Introduction to Trigonometry: Class 10 Maths · Chapter 8

Six ratios of a right triangle, their values at special angles, and the identities that link them

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

75 min
32 practice questions
20 MCQs

What you will learn in Introduction to Trigonometry

Name the opposite side, the adjacent side and the hypotenuse with respect to either acute angle of a right triangle.
Define sin, cos, tan, cosec, sec and cot of an acute angle, and explain why their values depend only on the angle and not on the size of the triangle.
Find all six trigonometric ratios of an angle when one of them is given, using the Pythagoras theorem.
Derive the ratios of 30°, 45° and 60° from simple triangles, explain the values at 0° and 90°, and say which ratios are not defined there.
Use the special values to evaluate expressions and to find unknown sides and angles of right triangles.
Prove sin²A + cos²A = 1, 1 + tan²A = sec²A and 1 + cot²A = cosec²A, state where each holds, and use them to prove other identities.

Original companion text aligned to NCERT Mathematics Class 10, Chapter 8, pp. 113–132. Not affiliated with or endorsed by NCERT.

8.1 Introduction

Look around and you will find right triangles everywhere, even where nobody has drawn one: in a leaning ladder, a kite string, your line of sight to a rooftop. You already know their most famous fact, the Pythagoras theorem. This chapter gives you a second, more powerful tool.

A kite on Makar Sankranti. Meera has let out a long string, pulled straight by the wind. The string, a vertical line dropped from the kite and a level line from her hand form a right triangle. Can she tell how high the kite is without climbing anything?
A painter's ladder. A ladder leans against the wall of a house. The ladder, the wall and the ground make a right triangle. If you know the ladder's length and how steeply it leans, can you say how high up the wall it reaches?
A mobile tower. You stand 40 m from the foot of a tower and look up at its top. Your line of sight, the tower and a level line from your eye make a right triangle. Can you find the height of the tower without leaving the spot?

3/3The heights marked ? are hard to measure directly. Trigonometry finds them from lengths and angles that are easy to measure.

Fig. 8.1Right triangles hidden in everyday scenes.NCERT Figs. 8.1–8.3 · p. 113

Three everyday scenes on level ground. On the left a child flies a kite on a straight string; in the middle a ladder leans against the wall of a house; on the right a person stands 40 m from a tall lattice mobile tower and looks up at its top. In the first step only the scenes are drawn. In the second step a shaded right triangle appears in each scene, with a right-angle mark and the acute angle at its lower corner marked: string, vertical drop and level line; ladder, wall and ground; line of sight, tower and level line. In the third step the unknown heights are marked with question marks and the 40 m distance is labelled.

In each case, a length that is hard to reach is one side of a right triangle whose other parts are easy to measure. The tools that connect these parts belong to trigonometry. The name joins the Greek trigōnon (a triangle, literally "three corners") and metron (a measure). So trigonometry is, quite literally, triangle measurement: it links the lengths of the sides of a triangle with the sizes of its angles.

Every GPS receiver, every surveyor measuring a plot of land and every engineer designing a bridge relies on trigonometry today. Its roots are ancient. Egyptian and Babylonian builders and sky-watchers used its ideas, and Greek and Indian astronomers made tables from it to estimate how far away the Moon and the planets are.

Timeline

Trigonometry through the ages

c. 1800 BCE
A Babylonian clay tablet
The tablet now called Plimpton 322 lists numbers that match right triangles with whole-number sides.
2nd century BCE
Hipparchus's chord table
The Greek astronomer Hipparchus listed the lengths of chords of a circle to study the sky.
499 CE
Āryabhaṭīya
Āryabhaṭa gave a table of half-chords, the ancestor of the sine we use today.
1600s
Short names
Writers in England began to use the short forms "sin" and "cos".

Three questions drive this chapter. Which ratios of sides belong to an angle of a right triangle, and why does the size of the triangle not matter? What exact values do these ratios take at $30^{\circ}$ , $45^{\circ}$ and $60^{\circ}$ , and what happens at the extremes $0^{\circ}$ and $90^{\circ}$ ? And which relations between the ratios, called identities, hold for every angle?

Quick check

1 of 2

In the ladder scene, which angle of the right triangle is

90^{\circ}

the angle between the ladder and the groundthe angle between the ladder and the wallthe angle between the wall and the groundthe angle at the top of the ladder

Show answer

Answer (C) the angle between the wall and the ground

The wall stands straight up from level ground, so the right angle is where the wall meets the ground. The ladder makes acute angles with both of them.

