Three equations hold for every acute angle: sin²A + cos²A = 1, 1 + tan²A = sec²A and 1 + cot²A = cosec²A. Prove them from Pythagoras, then use them to simplify expressions and to prove other identities step by step.
An equation such as $2x+1=7$ is true only for one value of $x$. An identity such as $(x+1)^2=x^2+2x+1$ is true for every value of $x$. That is why algebraic identities are so useful: you may replace one side by the other anywhere, whatever the numbers are. Trigonometry has identities too. They are true for every acute angle, whether it is $23^\circ$, $41^\circ$ or an angle nobody has measured yet.
In this section you will prove the three standard identities from the Pythagoras theorem, see them at work on a picture of squares, and learn the craft of proving a new identity: choose a side, rewrite everything in sines and cosines, apply an identity or factorise, and arrive at the other side.
The fundamental identity
📘 Definition 8.5 — Trigonometric identity
A trigonometric identity is an equation involving trigonometric ratios of an angle that is true for all values of the angle for which both sides are defined. (An equation that is true only for some angles, like $\sin\theta=\tfrac12$, is not an identity.)
📐 Theorem 8.6 — The fundamental identity
For every acute angle $A$,
$$\sin^2A+\cos^2A=1.$$
Proof. Take $\triangle ABC$ right-angled at $B$, with the acute angle $A$. By the Pythagoras theorem,
$$AB^2+BC^2=AC^2.$$
Divide every term by $AC^2$ (which is not zero):
$$\left(\frac{AB}{AC}\right)^2+\left(\frac{BC}{AC}\right)^2=1.$$
Now $\dfrac{AB}{AC}=\cos A$ and $\dfrac{BC}{AC}=\sin A$, so $\cos^2A+\sin^2A=1$. $\blacksquare$
The same result holds at $0^\circ$ and $90^\circ$ (check: $\sin^2 0^\circ+\cos^20^\circ=0+1$ and $\sin^290^\circ+\cos^290^\circ=1+0$). The identity is the Pythagoras theorem in disguise: if the hypotenuse has length $1$, the squares on the two legs have areas $\sin^2A$ and $\cos^2A$, and together they fill the square on the hypotenuse, which has area $1$.
Rearranged, it gives two useful "solved-for" forms:
$$\sin^2A=1-\cos^2A=(1-\cos A)(1+\cos A),\qquad\cos^2A=1-\sin^2A=(1-\sin A)(1+\sin A).$$
⚠️ Common mistakes
1. Different angles. The identity needs the same angle in both terms. For example $\sin^230^\circ+\cos^260^\circ=\tfrac14+\tfrac14=\tfrac12\ne1$.
2. Forgetting the square. $\sin A+\cos A$ is not $1$. For $A=40^\circ$ it is about $0.643+0.766=1.409$. Only the squares add to $1$.
3. Notation. $\sin^2A$ means $(\sin A)^2$. Never write $\sin A^2$.
4. Square roots. From $\cos^2A=1-\sin^2A$ you get $\cos A=\sqrt{1-\sin^2A}$ for an acute angle, because the cosine of an acute angle is positive. Do not write "$\pm$" here.
✏️ Example 1 — Writing every ratio in terms of $\sin A$
For an acute angle $A$, express $\cos A$, $\tan A$ and $\sec A$ in terms of $\sin A$. Then check your formulas when $\sin A=\dfrac35$.
💡 Show solution▶
Cosine From Theorem 8.6, $\cos^2A=1-\sin^2A$, so $\cos A=\sqrt{1-\sin^2A}$ (positive, since $A$ is acute).
Check For $\sin A=\dfrac35$: $\sqrt{1-\frac9{25}}=\dfrac45$, so $\cos A=\dfrac45$, $\tan A=\dfrac{3/5}{4/5}=\dfrac34$, $\sec A=\dfrac54$. These match the $3$-$4$-$5$ triangle ✓.
✏️ Example 2 — Using the identity without a calculator
Numerator The two angles are equal ($29^\circ$), so $\sin^229^\circ+\cos^229^\circ=1$ by Theorem 8.6.
