Secant, Cosecant and Cotangent

The three ratios, turned upside down.

Three reciprocals

The reciprocal of a number is 1 divided by it: the reciprocal of 4 is ¼, and the reciprocal of ½ is 2. Each of the three trigonometric ratios has a reciprocal with its own name.

The secant is the reciprocal of the cosine: sec θ = 1/cos θ. The cosecant is the reciprocal of the sine: cosec θ = 1/sin θ. The cotangent is the reciprocal of the tangent: cot θ = 1/tan θ. Since tan θ = sin θ / cos θ, turning it upside down gives cot θ = cos θ / sin θ as well.

Some books write csc θ for the cosecant. It is the same ratio.

The names cross over

The names do not pair up the way they look. The secant goes with the cosine, not with the sine, and the cosecant goes with the sine, not with the cosine. Only the cotangent pairs with the ratio it sounds like, the tangent.

Look at the third letter of each name. Sec has c, for cos; cosec has s, for sin; cot has t, for tan.

A value you know gives its reciprocal

To evaluate one of the three, find the ordinary ratio and take 1 over it. cos 60° = ½, so sec 60° = 1 ÷ ½ = 2. sin 30° = ½, so cosec 30° = 2. tan 45° = 1, so cot 45° = 1.

The same works for every exact value. sin 60° = √3/2, so cosec 60° = 2/√3, which is 2√3/3 once the denominator is rationalized. tan 30° = 1/√3, so cot 30° = √3. cos 45° = 1/√2, so sec 45° = √2.

In radians, with a calculator: cos 1.1 = 0.4536, so sec 1.1 = 1 / 0.4536 = 2.2046 to 4 decimal places; sin 0.3 = 0.2955, so cosec 0.3 = 3.3839; and cot 0.3 = 1 / tan 0.3 = 3.2327. A calculator has no sec key, so this is how all three are worked out.

Check a value by multiplying the pair: the product must be 1. sec 60° × cos 60° = 2 × ½ = 1.

xy

The gold curve is y = sec x and the plain curve is y = cos x, over one turn with one square across for every 90°. The two dots at 60° are cos 60° = ½ and sec 60° = 2. Where the cosine is 1 or −1, at 0°, 180° and 360°, the secant is the same; where the cosine shrinks toward 0, the secant grows without limit, and at 90° and 270°, the dashed lines, it is undefined.

Where they are undefined, and their size

A reciprocal is undefined where the ratio is 0, because 1 ÷ 0 has no value. cos 90° = 0, so sec 90° does not exist; sin 0° = 0 and sin 180° = 0, so cosec 0° and cosec 180° do not exist; and cot 0° does not exist either, because tan 0° = 0.

A sine or a cosine is always between −1 and 1, so 1 divided by it is never between −1 and 1. A secant or cosecant is always 1 or more, or −1 or less. A value such as sec θ = 0.5 is impossible.

Each reciprocal has the same sign as its ratio. Between 90° and 270° the cosine is negative, so the secant is negative there too: sec 120° = 1 / (−½) = −2.

Where the names come from

Draw the circle of radius 1 and the radius at angle θ. Draw the vertical line x = 1, which touches the circle at (1, 0); a line that touches a circle is called a tangent. Extend the radius until it meets that line.

This makes a large right triangle with the horizontal side 1. It is the triangle with sides cos θ, sin θ and 1, enlarged by the scale factor 1/cos θ. So its vertical side is sin θ/cos θ = tan θ, which is why the tangent has that name, and its long side, along the extended radius, is 1/cos θ = sec θ. That extended radius cuts through the circle, and a line that cuts a circle is called a secant.

tangent line1tan θ = 0.58sec θ = 1.15θ = 30°1 + tan²θ = 1 + 0.33 = 1.33sec²θ = 1.15² = 1.33

on the tangent line the ray rises tan θ = 0.58, and the cutting ray, the secant, is sec θ = 1.15 long: 1 + tan²θ = sec²θ is Pythagoras on this triangle

Turn the ray to 60° and read sec θ

At θ = 30° the radius, extended, meets the tangent line at a height of tan 30° = 0.58, shown in green, and the extended radius, in gold, is sec 30° = 1.15 long. The horizontal side is 1. Drag the angle up to 60°: the gold line becomes sec 60° = 2, twice the radius, because cos 60° = ½.

The usual mistakes

Pairing by the first letters. sec θ is not 1/sin θ; that is cosec θ. For sec 60°, the reciprocal of sin 60° gives 2/√3, which is cosec 60°.

Giving the ratio instead of its reciprocal. cos 60° is ½, but sec 60° is 2.

Reading cos⁻¹x as sec x. cos⁻¹ is the inverse cosine, the angle with a given cosine: cos⁻¹(0.5) = 60°. The reciprocal is written sec x, or (cos x)⁻¹ with brackets.

Expecting a value where the ratio is 0. sec 90° and cot 0° do not exist.

Guy ropes on a tent pole

In the application below, a pole 1.5 m tall is held by a rope at an angle θ to the ground. The pole is opposite θ, so the rope, the hypotenuse, is 1.5 / sin θ = 1.5 cosec θ, and the distance to the peg is 1.5 / tan θ = 1.5 cot θ.

