The Identity 1 + tan²θ = sec²θ

Divide the identity by ${\cos }^2\theta$cos²θ

Start from the identity ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1, which is Pythagoras on the circle of radius 1 and is true at every angle. An identity stays true when every term is divided by the same thing, as long as that thing is not 0.

Divide every term by ${\cos }^2\theta$cos²θ. The first term becomes $\frac{{\sin }^2\theta }{{\cos }^2\theta }$sin²θ/cos²θ, the middle term $\frac{{\cos }^2\theta }{{\cos }^2\theta }$cos²θ/cos²θ becomes 1, and the right side becomes $\frac{1}{{\cos }^2\theta }$1/cos²θ. So $\frac{{\sin }^2\theta }{{\cos }^2\theta }+1=\frac{1}{{\cos }^2\theta }$sin²θ/cos²θ + 1 = 1/cos²θ.

Now name each part. $\sin \theta /\cos \theta$sin θ / cos θ is $\tan \theta$tan θ, so $\frac{{\sin }^2\theta }{{\cos }^2\theta }$sin²θ/cos²θ is ${\tan }^2\theta$tan²θ. $1/\cos \theta$1 / cos θ is $\sec \theta$sec θ, so $\frac{1}{{\cos }^2\theta }$1/cos²θ is ${\sec }^2\theta$sec²θ. The identity becomes ${\tan }^2\theta +1={\sec }^2\theta$tan²θ + 1 = sec²θ, usually written $1+{\tan }^2\theta ={\sec }^2\theta$1 + tan²θ = sec²θ. It holds wherever $\cos \theta$cos θ is not 0, which is every angle except 90°, 270° and the angles that differ from them by a multiple of 180°.

Pythagoras on a larger triangle

There is a picture of the same fact. Extend the radius at angle $\theta$θ until it meets the vertical line x = 1. That makes a right triangle whose horizontal side is 1, whose vertical side is $\tan \theta$tan θ, and whose hypotenuse, the extended radius, is $\sec \theta$sec θ.

Pythagoras on that triangle gives $1^2+{\tan }^2\theta ={\sec }^2\theta$1² + tan²θ = sec²θ, the identity itself. It is the small triangle with sides $\cos \theta$cos θ, $\sin \theta$sin θ and 1, enlarged by the scale factor $\frac{1}{\cos }\theta$1/cos θ, and dividing by ${\cos }^2\theta$cos²θ is that enlargement done to the squares.

At $\theta =30^{\circ }$θ = 30° the triangle has a horizontal side of 1, a green vertical side tan 30° = 0.58 and a gold hypotenuse sec 30° = 1.15. The sums at the side show $1+{\tan }^2\theta =1.33$1 + tan²θ = 1.33 and ${\sec }^2\theta =1.33$sec²θ = 1.33. Drag the angle to 60°: $\tan 60^{\circ }=\sqrt{3}$tan 60° = √3 and sec 60° = 2, and $1+3=4=2^2$1 + 3 = 4 = 2².

Checking it

At 60°, $\tan 60^{\circ }=\sqrt{3}$tan 60° = √3 and sec 60° = 2, so $1+{\tan }^{2}60^{\circ }=1+3=4$1 + tan²60° = 1 + 3 = 4, and ${\sec }^{2}60^{\circ }=4$sec²60° = 4. At 30°, $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$tan 30° = 1/√3 and $\sec 30^{\circ }=\frac{2}{\sqrt{3}}$sec 30° = 2/√3, so $1+\frac{1}{3}=\frac{4}{3}$1 + 1/3 = 4/3 and ${\sec }^{2}30^{\circ }=\frac{4}{3}$sec²30° = 4/3.

In radians, at 0.3, ${\tan }^{2}0.3=0.0957$tan²0.3 = 0.0957 and ${\sec }^{2}0.3=1.0957$sec²0.3 = 1.0957, to 4 decimal places. At 1.1, ${\tan }^{2}1.1=3.8603$tan²1.1 = 3.8603 and ${\sec }^{2}1.1=4.8603$sec²1.1 = 4.8603. In both, the secant squared is exactly 1 more than the tangent squared.

The gold curve is $y={\sec }^{2}x$y = sec²x and the plain curve is $y={\tan }^{2}x$y = tan²x, over one turn with one square across for every 90°. The gold curve is the plain one lifted by exactly 1 at every x: at 0° and 180° they are 1 and 0, and the two dots at 60° are ${\tan }^{2}60^{\circ }=3$tan²60° = 3 and ${\sec }^{2}60^{\circ }=4$sec²60° = 4. Both are undefined at 90° and 270°, the dashed lines.

Dividing by ${\sin }^2\theta$sin²θ instead

Divide every term of ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1 by ${\sin }^2\theta$sin²θ instead. The first term becomes 1, $\frac{{\cos }^2\theta }{{\sin }^2\theta }$cos²θ/sin²θ is ${\cot }^2\theta$cot²θ, and $\frac{1}{{\sin }^2\theta }$1/sin²θ is ${\mathrm{cosec}}^2\theta$cosec²θ. So $1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta$1 + cot²θ = cosec²θ, which holds wherever $\sin \theta$sin θ is not 0.