Denominator $\sec^2\theta-\tan^2\theta=1$ for every acute $\theta$ (Theorem 8.7 below, rearranged), so the denominator is $1$ with $\theta=41^\circ$.
Answer $\dfrac11=1$. Neither $29^\circ$ nor $41^\circ$ is a standard angle; the identities do the work.
Interactive 1 — The Squares Always Add Up to 1
The right-angled triangle has hypotenuse $1$. Squares are built on all three sides: the red square on the vertical leg has area $\sin^2A$, the blue square on the horizontal leg has area $\cos^2A$, and the amber square on the hypotenuse has area $1$. Slide the angle $A$ (or press Play). The red and blue squares trade area, but the bar on the right — red plus blue — always stretches exactly as far as the amber one.
35°
$\sin^2A$—
$\cos^2A$—
$\sin^2A+\cos^2A$—
square on the hypotenuse—
Two more identities
The fundamental identity can be divided by $\cos^2A$ or by $\sin^2A$ to give two more. Both divisions are allowed because for an acute angle neither $\sin A$ nor $\cos A$ is zero.
📐 Theorem 8.7 — The secant identity
For every acute angle $A$,
$$1+\tan^2A=\sec^2A.$$
Proof. Divide both sides of $\sin^2A+\cos^2A=1$ by $\cos^2A$:
$$\frac{\sin^2A}{\cos^2A}+\frac{\cos^2A}{\cos^2A}=\frac1{\cos^2A}\quad\Longrightarrow\quad\tan^2A+1=\sec^2A.\qquad\blacksquare$$
(Directly from the triangle: divide $AB^2+BC^2=AC^2$ by $AB^2$.)
📐 Theorem 8.8 — The cosecant identity
For every acute angle $A$,
$$1+\cot^2A=\operatorname{cosec}^2A.$$
Proof. Divide both sides of $\sin^2A+\cos^2A=1$ by $\sin^2A$:
$$1+\frac{\cos^2A}{\sin^2A}=\frac{1}{\sin^2A}\quad\Longrightarrow\quad1+\cot^2A=\operatorname{cosec}^2A.\qquad\blacksquare$$
Useful forms. Each identity can be solved for any of its terms:
$$\sec^2A-\tan^2A=1,\qquad\tan^2A=\sec^2A-1,\qquad\operatorname{cosec}^2A-\cot^2A=1,\qquad\cot^2A=\operatorname{cosec}^2A-1.$$
Factorising the "difference of squares" forms gives $(\sec A-\tan A)(\sec A+\tan A)=1$ and $(\operatorname{cosec}A-\cot A)(\operatorname{cosec}A+\cot A)=1$. These products are the key to many "conditional" problems: if you know one factor, the other is its reciprocal.
✏️ Example 3 — A first proof
Prove that $(1-\cos^2A)\sec^2A=\tan^2A$.
💡 Show solution▶
Start with LHS $\text{LHS}=(1-\cos^2A)\sec^2A$.
Identity $1-\cos^2A=\sin^2A$ (Theorem 8.6), and $\sec^2A=\dfrac1{\cos^2A}$. So $\text{LHS}=\sin^2A\times\dfrac1{\cos^2A}=\dfrac{\sin^2A}{\cos^2A}$.
🧭 Problem-solving strategy — proving a trigonometric identity
Choose the side that looks more complicated and work only on that side, until it becomes the other side. (Write "LHS = …" at the start.)
If you are not sure what to do, rewrite everything in terms of $\sin$ and $\cos$: $\tan=\frac{\sin}{\cos}$, $\sec=\frac1{\cos}$, $\operatorname{cosec}=\frac1{\sin}$, $\cot=\frac{\cos}{\sin}$.
Look for the chance to use an identity: $\sin^2+\cos^2$, $1+\tan^2$, $1+\cot^2$, or a difference of squares $1-\sin^2A=(1-\sin A)(1+\sin A)$.
