Trigonometric Identities and How to Prove Them, Class 10

The three iden­ti­ties and their proofs#

Iden­tity 1: sin² A + cos² A = 1#

Let $\triangle ABC$ be right-angled at $B$. By Pythago­ras' the­o­rem,

$$AB^2 + BC^2 = AC^2.$$

Divide every term by $AC^2$ (which is not zero):

$$\displaystyle \left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = 1 \quad \Longrightarrow \quad \cos^2 A + \sin^2 A = 1.$$

This proves the iden­tity for every acute angle $A$. For the end points, $\sin^2 0^\circ + \cos^2 0^\circ = 0 + 1 = 1$ and $\sin^2 90^\circ + \cos^2 90^\circ = 1 + 0 = 1$, so it holds for all $A$ with $0^\circ \le A \le 90^\circ$.

Iden­tity 2: 1 + tan² A = sec² A#

Divide $AB^2 + BC^2 = AC^2$ by $AB^2$ instead:

$$\displaystyle 1 + \left(\frac{BC}{AB}\right)^2 = \left(\frac{AC}{AB}\right)^2 \quad \Longrightarrow \quad 1 + \tan^2 A = \sec^2 A.$$

The divi­sion needs $AB \ne 0$, that is $\cos A \ne 0$. Since $\tan 90^\circ$ and $\sec 90^\circ$ are not defined, this iden­tity holds for $0^\circ \le A \lt 90^\circ$. At $A = 0^\circ$ both sides equal 1.

Iden­tity 3: 1 + cot² A = cosec² A#

Divide $AB^2 + BC^2 = AC^2$ by $BC^2$:

$$\displaystyle \left(\frac{AB}{BC}\right)^2 + 1 = \left(\frac{AC}{BC}\right)^2 \quad \Longrightarrow \quad 1 + \cot^2 A = \operatorname{cosec}^2 A.$$

This needs $BC \ne 0$, that is $\sin A \ne 0$, so it holds for $0^\circ \lt A \le 90^\circ$. At $A = 90^\circ$ both sides equal 1.

Use­ful rearrange­ments#

$$\begin{aligned} \sin^2 A &= 1 - \cos^2 A, & \cos^2 A &= 1 - \sin^2 A, \\ \sec^2 A - \tan^2 A &= 1, & \operatorname{cosec}^2 A - \cot^2 A &= 1. \end{aligned}$$

The last two fac­torise, which is a pow­er­ful trick: $(\sec A - \tan A)(\sec A + \tan A) = 1$, so $\displaystyle \sec A - \tan A = \frac{1}{\sec A + \tan A}$.

Express­ing every ratio in terms of one#

In terms of sin A#

For an acute angle, all ratios are pos­i­tive, so we take pos­i­tive square roots:

$$\displaystyle \cos A = \sqrt{1 - \sin^2 A}, \qquad \tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}, \qquad \sec A = \frac{1}{\sqrt{1 - \sin^2 A}},$$

$$\displaystyle \operatorname{cosec} A = \frac{1}{\sin A}, \qquad \cot A = \frac{\sqrt{1 - \sin^2 A}}{\sin A}.$$

In terms of tan A#

From $\sec^2 A = 1 + \tan^2 A$ we get $\displaystyle \cos A = \frac{1}{\sqrt{1 + \tan^2 A}}$, and then $\displaystyle \sin A = \tan A \cos A = \frac{\tan A}{\sqrt{1 + \tan^2 A}}$.

A numer­i­cal use#

If $\displaystyle \cos A = \frac{5}{13}$ for an acute angle $A$, then

$$\displaystyle \sin A = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}, \qquad \tan A = \frac{\sin A}{\cos A} = \frac{12}{5}.$$

This gives the same answer as draw­ing a tri­an­gle, with­out need­ing the draw­ing.

A strat­egy for prov­ing iden­ti­ties#

1. Start with the side that looks more com­pli­cated. Sim­pli­fy­ing is eas­ier than com­pli­cat­ing.
2. If the expres­sion mixes sev­eral ratios, write every­thing in terms of $\sin$ and $\cos$.
3. Com­bine frac­tions over a com­mon denom­i­na­tor, expand brack­ets, or mul­ti­ply by a con­ju­gate such as $1 + \sin A$ when you see $1 - \sin A$.
4. Look for $\sin^2 A + \cos^2 A$, $\sec^2 A - \tan^2 A$ or $\operatorname{cosec}^2 A - \cot^2 A$ and replace it by 1.
5. Sim­plify until you reach the other side exactly, and end with "= RHS".
6. If you get stuck, sim­plify the other side as well and show that both reduce to the same expres­sion.