8.2 Trigonometric Ratios

Take a right triangle $A B C$ with the right angle at $B$ . Its other two angles, $∠ A$ and $∠ C$ , are acute, and together they make $90^{\circ}$ . Fix your attention on $∠ A$ , written simply as $A$ . Each side now gets a name that says where it sits with respect to $A$ .

The hypotenuse $A C$ lies across from the right angle. It is the longest side, and its name never changes.
The side opposite to $A$ is $B C$ . It is the only side that does not touch $A$ .
The side adjacent to $A$ is $A B$ . Together with the hypotenuse, it forms the angle $A$ .

1/2From $A$ : $B C$ is opposite, $A B$ is adjacent and $A C$ is the hypotenuse.

Fig. 8.2Opposite and adjacent depend on the angle you look from.NCERT Figs. 8.4, 8.5 · p. 114

Triangle ABC has its right angle at B, with AB along the bottom and BC upright. In the first step angle A is shaded; BC is labelled opposite A, AB is labelled adjacent to A, and AC is labelled hypotenuse. In the second step angle C is shaded instead; now AB is labelled opposite C and BC adjacent to C, while AC is still the hypotenuse. In both steps the opposite side is orange and the adjacent side blue, so the two shorter sides swap colours as well as names.

If you look from $∠ C$ instead, the two shorter sides swap names: $A B$ becomes the side opposite to $C$ , and $B C$ the side adjacent to $C$ . So "opposite" and "adjacent" make sense only after you say which angle you are looking from.

The six ratios

A ratio of two sides of a right triangle, taken with respect to one of its acute angles, is called a trigonometric ratio of that angle. Three sides can be paired in six ways, and each way has a name.

Table 8.1The six trigonometric ratios of

∠ A

△ A B C

, right-angled at

B

NCERT p. 115

Ratio (name)	Which sides	In △ABC
$\sin A$ (sine)	opposite ÷ hypotenuse	$\frac{B C}{A C}$
$\cos A$ (cosine)	adjacent ÷ hypotenuse	$\frac{A B}{A C}$
$\tan A$ (tangent)	opposite ÷ adjacent	$\frac{B C}{A B}$
$cosec A$ (cosecant)	hypotenuse ÷ opposite	$\frac{A C}{B C}$
$\sec A$ (secant)	hypotenuse ÷ adjacent	$\frac{A C}{A B}$
$\cot A$ (cotangent)	adjacent ÷ opposite	$\frac{A B}{B C}$

The terms sine, cosine and tangent are the ones you will use most. The last three ratios in the table are the reciprocals of the first three:

\begin{aligned} cosec A & = \frac{1}{\sin A} \\ \sec A & = \frac{1}{\cos A} \\ \cot A & = \frac{1}{\tan A} \end{aligned}

Dividing the top and the bottom of $\frac{B C}{A B}$ by the hypotenuse $A C$ links the tangent to the sine and the cosine:

\tan A = \frac{\sin A}{\cos A}

(8.1)

\cot A = \frac{\cos A}{\sin A}

(8.2)

A word of warning about the symbols. In $3 x$ you can split off the 3, but in $\sin A$ you cannot split "sin" from $A$ : the whole symbol names one number, the sine of the angle $A$ . Writing "sin" alone is like writing a square-root sign with nothing under it. The same goes for $\cos A$ , $\tan A$ and the rest.

From history

Where the word "sine" comes from

Around 499 CE, Āryabhaṭa used the half-chord of a circle in his astronomy. He called it ardha-jyā (अर्धज्या, "half a bow-string"), later shortened to jyā or jīvā. Arab scholars who took up his work wrote the word as jība. When their books were turned into Latin in the twelfth century, the word was taken for a similar Arabic word meaning a fold or a pocket, and it became the Latin sinus. From sinus we get "sine".

Indian astronomers also tabulated the sine of the other acute angle of the triangle, under the name koṭi-jyā; that is our cosine. The short forms came many centuries later. The English astronomer Edmund Gunter (1581–1626) is credited with the short form "sin", and "cos" first appeared in 1674, in the work of Sir Jonas Moore.

Ratios depend on the angle, not on the size

Does the sine of an angle change if we draw a bigger triangle with the same angle? Look at Fig. 8.3. Points $D$ , $E$ and $C$ lie on one arm of $∠ A$ , and the perpendiculars $D F$ , $E G$ and $C B$ fall on the other arm.

4/4Each ratio is about 0.47: it belongs to the angle $A$ , not to any one triangle.

Fig. 8.3Same angle, same ratio:

\frac{8}{17} = \frac{16}{34} = \frac{24}{51} .