With fractions: combine them over a common denominator, or multiply top and bottom by a "conjugate" such as $(1-\cos A)$ to create $1-\cos^2A=\sin^2A$.
Stop when both sides agree. Write $\blacksquare$ and state which identity you used at each step.
✏️ Example 4 — Squares of a difference
Prove that $(\operatorname{cosec}\theta-\cot\theta)^2=\dfrac{1-\cos\theta}{1+\cos\theta}$.
💡 Show solution▶
Sines and cosines $\text{LHS}=\left(\dfrac1{\sin\theta}-\dfrac{\cos\theta}{\sin\theta}\right)^2=\dfrac{(1-\cos\theta)^2}{\sin^2\theta}$.
Identity $\sin^2\theta=1-\cos^2\theta=(1-\cos\theta)(1+\cos\theta)$. So $\text{LHS}=\dfrac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)}$.
Cancel Cancelling one factor $(1-\cos\theta)$ (not zero for acute $\theta$): $\text{LHS}=\dfrac{1-\cos\theta}{1+\cos\theta}=\text{RHS}$. $\blacksquare$
Check At $\theta=60^\circ$: LHS $=\left(\tfrac{2}{\sqrt3}-\tfrac1{\sqrt3}\right)^2=\tfrac13$; RHS $=\dfrac{1/2}{3/2}=\tfrac13$ ✓.
✏️ Example 5 — Adding two fractions
Prove that $\dfrac{\sin\theta}{1+\cos\theta}+\dfrac{1+\cos\theta}{\sin\theta}=2\operatorname{cosec}\theta$.
💡 Show solution▶
Common denominator $\text{LHS}=\dfrac{\sin^2\theta+(1+\cos\theta)^2}{\sin\theta\,(1+\cos\theta)}$.
Expand The numerator is $\sin^2\theta+1+2\cos\theta+\cos^2\theta=(\sin^2\theta+\cos^2\theta)+1+2\cos\theta=2+2\cos\theta=2(1+\cos\theta)$.
Pick one of the three identities and slide the angle $A$ (or press Play, which also steps through the identities). The left side is drawn as a stacked bar, the right side as a single green bar. Whatever the angle, the two bars have exactly the same length and the difference is $0$. A true identity is true for every angle, not just a few.
40°
left side—
right side—
left − right—
angle—
Exercises
I
Using the identities. In every part the angles are acute.
8 $\sin^4A-\cos^4A=(\sin^2A-\cos^2A)(\sin^2A+\cos^2A)=\sin^2A-\cos^2A$. Replacing $\cos^2A$ by $1-\sin^2A$: $\sin^2A-(1-\sin^2A)=2\sin^2A-1$.
III
Conditional problems and applications.
1 If $\cos\theta+\cos^2\theta=1$, show that $\sin^2\theta+\sin^4\theta=1$.
2 If $\sec\theta+\tan\theta=2$, find $\sec\theta-\tan\theta$, and then $\sec\theta$, $\tan\theta$ and $\sin\theta$.
3 If $\operatorname{cosec}\theta-\cot\theta=\dfrac13$, find $\operatorname{cosec}\theta$, $\cot\theta$ and $\sin\theta$.
4 If $\tan A=\dfrac5{12}$, find $\sin A+\cos A$ and $\sin A\cos A$.
5 If $a\cos\theta+b\sin\theta=m$ and $a\sin\theta-b\cos\theta=n$, prove that $a^2+b^2=m^2+n^2$.
6 If $x=a\sec\theta$ and $y=b\tan\theta$ ($a,b\ne0$), prove that $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$.
7 A cabin of a Ferris wheel has position $(x,y)=(10\cos\theta,\ 10\sin\theta)$ metres relative to the axle. Show that the cabin is always $10$ m from the axle.
8 A calculator gives $\sin40^\circ=0.6428$ and $\cos40^\circ=0.7660$ (to 4 d.p.). Check $\sin^240^\circ+\cos^240^\circ$ and explain any small difference from $1$.