Worked proofs#

Proof 1: a one-step sub­sti­tu­tion#

Prove that $\tan^2 A\,(1 - \sin^2 A) = \sin^2 A$ for $0^\circ \le A \lt 90^\circ$.

Proof.

$$\displaystyle \text{LHS} = \tan^2 A\,(1 - \sin^2 A) = \frac{\sin^2 A}{\cos^2 A} \cdot \cos^2 A = \sin^2 A = \text{RHS}.$$

Proof 2: con­vert­ing to sin and cos#

Prove that $\operatorname{cosec} A\,(1 - \cos A)(\operatorname{cosec} A + \cot A) = 1$ for $0^\circ \lt A \le 90^\circ$.

Proof. Write $\displaystyle \operatorname{cosec} A = \frac{1}{\sin A}$ and $\displaystyle \cot A = \frac{\cos A}{\sin A}$:

$$\displaystyle \begin{aligned} \text{LHS} &= \frac{1}{\sin A}\,(1 - \cos A) \cdot \frac{1 + \cos A}{\sin A} \\ &= \frac{1 - \cos^2 A}{\sin^2 A} = \frac{\sin^2 A}{\sin^2 A} = 1 = \text{RHS}. \end{aligned}$$

Proof 3: adding two frac­tions#

Prove that $\displaystyle \frac{1 - \cos \theta}{\sin \theta} + \frac{\sin \theta}{1 - \cos \theta} = 2 \operatorname{cosec} \theta$ for $0^\circ \lt \theta \le 90^\circ$.

Proof. Take the com­mon denom­i­na­tor $\sin \theta\,(1 - \cos \theta)$, which is not zero because $\theta \ne 0^\circ$:

$$\displaystyle \begin{aligned} \text{LHS} &= \frac{(1 - \cos \theta)^2 + \sin^2 \theta}{\sin \theta\,(1 - \cos \theta)} = \frac{1 - 2\cos \theta + \cos^2 \theta + \sin^2 \theta}{\sin \theta\,(1 - \cos \theta)} \\ &= \frac{1 + 1 - 2\cos \theta}{\sin \theta\,(1 - \cos \theta)} = \frac{2(1 - \cos \theta)}{\sin \theta\,(1 - \cos \theta)} \\ &= \frac{2}{\sin \theta} = 2\operatorname{cosec} \theta = \text{RHS}. \end{aligned}$$

Proof 4: dif­fer­ence of squares#

Prove that $\sin^4 A - \cos^4 A = 2\sin^2 A - 1$.

Proof. Treat the left side as $x^2 - y^2$ with $x = \sin^2 A$ and $y = \cos^2 A$:

$$\begin{aligned} \text{LHS} &= (\sin^2 A - \cos^2 A)(\sin^2 A + \cos^2 A) = (\sin^2 A - \cos^2 A) \times 1 \\ &= \sin^2 A - (1 - \sin^2 A) = 2\sin^2 A - 1 = \text{RHS}. \end{aligned}$$

Proof 5: ratio­nal­is­ing under a square root#

Prove that $\displaystyle \sqrt{\frac{\sec A - 1}{\sec A + 1}} = \frac{\sin A}{1 + \cos A}$ for an acute angle $A$.

Proof. First mul­ti­ply numer­a­tor and denom­i­na­tor inside the root by $\cos A$, then by $1 + \cos A$:

$$\displaystyle \begin{aligned} \text{LHS} &= \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \sqrt{\frac{(1 - \cos A)(1 + \cos A)}{(1 + \cos A)^2}} = \sqrt{\frac{1 - \cos^2 A}{(1 + \cos A)^2}} \\ &= \sqrt{\frac{\sin^2 A}{(1 + \cos A)^2}} = \frac{\sin A}{1 + \cos A} = \text{RHS}. \end{aligned}$$

Tak­ing the pos­i­tive root is cor­rect because $\sin A$ and $1 + \cos A$ are both pos­i­tive for an acute angle.

Proof 6: a squared sum#

Prove that $\displaystyle (\sec \theta + \tan \theta)^2 = \frac{1 + \sin \theta}{1 - \sin \theta}$ for $0^\circ \le \theta \lt 90^\circ$.