Worked example: The Guy Ropes That Hold Up a Tent Pole

Question A tent pole stands 1.5 m tall on level ground. A straight guy rope runs from its top to a peg in the ground, making an angle θ with the ground. (a) Show that the rope is 1.5cosecθ m long and the peg is 1.5cotθ m from the foot of the pole, and find both when θ = π3. (b) On the other side of the pole the camper uses a rope 3 m long. At what angle does it meet the ground, and how far from the pole is its peg? Give exact answers, then answers to 2 decimal places.

  1. 1.The pole is opposite the angle θ and the rope is the hypotenuse, so sinθ = 1.5L for a rope of length L. Then L = 1.5sinθ = 1.5cosecθ.

    pi/31.5 mLsin θ = 1.5/L, so L = 1.5 cosec θ
    pi/31.5 mLsin θ = 1.5/L, so L = 1.5 cosec θ
    The pole is opposite θ: sinθ = 1.5L, so L = 1.5cosecθ.
  2. 2.The distance d from the foot of the pole to the peg is adjacent to θ, so tanθ = 1.5d, and d = 1.5tanθ = 1.5cotθ.

    pi/31.5 mLdsin θ = 1.5/L, so L = 1.5 cosec θtan θ = 1.5/d, so d = 1.5 cot θ
    pi/31.5 mLdsin θ = 1.5/L, so L = 1.5 cosec θtan θ = 1.5/d, so d = 1.5 cot θ
    The distance to the peg is adjacent to θ: tanθ = 1.5d, so d = 1.5cotθ.
  3. 3.(a) At θ = π3, cosecπ3 = 2√3 and cotπ3 = 1√3. So L = 3√3 = √3, which is 1.73 m, and d = 1.5√3 = √32, which is 0.87 m.

    pi/31.5 m1.73 m0.87 msin θ = 1.5/L, so L = 1.5 cosec θtan θ = 1.5/d, so d = 1.5 cot θθ = pi/3: L = √3 = 1.73 m, d = √3/2 = 0.87 m
    pi/31.5 m1.73 m0.87 msin θ = 1.5/L, so L = 1.5 cosec θtan θ = 1.5/d, so d = 1.5 cot θθ = pi/3: L = √3 = 1.73 m, d = √3/2 = 0.87 m
    (a) At π3: L = 1.5 × 2√3 = √3 ≈ 1.73 m and d = √32 ≈ 0.87 m.
  4. 4.With a 3 m rope, 1.5cosecθ = 3, so cosecθ = 2 and sinθ = 12. The angle is inside a right-angled triangle, so it is acute, and θ = π6.

    pi/31.5 m1.73 m0.87 mpi/63 msin θ = 1.5/L, so L = 1.5 cosec θtan θ = 1.5/d, so d = 1.5 cot θθ = pi/3: L = √3 = 1.73 m, d = √3/2 = 0.87 m1.5 cosec θ = 3: sin θ = 1/2, θ = pi/6
    pi/31.5 m1.73 m0.87 mpi/63 msin θ = 1.5/L, so L = 1.5 cosec θtan θ = 1.5/d, so d = 1.5 cot θθ = pi/3: L = √3 = 1.73 m, d = √3/2 = 0.87 m1.5 cosec θ = 3: sin θ = 1/2, θ = pi/6
    For a 3 m rope, cosecθ = 2, so sinθ = 12 and θ = π6.
  5. 5.(b) The rope meets the ground at π6, which is 30°, and its peg is 1.5cotπ6 = 1.5√3 m from the pole, which is 2.60 m. Check: 1.52 + (1.5√3)2 = 2.25 + 6.75 = 9 = 32.

    pi/31.5 m1.73 m0.87 mpi/63 m2.60 msin θ = 1.5/L, so L = 1.5 cosec θtan θ = 1.5/d, so d = 1.5 cot θθ = pi/3: L = √3 = 1.73 m, d = √3/2 = 0.87 m1.5 cosec θ = 3: sin θ = 1/2, θ = pi/6d = 1.5 cot(pi/6) = 1.5√3 = 2.60 m
    pi/31.5 m1.73 m0.87 mpi/63 m2.60 msin θ = 1.5/L, so L = 1.5 cosec θtan θ = 1.5/d, so d = 1.5 cot θθ = pi/3: L = √3 = 1.73 m, d = √3/2 = 0.87 m1.5 cosec θ = 3: sin θ = 1/2, θ = pi/6d = 1.5 cot(pi/6) = 1.5√3 = 2.60 m
    (b) The peg is 1.5cotπ6 = 1.5√3 ≈ 2.60 m from the pole.

Answer: (a) the rope is √3 m, which is 1.73 m, and the peg is √32 m, which is 0.87 m, from the pole; (b) π6, which is 30°, with the peg 1.5√3 m, which is 2.60 m, from the pole

Common mistakes

  • Taking cosecθ to be 1cosθ. That is the secant; the cosecant is 1sinθ, and with the pole opposite the angle it is the sine that links the pole to the rope.
  • Giving θ = 5π6 as a second answer, because sin5π6 = 12 too. An angle inside a right-angled triangle is acute, so only π6 fits.

More radians and trigonometric identities problems, worked step by step →

Practice Secant, Cosecant and Cotangent in the app