💡 Show solutions▶
1 $\cos\theta=1-\cos^2\theta=\sin^2\theta$. So $\sin^2\theta+\sin^4\theta=\sin^2\theta+(\sin^2\theta)^2=\cos\theta+\cos^2\theta=1$.
2 $(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1$, so $\sec\theta-\tan\theta=\dfrac12$. Adding: $2\sec\theta=\dfrac52$, so $\sec\theta=\dfrac54$; subtracting: $2\tan\theta=\dfrac32$, so $\tan\theta=\dfrac34$. Then $\cos\theta=\dfrac45$ and $\sin\theta=\tan\theta\cos\theta=\dfrac35$. (Check: $\tfrac{25}{16}-\tfrac9{16}=1$ ✓.)
3 $(\operatorname{cosec}\theta-\cot\theta)(\operatorname{cosec}\theta+\cot\theta)=1$, so $\operatorname{cosec}\theta+\cot\theta=3$. Adding: $2\operatorname{cosec}\theta=\dfrac{10}3$, so $\operatorname{cosec}\theta=\dfrac53$; subtracting: $2\cot\theta=\dfrac83$, so $\cot\theta=\dfrac43$. Hence $\sin\theta=\dfrac35$.
6 $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=\sec^2\theta-\tan^2\theta=1$ by Theorem 8.7.
7 Distance $=\sqrt{x^2+y^2}=\sqrt{100\cos^2\theta+100\sin^2\theta}=\sqrt{100(\cos^2\theta+\sin^2\theta)}=\sqrt{100}=10$ m.
8 $0.6428^2=0.41319$ and $0.7660^2=0.58676$; the sum is $0.99995\approx1.0000$. The tiny difference comes from rounding the calculator values to 4 decimal places.
IV
Challenge.
1 Prove that $\sin^6A+\cos^6A=1-3\sin^2A\cos^2A$.
2 If $\tan\theta+\sec\theta=p$, prove that $\dfrac{p^2-1}{p^2+1}=\sin\theta$.
3 Prove that $(1+\cot A-\operatorname{cosec}A)(1+\tan A+\sec A)=2$.
💡 Show solutions▶
1 Use $a^3+b^3=(a+b)^3-3ab(a+b)$ with $a=\sin^2A$, $b=\cos^2A$, $a+b=1$: $\sin^6A+\cos^6A=1-3\sin^2A\cos^2A$.
2 $p=\dfrac{\sin\theta+1}{\cos\theta}$, so $p^2=\dfrac{(1+\sin\theta)^2}{1-\sin^2\theta}=\dfrac{1+\sin\theta}{1-\sin\theta}$. Then $\dfrac{p^2-1}{p^2+1}=\dfrac{(1+\sin\theta)-(1-\sin\theta)}{(1+\sin\theta)+(1-\sin\theta)}=\dfrac{2\sin\theta}{2}=\sin\theta$.
3 In sines and cosines: $\left(\dfrac{\sin A+\cos A-1}{\sin A}\right)\left(\dfrac{\cos A+\sin A+1}{\cos A}\right)=\dfrac{(\sin A+\cos A)^2-1}{\sin A\cos A}=\dfrac{2\sin A\cos A}{\sin A\cos A}=2$.
How this is examined: Board papers regularly contain a "prove that" identity (2–3 marks) built on $\sin^2+\cos^2=1$, plus 1-mark MCQs and assertion–reason items that test whether you can simplify a short expression. Start from the more complicated side, write the identity you use at each step, and finish with "= RHS". JEE Main uses the identities in value-finding problems, often with $(\sin\theta+\cos\theta)^2=1+2\sin\theta\cos\theta$.
Trap (C) comes from using only the first identity and forgetting the second factor. Answer: (B).
E2.Board · 1 markAssertion (A): $\sin40^\circ+\cos40^\circ=1$. Reason (R): $\sin^2\theta+\cos^2\theta=1$ for every acute angle $\theta$. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
💡 Show solution▶
Check R This is Theorem 8.6, true.