NCERT Fig. 8.6 · p. 116

An angle A with points D, E and C on its sloping arm and perpendiculars DF, EG and CB dropped to its horizontal arm, making right triangles ADF, AEG and ACB. Their opposite sides are 8, 16 and 24 units and their hypotenuses are 17, 34 and 51 units. Step one shades the small triangle and shows DF ÷ AD = 8 ÷ 17. Step two shades the middle triangle and adds EG ÷ AE = 16 ÷ 34. Step three shades the large triangle and adds CB ÷ AC = 24 ÷ 51. The last step shows all three triangles and notes that each ratio is about 0.47.

The triangles $A D F$ , $A E G$ and $A C B$ share the angle $A$ , and each has a right angle. So, by the AA similarity criterion of Chapter 6, they are similar, and their matching sides are in proportion. This gives

\frac{D F}{A D} = \frac{E G}{A E} = \frac{C B}{A C}

Each of these fractions is the side opposite $A$ divided by the hypotenuse, so each one is $\sin A$ . In the same way $\frac{A F}{A D} = \frac{A G}{A E} = \frac{A B}{A C},$ so $\cos A$ is the same in all three triangles, and so is every other ratio. The value of a trigonometric ratio depends only on the angle, not on how big the triangle is. That is why it makes sense to talk about "the sine of $35^{\circ}$ " at all.

Try this

Measure the sine of 35° yourself

You’ll need

a ruler and a protractor
a set square
a sharp pencil

Predict first

You will mark points 5 cm, 8 cm and 10 cm from the corner along one arm of a

35^{\circ}

angle and drop perpendiculars. Which point gives the largest value of opposite ÷ hypotenuse?

the 5 cm pointthe 8 cm pointthe 10 cm pointall give about the same value

Draw a ray $A X$ and, with the protractor, an angle $∠ X A Y = 35^{\circ}$ .
On $A Y$ , mark points 5 cm, 8 cm and 10 cm from $A$ .
From each point, drop a perpendicular to $A X$ using the set square.
Measure each perpendicular and divide it by the distance of its point from $A$ .

Table 8.2

Hypotenuse	Opposite	Ratio
5 cm
8 cm
10 cm

Your entries are saved automatically.

InteractiveRight-triangle explorer

A right triangle ABC with the right angle at B. You can drag the corner C, or use two sliders: one sets angle A from 0° to 90°, the other changes only the hypotenuse, from 3 cm to 10 cm. The opposite side, adjacent side and hypotenuse are colour-coded and labelled with their lengths. A table lists all six ratios with the sides they use and their values, with exact forms such as 1/2 or √3/2 at 30°, 45° and 60°, and "not defined" where a ratio would divide by zero at 0° or 90°. Buttons switch the view to angle C, jump to the special angles, and Play sweeps the angle from 0° to 90°. A line below checks that sin² + cos² = 1.

Fig. 8.4Drag

C

to change the angle, or change only the size and watch the ratios stay put.

Quick check

1 of 4

△ P Q R

, right-angled at

Q

, which fraction is

\tan R

\frac{P Q}{Q R}

\frac{Q R}{P Q}

\frac{P Q}{P R}

\frac{Q R}{P R}

Show answer

Answer (A) $\frac{P Q}{Q R}$

From

R

, the opposite side is

P Q

and the adjacent side is

Q R

, so

\tan R = \frac{P Q}{Q R}

. The tempting

\frac{Q R}{P Q}

\tan P

: it looks from the wrong angle.

From one ratio to all six

Knowing one ratio of an acute angle is enough to find all six. Say you are told only that $\cos A = \frac{2}{3}$ , where $∠ B = 90^{\circ}$ in $△ A B C$ .

Draw any right triangle in which the adjacent side and the hypotenuse are in the ratio $2 : 3$ . Since the size does not matter, call them $A B = 2 k$ and $A C = 3 k$ for some positive number $k$ . The missing side comes from the Pythagoras theorem:

\begin{aligned} B C^{2} & = A C^{2} - A B^{2} \\ = 9 k^{2} - 4 k^{2} \\ = 5 k^{2} \end{aligned}

So $B C = \sqrt{5} k$ . We take only the positive root, because a length cannot be negative. Now every ratio follows from the definitions: $\sin A = \frac{\sqrt{5}}{3}$ , $\tan A = \frac{\sqrt{5}}{2}$ , $cosec A = \frac{3}{\sqrt{5}}$ , $\sec A = \frac{3}{2}$ and $\cot A = \frac{2}{\sqrt{5}}$ . Notice that $k$ cancels every time, as it must.

