Introduction to TrigonometryClass 10 Maths, Chapter 8

This chapter introduces six ratios linking the sides of a right triangle to its acute angles, their values at the standard angles, and the identities that connect them. After revising it here, you should be able to find any ratio from a given one, use the standard angle table, and prove simple identities.

Follows the Class 10 Maths textbook, Chapter 8. Made with Classly and checked by our team.

Cheat sheet

Introduction to Trigonometry on one page, for a last look before a test.

Introduction to Trigonometry

Cheat sheetClass 10 · Mathematics

⚑ Panic box

Core identity — $\sin^2 A+\cos^2 A=1$

Other identities — $1+\tan^2 A=\sec^2 A$; $1+\cot^2 A=\cosec^2 A$

Triangle sides — label opposite and adjacent from the chosen angle; hypotenuse is always opposite $90^\circ$

Special angles — know the $0^\circ,30^\circ,45^\circ,60^\circ,90^\circ$ table; check undefined entries

1. Ratios & side relationships

RatioDefinition in a right triangleReciprocal / link
$\sin A$opposite / hypotenuse$\cosec A=1/\sin A$
$\cos A$adjacent / hypotenuse$\sec A=1/\cos A$
$\tan A$opposite / adjacent$\cot A=1/\tan A=\cos A/\sin A$
  • Side labels — opposite and adjacent depend on the chosen acute angle; hypotenuse does not.
  • Same angle, same ratios — changing the right triangle's size does not change its ratios (similar triangles).
  • From one ratio to the rest — treat a given side ratio as lengths in scale $k$; use Pythagoras for the third side, take the positive length, then form each ratio.

2. Special-angle values

$A$$0^\circ$$30^\circ$$45^\circ$$60^\circ$$90^\circ$
$\sin A$$0$$\frac12$$\frac{1}{\sqrt2}$$\frac{\sqrt3}{2}$$1$
$\cos A$$1$$\frac{\sqrt3}{2}$$\frac{1}{\sqrt2}$$\frac12$$0$
$\tan A$$0$$\frac{1}{\sqrt3}$$1$$\sqrt3$undefined
$\cosec A$undefined$2$$\sqrt2$$\frac{2}{\sqrt3}$$1$
$\sec A$$1$$\frac{2}{\sqrt3}$$\sqrt2$$2$undefined
$\cot A$undefined$\sqrt3$$1$$\frac{1}{\sqrt3}$$0$
  • As $A$ rises from $0^\circ$ to $90^\circ$ — $\sin A$ rises from $0$ to $1$; $\cos A$ falls from $1$ to $0$.
  • Range check — $\sin A,\cos A\leq 1$; $\sec A\geq1$ for $0^\circ\leq A<90^\circ$; $\cosec A\geq1$ for $0^\circ<A\leq90^\circ$.

3. Identities & conditions

IdentityValid anglesUseful rearrangement
$\sin^2 A+\cos^2 A=1$$0^\circ\leq A\leq90^\circ$$\sin^2 A=1-\cos^2 A$; $\cos^2 A=1-\sin^2 A$
$1+\tan^2 A=\sec^2 A$$0^\circ\leq A<90^\circ$$\sec^2 A-\tan^2 A=1$
$1+\cot^2 A=\cosec^2 A$$0^\circ<A\leq90^\circ$$\cosec^2 A-\cot^2 A=1$
  • Identity — an equation true for every angle in the stated domain where its ratios are defined.
  • Proving an identity — rewrite ratios using $\sin$ and $\cos$, or convert to the ratios in the identity you need; simplify one side to the other.
  • Known ratio → other ratios — use reciprocal links and the identities; for an acute angle, choose positive square roots.

4. Finding a side or angle

If you needThen use
Opposite and hypotenuse$\sin A$
Adjacent and hypotenuse$\cos A$
Opposite and adjacent$\tan A$
A missing side in a right trianglePythagoras' theorem, or a ratio containing the known and unknown sides
A special angle from a ratioMatch the ratio in the table; for acute angles, equal sine or cosine values imply equal angles

5. Traps

  • Opposite/adjacent are fixed triangle labels → identify them again whenever the reference angle changes.
  • $\cosec A=\sin^{-1}A$ → $\cosec A=1/\sin A$; inverse sine is different notation.
  • $\sin A$ means $\sin\times A$ → $\sin A$ names the sine of angle $A$.
  • $\tan A$ is always less than $1$ → it equals $1$ at $45^\circ$ and exceeds $1$ above $45^\circ$ up to $90^\circ$.
  • All six ratios exist at $0^\circ$ and $90^\circ$ → division by zero makes $\cosec 0^\circ$, $\cot 0^\circ$, $\tan 90^\circ$, and $\sec 90^\circ$ undefined.
  • $\sin^2 A=(\sin A)^2$ means $\sin(A^2)$ → the square applies to the ratio, not the angle.