Check A The identity is about the squares. $\sin40^\circ+\cos40^\circ\approx0.643+0.766=1.409\ne1$ (as $(\sin\theta+\cos\theta)^2=1+2\sin\theta\cos\theta\gt1$). A is false.
Answer (D).
E3.Board · 3 marks Using a right-angled triangle, prove that $\sin^2A+\cos^2A=1$ for an acute angle $A$, and deduce that $1+\tan^2A=\sec^2A$.
💡 Show solution▶
Figure and Pythagoras(1 mark) In $\triangle ABC$, right-angled at $B$: $AB^2+BC^2=AC^2$.
Divide(1 mark) Dividing by $AC^2$: $\left(\dfrac{AB}{AC}\right)^2+\left(\dfrac{BC}{AC}\right)^2=1$, i.e. $\cos^2A+\sin^2A=1$.
Deduce(1 mark) Divide by $\cos^2A$: $\dfrac{\sin^2A}{\cos^2A}+1=\dfrac1{\cos^2A}$, i.e. $\tan^2A+1=\sec^2A$. $\blacksquare$
E4.Board · 2 marks Prove that $(\operatorname{cosec}\theta-\cot\theta)(1+\cos\theta)=\sin\theta$.
E6.JEE Main If $\sin\theta+\cos\theta=\dfrac75$ for an acute angle $\theta$, then $625\left(\sin^4\theta+\cos^4\theta\right)=$ (A) $337$ (B) $337.5$ (C) $288$ (D) $625$
💡 Show solution▶
Product Square the given: $\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta=\dfrac{49}{25}$, so $1+2\sin\theta\cos\theta=\dfrac{49}{25}$ and $\sin\theta\cos\theta=\dfrac{12}{25}$.
🔬 Research–Teaching NexusMini project · optional, about 40 minutes
Research question. A true identity holds for every angle; a false "identity" fails for most. Can a spreadsheet tell the two apart — and how big are the differences you see because of rounding?
Why it matters. Engineers and programmers test a formula numerically before they trust it, and software has to cope with rounding errors that are tiny but never exactly zero. A numerical check can never prove an identity for all angles, but it can quickly expose a false one — which is why a proof is still needed.
What to do.
In a spreadsheet, put the angles $5^\circ,10^\circ,\dots,85^\circ$ in column A. In column B convert to radians ($=\text{A2}\times\text{PI()}/180$).
In column C compute $\sin^2A+\cos^2A$, in column D compute $1+\tan^2A-\sec^2A$ (use $1/\cos$ for secant) and in column E compute $1+\cot^2A-\operatorname{cosec}^2A$. Format with 12 decimal places.
Find the largest absolute deviation from $1$ (or $0$) in each column. Explain in a sentence why it is not exactly zero.
Now test two false "identities": $\sin A+\cos A=1$ and $\sin^2A+\cos^2B=1$ (with $B=A+10^\circ$). For which angles, if any, do they come out true?
Plot the first true identity and the first false one on the same chart.
Tools: A spreadsheet (Excel, LibreOffice or Google Sheets) or a calculator
Hand in: Your spreadsheet or a printout of the table, the chart, the largest deviations, and a short paragraph on what a numerical check can and cannot show.
Going further. Find, by trial, the angle between $0^\circ$ and $90^\circ$ at which $\sin A+\cos A$ is largest. What is its value, and which standard angle is it?
Teacher/parent note: spreadsheet trigonometric functions work in radians, hence the conversion in column B. The largest deviations are of the order $10^{-16}$, a floating-point effect. The final question leads to $\sin A+\cos A=\sqrt2$ at $A=45^\circ$ and is a friendly preview of ideas met in Class 11.
🧪 Self-check quiz
0/5 correct
Q1. For an acute angle $A$, $1+\tan^2A$ equals
Q2. $(1-\sin^2\theta)\sec^2\theta=$
Q3. The value of $\sec^227^\circ-\tan^227^\circ$ is
Q4. Which of the following is not an identity?
Q5. If $\operatorname{cosec}\theta-\cot\theta=\dfrac12$, then $\operatorname{cosec}\theta+\cot\theta=$