = 2, cot 3 0^{\circ} = 3 \approx 1.7321$
Simplifying with two identities 2 steps
Simplify (1 − cos²θ)(1 + cot²θ).
Show solution
1. Each bracket is one side of an identity; replace it by the other side:
  
  $(1 - cos^{2} θ) (1 + cot^{2} θ) = sin^{2} θ \cdot csc^{2} θ$
2. and csc θ = 1/sin θ:
  
  $sin^{2} θ \cdot \frac{1}{sin ^{2} θ} = 1$
Answer

1, at every angle that is not a multiple of 180°, where cot θ is undefined.

$(1 - cos^{2} θ) (1 + cot^{2} θ) = 1, θ \neq = k \cdot 18 0^{\circ}$
A proof from the factored identity 3 steps
Show that 1/(sec θ − tan θ) = sec θ + tan θ, and check it at θ = 60°.

Open this problem in the tool
Show solution
1. Multiply top and bottom by sec θ + tan θ:
  
  $\frac{1}{sec θ - tan θ} = \frac{sec θ + tan θ}{sec ^{2} θ - tan ^{2} θ}$
2. By 1 + tan²θ = sec²θ the denominator is 1:
  
  $= \frac{sec θ + tan θ}{1} = sec θ + tan θ$
3. At 60°, sec θ = 2 and tan θ = √3:
  
  $\frac{1}{2 - 3} = \frac{2 + 3}{( 2 - 3 ) ( 2 + 3 )} = \frac{2 + 3}{4 - 3} = 2 + 3$
Answer

Both sides are equal wherever they are defined; at 60° both are 2 + √3 ≈ 3.7321.

$\frac{1}{sec 6 0 ^{\circ} - tan 6 0 ^{\circ}} = 2 + 3 \approx 3.7321$
A value no angle has 2 steps
Find every angle with sec θ = 1/2.

Open this problem in the tool
Show solution
1. sec θ = 1/2 would mean
  
  $cos θ = \frac{1}{sec θ} = 2$
2. but cos θ is a coordinate of a point on a circle of radius 1, so |cos θ| ≤ 1, and therefore
  
  $∣ sec θ ∣ = \frac{1}{∣ cos θ ∣} \geq 1$
Answer

There is none: sec θ and csc θ never take a value strictly between −1 and 1. The preset refuses 1/2 and keeps θ at 60°, where the reciprocal runs the other way, cos θ = 1/2 and sec θ = 2.

${θ : sec θ = \frac{1}{2}} = \emptyset$
Removing the parameter 2 steps
A curve is given by x = 3 sec t, y = 2 tan t. Find its Cartesian equation.
Show solution
1. Isolate the two functions:
  
  $sec t = \frac{x}{3}, tan t = \frac{y}{2}$
2. and put them into sec²t − tan²t = 1:
  
  $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$
Answer

x²/9 − y²/4 = 1, a hyperbola. Since |sec t| ≥ 1, every point on it has |x| ≥ 3.

$\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1, ∣ x ∣ \geq 3$

Three identities are one triangle at three sizes 🖖

Triangle OMP has legs |cos θ| and |sin θ| and a hypotenuse of 1, and Pythagoras on it is sin²θ + cos²θ = 1. Enlarge it by the factor 1/|cos θ| and its sides become 1, |tan θ| and |sec θ|, the sides of triangle OAT, which stands on the tangent line x = 1. Pythagoras on OAT reads 1 + tan²θ = sec²θ. Enlarge by 1/|sin θ| instead and the sides become 1, |cot θ| and |csc θ|, those of OBS on the line y = 1, which gives 1 + cot²θ = csc²θ. In algebra the enlargement is dividing the first identity by cos²θ or by sin²θ, so one identity is worth remembering and the other two are a line of working. The picture also shows where each one fails: at 90° the line OP runs parallel to x = 1 and OAT has no third corner. Thomas Fincke introduced the terms tangent and secant in his Geometriae rotundi of 1583; in Latin they mean touching and cutting, and AT lies on a line that touches the circle while OT lies on one that cuts through it.

√(1 − sin²θ) is not cos θ 🖖

Press sin θ = 3/5, quadrant II. The identity gives cos²θ = 1 − 0.36 = 0.64, and 0.64 has two square roots. The root sign means the positive one, so √(1 − sin²θ) is |cos θ|, and writing cos θ = √(1 − sin²θ) is wrong at every angle in quadrants II and III. Here cos θ = −0.8. The identity fixes the size and only the quadrant fixes the sign, which is why the question has to say where θ lies, as “θ is obtuse” does. With the value still 0.6, click I: θ moves to 36.8699° and cos θ becomes +0.8. Click III and the page refuses, because sine is negative there.

Near 90°, sec θ and tan θ are huge and almost equal 🖖

Press θ = 89°: tan θ = 57.29 and sec θ = 57.2987, only 0.0087 apart. The second identity is why, this close to 90°, they cannot drift apart. Written as sec²θ − tan²θ = 1 and factored, it says (sec θ − tan θ)(sec θ + tan θ) = 1, so the gap is exactly 1 divided by the sum, and 1/114.5887 ≈ 0.0087. Drag θ closer to 90° and the sum grows without bound while the gap shrinks towards 0; at 90° itself both functions are undefined. The factored form is also the move behind a standard proof: 1/(sec θ − tan θ) = sec θ + tan θ, because multiplying top and bottom by sec θ + tan θ leaves a denominator of 1.

Learning path

Trigonometric identities and equations

References (2)

The reciprocal identities and the three Pythagorean identities, sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ and 1 + cot²θ = csc²θ (OpenStax lists the last two in the other order): J. Abramson, Precalculus 2e, OpenStax (2021), section 7.1, “Simplifying and Verifying Trigonometric Identities”.
Insight block 1: Fincke's Geometriae rotundi (1583) “introduces the terms ‘tangents’ and ‘secants’”: J. J. O'Connor and E. F. Robertson, “Thomas Fincke”, MacTutor History of Mathematics Archive, University of St Andrews (last updated January 2012).

By the 42Math team · Reviewed against published sources. Corrections welcome.