Proof. Con­vert the left side to sine and cosine:

$$\displaystyle \begin{aligned} \text{LHS} &= \left(\frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}\right)^2 = \frac{(1 + \sin \theta)^2}{\cos^2 \theta} = \frac{(1 + \sin \theta)^2}{1 - \sin^2 \theta} \\ &= \frac{(1 + \sin \theta)^2}{(1 + \sin \theta)(1 - \sin \theta)} = \frac{1 + \sin \theta}{1 - \sin \theta} = \text{RHS}. \end{aligned}$$

Can­celling $1 + \sin \theta$ is allowed because it is at least 1, never zero. The denom­i­na­tor $1 - \sin \theta$ is not zero because $\theta \ne 90^\circ$, which is also what makes $\sec \theta$ and $\tan \theta$ defined.

Com­mon mis­takes#

• Trans­pos­ing terms from one side to the other as if solv­ing an equa­tion. In a proof, work on one side at a time.
• Writ­ing $\sin^2 A + \cos^2 A = 1$ as $\sin A + \cos A = 1$. The squares mat­ter: at $45^\circ$, $\sin A + \cos A = \sqrt{2}$.
• Expand­ing $(1 - \cos \theta)^2$ as $1 - \cos^2 \theta$ and los­ing the mid­dle term $-2\cos\theta$.
• Can­celling a term that is added, not mul­ti­plied: $\displaystyle \frac{1 + \sin A}{\sin A}$ is not $1$.
• Using $1 + \tan^2 A = \sec^2 A$ at $A = 90^\circ$, where nei­ther side is defined.
• Stop­ping one step short, for exam­ple at $\displaystyle \frac{1}{\sin \theta} \cdot 2$, with­out writ­ing it in the exact form asked for.

Try these#

1. If $\sec A + \tan A = 3$, find $\sec A - \tan A$ and $\sin A$. Answer: $\displaystyle \sec A - \tan A = \frac{1}{3}$, $\displaystyle \sin A = \frac{4}{5}$.
2. Sim­plify $\displaystyle \frac{1 + \cot^2 A}{1 + \tan^2 A}$ for $0^\circ \lt A \lt 90^\circ$. Answer: $\cot^2 A$.
3. If $\displaystyle \sin \theta = \frac{3}{5}$ and $\theta$ is acute, use an iden­tity to find $\sec \theta$. Answer: $\displaystyle \frac{5}{4}$.
4. Find the value of $7\operatorname{cosec}^2 A - 7\cot^2 A$. Answer: $7$.
5. Find the value of $\sin^2 25^\circ + \cos^2 25^\circ$. Answer: $1$.
6. Sim­plify $\cos \theta \tan \theta \operatorname{cosec} \theta$. Answer: $1$.

Key terms#

Iden­tity

An equa­tion that is true for every value of the angle for which both sides are defined.

Pythagorean iden­ti­ties

The three iden­ti­ties sin² A + cos² A = 1, 1 + tan² A = sec² A and 1 + cot² A = cosec² A.

LHS and RHS

The left-hand side and right-hand side of the state­ment being proved.

Con­ju­gate

The part­ner expres­sion, such as 1 + sin A for 1 − sin A, whose prod­uct gives a dif­fer­ence of squares.

Com­mon denom­i­na­tor

A sin­gle denom­i­na­tor used to add or sub­tract frac­tions, often a prod­uct of the sep­a­rate denom­i­na­tors.

Domain of an iden­tity

The angles for which every ratio in it is defined, for exam­ple 0° ≤ A < 90° for 1 + tan² A = sec² A.

Com­mon ques­tions#

Can I work on both sides at once?#

Yes, as long as you sim­plify each side sep­a­rately and show that both reach the same expres­sion. What you must not do is move terms across the equals sign.

Why do we always con­vert to sin and cos?#

Because sine and cosine are linked by the sim­plest iden­tity, and every other ratio can be writ­ten using them. It is not the only route, but it almost always works.

Is the iden­tity true for obtuse angles too?#

Yes, but the Class 10 syl­labus only defines the ratios for angles from 0° to 90°, so state and use them in that range.

How do I know when to mul­ti­ply by a con­ju­gate?#

When you see $1 - \sin A$, $1 + \cos A$ or $\sec A - \tan A$ in a denom­i­na­tor or under a root, mul­ti­ply­ing by its con­ju­gate cre­ates a squared ratio that an iden­tity can replace.

How much work­ing should I show in a 3-mark proof?#

Show every sub­sti­tu­tion and every use of an iden­tity on its own line, and fin­ish with "= RHS". Exam­in­ers look for the iden­tity you used at each step.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Loney, S. L. Plane Trigonom­e­try. Cam­bridge Uni­ver­sity Press.
3. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
4. Stew­art, J., Redlin, L. and Wat­son, S. Pre­cal­cu­lus: Math­e­mat­ics for Cal­cu­lus. Cen­gage Learn­ing.