Example 8.1NCERT Example 1 · p. 118

A ramp that rises 12 cm for every 35 cm

A ramp at a school gate rises 12 cm for every 35 cm it runs along the ground. So the angle $A$ that it makes with the ground has $\tan A = \frac{12}{35}$ . Find the other five trigonometric ratios of $A$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 8.2NCERT Example 2 · p. 118

Equal tangents mean equal angles

Two acute angles, $∠ A$ and $∠ P$ , have the same tangent. Prove that the angles are equal.

Try it yourself first. Then reveal the solution one step at a time.

The same is true for every ratio: if two acute angles have equal sines, or equal cosines, the angles are equal. You will prove the cosine case in Practice 8.1.

Example 8.3NCERT Example 3 · p. 119

The third side by a difference of squares

The hypotenuse $P R$ of a right triangle $P Q R$ ( $∠ Q = 90^{\circ}$ ) is 65 units long and the side $Q R$ is 33 units. Let $θ$ be the angle at $R$ . Work out (i) $\sin^{2} θ + \cos^{2} θ$ and (ii) $\sin^{2} θ - \cos^{2} θ$ .

Try it yourself first. Then reveal the solution one step at a time.

Part (i) came out as exactly 1. That is no accident: in Section 8.4 you will prove that $\sin^{2} θ + \cos^{2} θ = 1$ for every angle $θ$ .

Example 8.4NCERT Example 4 · p. 120

A square napkin folded in half

A square napkin is folded along a diagonal to make $△ A B C$ with $∠ B = 90^{\circ}$ and $A B = B C$ . Show that $\sin A = \cos A$ , and find the value of $\sin^{2} A + 2 \sin A \cos A$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 8.5NCERT Example 5 · p. 120

One side and a difference

In a right triangle $P Q R$ with $∠ Q = 90^{\circ}$ , the side $P Q$ is 28 cm and the hypotenuse is 8 cm longer than $Q R$ . Find $\sin R$ and $\cos R$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

\sin A = \frac{5}{13}

for an acute angle

A

, what is

\cot A

\frac{5}{12}

\frac{12}{5}

\frac{13}{12}

\frac{12}{13}

Show answer

Answer (B) $\frac{12}{5}$

Opposite

= 5 k

and hypotenuse

= 13 k

, so the adjacent side is

12 k

. Then

\cot A

= adjacent ÷ opposite

= \frac{12}{5}

. The answer

\frac{5}{12}

\tan A

, the same fraction upside down.

8.3 Trigonometric Ratios of Some Specific Angles

A handful of angles appear in board papers far more often than any others: $30^{\circ}$ , $45^{\circ}$ and $60^{\circ}$ , together with the two extremes $0^{\circ}$ and $90^{\circ}$ . The first three come from shapes you can draw with a compass and ruler, and their ratios have exact values that you should know by heart.

Ratios of 45°

Cut a square along a diagonal (Fig. 8.5). Each half is a right triangle whose two shorter sides are equal, so its two acute angles are equal too. As they add up to $90^{\circ}$ , each of them is $45^{\circ}$ .

3/3The diagonal is $\sqrt{2} a$ , so $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$ and $\tan 45^{\circ} = 1$ .

Fig. 8.5Half a square gives the ratios of

45^{\circ}

.NCERT Fig. 8.14 · p. 122

A square of side a with a dashed diagonal AC. In the first step the whole square is shown. In the second step the half triangle ABC, right-angled at B, is shaded and its two acute angles are marked 45°, while the other half fades. In the third step the diagonal is labelled root 2 times a, and the ratios sin 45° = 1/√2, cos 45° = 1/√2 and tan 45° = 1 appear beside the triangle.

DerivationRatios of 45°

Let the half-square be $△ A B C$ with its right angle at $B$ , so that $∠ A = ∠ C = 45^{\circ}$ . Sides opposite equal angles are equal, so $A B = B C$ ; call each of them $a$ . By the Pythagoras theorem,

\begin{aligned} A C^{2} & = a^{2} + a^{2} \\ = 2 a^{2} \\ A C & = \sqrt{2} a \end{aligned}

Now read off the ratios from the definitions; the $a$ cancels every time:

Ratio	Sides	Value
$\sin 45^{\circ}$	$\frac{B C}{A C} = \frac{a}{\sqrt{2} a}$	$\frac{1}{\sqrt{2}}$
$\cos 45^{\circ}$	$\frac{A B}{A C} = \frac{a}{\sqrt{2} a}$	$\frac{1}{\sqrt{2}}$
$\tan 45^{\circ}$	$\frac{B C}{A B} = \frac{a}{a}$	$1$

Their reciprocals are $cosec 45^{\circ} = \sqrt{2}$ , $\sec 45^{\circ} = \sqrt{2}$ and $\cot 45^{\circ} = 1$ .