Flashcards

10 cards on Introduction to Trigonometry. Flip them here, or read them all below.

Introduction to Trigonometry flashcards1 of 10

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All 10 cards

  1. 1

    In a right triangle, $\sin A=$

    $\sin A=\frac{\text{opposite}}{\text{hypotenuse}}$

  2. 2

    In a right triangle, $\cos A=$

    $\cos A=\frac{\text{adjacent}}{\text{hypotenuse}}$

  3. 3

    In a right triangle, $\tan A=$

    $\tan A=\frac{\text{opposite}}{\text{adjacent}}$

  4. 4

    The reciprocal of $\sin A$ is

    $\cosec A=\frac{1}{\sin A}$

  5. 5

    Complete the fundamental identity: $\sin^2 A+\cos^2 A=$

    $\sin^2 A+\cos^2 A=1$

  6. 6

    Complete the identity: $1+\tan^2 A=$

    $1+\tan^2 A=\sec^2 A$

  7. 7

    What is $\sin 30^\circ$?

    $\frac{1}{2}$

  8. 8

    What is $\sin 45^\circ$?

    $\frac{\sqrt{2}}{2}$

  9. 9

    What is $\tan 45^\circ$?

    11

  10. 10

    Why does $\sin A$ stay the same in a bigger or smaller right triangle with the same angle $A$?

    The triangles are similar, so their corresponding sides are in proportion and the ratio does not change.

Quiz

5 questions, each with the reason behind its answer.

1 of 5

In a right triangle, $\tan \theta=\frac{5}{12}$. If $\theta$ is acute, what is $\sec \theta$?

Answer key
  1. 1

    In a right triangle, $\tan \theta=\frac{5}{12}$. If $\theta$ is acute, what is $\sec \theta$?

    Answer: A. $\frac{13}{12}$

    Take the opposite side as $5k$ and the adjacent side as $12k$. Then the hypotenuse is $\sqrt{25k^2+144k^2}=13k$, so $\sec\theta=\frac{13}{12}$.

  2. 2

    For an acute angle $\theta$, what is the value of $\frac{1-\cos^2\theta}{\sin^2\theta}$?

    Answer: B. 11

    Using $\sin^2\theta+\cos^2\theta=1$, the numerator $1-\cos^2\theta$ equals $\sin^2\theta$. The quotient is therefore $1$.

  3. 3

    Evaluate $\sin 60^\circ\cos 30^\circ+\cos 60^\circ\sin 30^\circ$.

    Answer: A. 11

    Substituting the standard values gives $\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{1}{2}\cdot\frac{1}{2}=\frac{3}{4}+\frac{1}{4}=1$.

  4. 4

    What is the value of $\cos^2 45^\circ+\sin^2 30^\circ$?

    Answer: C. $\frac{3}{4}$

    Since $\cos 45^\circ=\frac{1}{\sqrt{2}}$ and $\sin 30^\circ=\frac{1}{2}$, the sum of their squares is $\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$.

  5. 5

    For any two right triangles containing the same acute angle, the sine of that angle has the same value.

    Answer: A. True

    Two right triangles with the same acute angle are similar, so their sides are in the same proportion. The ratio opposite / hypotenuse is therefore the same, and so is the sine.

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Questions about Introduction to Trigonometry

What are the most important topics in Class 10 Maths Chapter 8?

Focus on the definitions of sine, cosine and tangent, the values at $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$ and $90^\circ$, and the three identities. Most exam questions ask you to find one ratio from another, evaluate an expression using the table, or prove an identity.

How do I remember the trigonometry table?

Write the sine row as $\frac{\sqrt0}{2}, \frac{\sqrt1}{2}, \frac{\sqrt2}{2}, \frac{\sqrt3}{2}, \frac{\sqrt4}{2}$ for $0^\circ$ to $90^\circ$. The cosine row is the same list in reverse. Tangent is sine divided by cosine, and the other three ratios are reciprocals.

Why do the trigonometric ratios depend only on the angle?

Right triangles with the same acute angle are similar, so their sides are in proportion. That means opposite / hypotenuse, and every other ratio, stays the same however large the triangle is. This is why a ratio can be tied to an angle alone.

Is $\sin^2 A$ the same as $\sin A^2$, and is $\cosec A$ the same as $\sin^{-1} A$?

$\sin^2 A$ means $(\sin A)^2$, the square of the ratio, not the sine of $A^2$. Also, $\cosec A$ is $\frac{1}{\sin A}$, while $\sin^{-1}A$ is the inverse sine, which is a different thing. You can make your own flashcards or quiz for any chapter in Classly to practise these.