Check at 30°: $\cot 30^{\circ }=\sqrt{3}$cot 30° = √3 and cosec 30° = 2, so $1+3=4=2^2$1 + 3 = 4 = 2². At 0.3 radians, ${\cot }^{2}0.3=10.4505$cot²0.3 = 10.4505 and ${\mathrm{cosec}}^{2}0.3=11.4505$cosec²0.3 = 11.4505.

The two identities match by name: tan goes with sec, and cot goes with cosec. Both are better derived in one line than memorized.

A secant from a tangent

Suppose $\theta$θ is acute and $\tan \theta =\frac{3}{4}$tan θ = 3/4. Then ${\sec }^2\theta =1+{\tan }^2\theta =1+\frac{9}{16}=\frac{25}{16}$sec²θ = 1 + tan²θ = 1 + 9/16 = 25/16, so $\sec \theta =\frac{5}{4}$sec θ = 5/4 or $-\frac{5}{4}$−5/4. An acute angle has a positive cosine, so its secant is positive too: $\sec \theta =\frac{5}{4}$sec θ = 5/4. Then $\cos \theta =\frac{4}{5}$cos θ = 4/5, the reciprocal.

Check with a triangle. With 3 opposite $\theta$θ and 4 adjacent, the hypotenuse is $\sqrt{9+16}=5$√(9 + 16) = 5, so $\cos \theta =\frac{4}{5}$cos θ = 4/5 and $\sec \theta =\frac{5}{4}$sec θ = 5/4. On a calculator, $\theta =36.87^{\circ }$θ = 36.87° and $1/\cos 36.87^{\circ }=1.25$1 / cos 36.87° = 1.25.

The cotangent form works the same way. If $\theta$θ is acute and $\cot \theta =\frac{4}{3}$cot θ = 4/3, then ${\mathrm{cosec}}^2\theta =1+\frac{16}{9}=\frac{25}{9}$cosec²θ = 1 + 16/9 = 25/9, so $\mathrm{cosec}\theta =\frac{5}{3}$cosec θ = 5/3 and $\sin \theta =\frac{3}{5}$sin θ = 3/5.

An equation made quadratic

Solve ${\tan }^2\theta +\sec \theta =1$tan²θ + sec θ = 1 for $0^{\circ }\le \theta \le 360^{\circ }$0° ≤ θ ≤ 360°. The equation mixes $\tan \theta$tan θ and $\sec \theta$sec θ, so replace ${\tan }^2\theta$tan²θ with ${\sec }^2\theta -1$sec²θ − 1: ${\sec }^2\theta -1+\sec \theta =1$sec²θ − 1 + sec θ = 1, which is ${\sec }^2\theta +\sec \theta -2=0$sec²θ + sec θ − 2 = 0.

That is a quadratic in $\sec \theta$sec θ, and it factors: $\left(\sec \theta +2\right)\left(\sec \theta -1\right)=0$(sec θ + 2)(sec θ − 1) = 0. So $\sec \theta =1$sec θ = 1, which means $\cos \theta =1$cos θ = 1 and $\theta =0^{\circ }$θ = 0° or 360°; or $\sec \theta =-2$sec θ = −2, which means $\cos \theta =-\frac{1}{2}$cos θ = −½ and $\theta =120^{\circ }$θ = 120° or 240°.

Check $\theta =120^{\circ }$θ = 120°: ${\tan }^{2}120^{\circ }={\left(-\sqrt{3}\right)}^2=3$tan²120° = (−√3)² = 3 and sec 120° = −2, and 3 + (−2) = 1. Check $\theta =0^{\circ }$θ = 0°: 0 + 1 = 1.

The usual mistakes

Adding 1 before squaring. With $\tan \theta =\frac{3}{4}$tan θ = 3/4, $1+\frac{3}{4}=\frac{7}{4}$1 + 3/4 = 7/4 is not ${\sec }^2\theta$sec²θ; square first to get $\frac{9}{16}$9/16, then add 1.

Turning the answer upside down. $\frac{4}{5}$4/5 is $\cos \theta$cos θ; $\sec \theta$sec θ is its reciprocal, $\frac{5}{4}$5/4.

Pairing the wrong functions. Dividing by ${\cos }^2\theta$cos²θ gives tan and sec; dividing by ${\sin }^2\theta$sin²θ gives cot and cosec. $1+{\tan }^2\theta$1 + tan²θ is ${\sec }^2\theta$sec²θ, not ${\mathrm{cosec}}^2\theta$cosec²θ.

Moving the 1 to the wrong side. ${\sec }^2\theta -1$sec²θ − 1 is ${\tan }^2\theta$tan²θ, not $1+{\tan }^2\theta$1 + tan²θ.

A funicular railway

In the application below, a railway rises 3 m for every 4 m across, so $\tan \theta =\frac{3}{4}$tan θ = 3/4, and its line is 240 m long on the map. The length of track is $240\sec \theta$240 sec θ, and $1+{\tan }^2\theta ={\sec }^2\theta$1 + tan²θ = sec²θ gives $\sec \theta =\frac{5}{4}$sec θ = 5/4 without finding $\theta$θ.