Two new identities fall straight out of $\sin^2\theta+\cos^2\theta=1$ — and together they turn awkward equations in sec, cosec and cot into ordinary quadratics.
9 min read3 worked examples
Two new identities
The Pythagorean identity $\sin^2\theta+\cos^2\theta\equiv1$ is the source of two further identities, obtained by dividing through by $\cos^2\theta$ or by $\sin^2\theta$:
Divide every term by $\cos^2\theta$: $\dfrac{\sin^2\theta}{\cos^2\theta}+\dfrac{\cos^2\theta}{\cos^2\theta}\equiv\dfrac{1}{\cos^2\theta}$.
Dividing an identity by the same (non-zero) quantity keeps it true.
3
$\dfrac{\sin\theta}{\cos\theta}=\tan\theta$, so this becomes $\tan^2\theta + 1 \equiv \sec^2\theta$.
Recognise tan θ and sec θ from their definitions.
4
Dividing the original identity by $\sin^2\theta$ instead gives $\cot^2\theta+1\equiv\csc^2\theta$ by the same reasoning, since $\dfrac{\cos\theta}{\sin\theta}=\cot\theta$.
Symmetric argument, dividing by sin²θ this time.
▪ derived
Worth memorising
$1+\tan^2\theta\equiv\sec^2\theta$ and $1+\cot^2\theta\equiv\csc^2\theta$ are used constantly to turn an equation mixing two reciprocal functions into a quadratic in just one of them — exactly the same trick as substituting $\cos^2\theta=1-\sin^2\theta$ in ordinary trig equations.
Worked example 1
A short warm-up proof
Prove that $\dfrac{1}{\sec^2\theta}+\dfrac{1}{\csc^2\theta}\equiv1$.
MethodRewrite in terms of sine and cosine
1
$\dfrac{1}{\sec^2\theta}=\cos^2\theta$ and $\dfrac{1}{\csc^2\theta}=\sin^2\theta$, by the reciprocal definitions.
Flip each reciprocal function back to cosine and sine.
2
So the left-hand side becomes $\cos^2\theta+\sin^2\theta$.
Substitute both rewritten terms.
3
$\cos^2\theta+\sin^2\theta\equiv1$, which is the right-hand side.
The Pythagorean identity.
▪ proved
Worked example 2
Solving a quadratic-in-tan equation
Solve $2\sec^2x - \tan x = 5$ for $0 \le x < 360°$.
MethodSubstitute the identity to get a quadratic in one function
1
Replace $\sec^2x$ using $\sec^2x\equiv1+\tan^2x$: $2(1+\tan^2x)-\tan x=5$.
The equation mixes sec and tan; the identity rewrites everything in tan alone.
2
Expand and rearrange: $2\tan^2x-\tan x-3=0$.
Collect all terms on one side.
3
Factorise: $(2\tan x-3)(\tan x+1)=0$, so $\tan x=\dfrac32$ or $\tan x=-1$.
An ordinary quadratic, now solved by factorising.
4
$\tan x=\dfrac32$: principal value $56.3°$, plus $180°$ gives $x=56.3°$ or $236.3°$ (3 s.f.).
Tan has period 180°, so add 180° to get the second solution in range.
5
$\tan x=-1$: principal value $-45°$; adding $180°$ and $360°$ gives $x=135°$ or $315°$.
Shift the principal value into the required interval using the 180° period.
6
$x=56.3°,\ 135°,\ 236.3°,\ 315°$ (3 s.f.).
All four solutions in $0\le x<360°$.
▪ solved
Examiner's eye
Always check which quadrants your chosen substitution has silently ruled out. Squaring or using $\sec^2x\equiv1+\tan^2x$ doesn't lose solutions here since the identity holds for every $x$ where $\tan x$ is defined — but it's still worth substituting a solution back into the original equation to confirm.
Worked example 3
A harder identity proof · Challenge
Prove that $\sec^4\theta-\tan^4\theta\equiv\sec^2\theta+\tan^2\theta$.
Spotting a difference of two squares is one of the most common ways an A-level identity proof "unlocks" — the same pattern appears throughout this course whenever an expression contains both $\sec^2$ and $\tan^2$ (or $\csc^2$ and $\cot^2$) terms of matching degree.
Check yourself
Solve $\cot^2x+3\csc x+3=0$ for $0\le x<360°$. (Hint: rewrite cot²x using the identity involving csc²x, then factorise.)
Method
1
Replace $\cot^2x$ using $\cot^2x\equiv\csc^2x-1$: $\csc^2x-1+3\csc x+3=0$.
Turns the equation into one function, cosec, alone.
2
Simplify: $\csc^2x+3\csc x+2=0$.
Collect like terms.
3
Factorise: $(\csc x+1)(\csc x+2)=0$, so $\csc x=-1$ or $\csc x=-2$.