Ratios of 30° and 60°

For these two angles, start with an equilateral triangle $P Q R$ of side $2 a$ , in which every angle is $60^{\circ}$ . Drop the perpendicular $P S$ from $P$ to $Q R$ (Fig. 8.6).

4/5From $P$ : opposite $a$ , adjacent $\sqrt{3} a$ , hypotenuse $2 a$ .

Fig. 8.6Half an equilateral triangle gives the ratios of

30^{\circ}

and

60^{\circ}

.NCERT Fig. 8.15 · p. 122

An equilateral triangle PQR with sides 2a and all angles 60°. In the second step the altitude PS is drawn to the midpoint S of QR, so QS = SR = a, and the top angle splits into two 30° angles. In the third step the half triangle PQS is shaded, with PQ = 2a, QS = a and the altitude PS = root 3 times a. In the fourth step the 30° angle at P is marked and the ratios sin 30° = 1/2, cos 30° = √3/2 and tan 30° = 1/√3 appear. In the fifth step the 60° angle at Q is marked and the ratios sin 60° = √3/2, cos 60° = 1/2 and tan 60° = √3 appear.

DerivationRatios of 30° and 60°

Triangles $P Q S$ and $P R S$ are right-angled at $S$ . Their hypotenuses $P Q$ and $P R$ are equal, and they share the side $P S$ , so they are congruent by RHS. Hence $Q S = S R = a$ and $∠ Q P S = ∠ R P S = 30^{\circ}$ .

Now work in the right triangle $P Q S$ , where $∠ Q = 60^{\circ}$ , $∠ Q P S = 30^{\circ}$ , $P Q = 2 a$ and $Q S = a$ . By the Pythagoras theorem,

\begin{aligned} P S^{2} & = (2 a)^{2} - a^{2} \\ = 3 a^{2} \\ P S & = \sqrt{3} a \end{aligned}

From $∠ Q P S = 30^{\circ}$ , the opposite side is $Q S = a$ and the adjacent side is $P S = \sqrt{3} a$ . From $∠ Q = 60^{\circ}$ in the same triangle, the two shorter sides swap roles: the opposite side is $P S$ and the adjacent side is $Q S$ . The hypotenuse is $P Q = 2 a$ for both.

Ratio	Sides	Value
$\sin 30^{\circ}$	$\frac{a}{2 a}$	$\frac{1}{2}$
$\cos 30^{\circ}$	$\frac{\sqrt{3} a}{2 a}$	$\frac{\sqrt{3}}{2}$
$\tan 30^{\circ}$	$\frac{a}{\sqrt{3} a}$	$\frac{1}{\sqrt{3}}$
$\sin 60^{\circ}$	$\frac{\sqrt{3} a}{2 a}$	$\frac{\sqrt{3}}{2}$
$\cos 60^{\circ}$	$\frac{a}{2 a}$	$\frac{1}{2}$
$\tan 60^{\circ}$	$\frac{\sqrt{3} a}{a}$	$\sqrt{3}$

The reciprocal ratios follow at once; they are collected in Table 8.4 below. For example, $cosec 30^{\circ}$ is 2, because $\sin 30^{\circ}$ is one half.

Ratios of 0° and 90°

A right triangle cannot really have an acute angle of $0^{\circ}$ or $90^{\circ}$ , so the definitions cannot be used directly. Instead, we watch what happens to the ratios as an angle gets very close to these values, and define the ratios to fit.

Picture a ladder of fixed length leaning against a wall (Fig. 8.7). The ladder is the hypotenuse. The height it reaches on the wall is the side opposite the angle $A$ it makes with the floor, and the distance of its foot from the wall is the adjacent side. So $\sin A = \frac{height}{ladder}$ and $\cos A = \frac{base}{ladder}$ .

2/4At $30^{\circ}$ the height is exactly half the ladder: $\sin 30^{\circ} = 0.5$ .

Fig. 8.7As the ladder stands up,

\sin A

grows towards 1 and

\cos A

shrinks towards 0.NCERT Figs. 8.16–8.18 · p. 123

A ladder of fixed length leans against a wall. The height it reaches on the wall is marked in orange and labelled height, and its base along the floor is marked in blue and labelled base. A readout beside the ladder shows the angle A with the floor, sin A and cos A. Step one: A = 10°, the ladder almost flat, sin A = 0.17 and cos A = 0.98. Step two: A = 30°, sin A = 0.50 and cos A = 0.87. Step three: A = 60°, sin A = 0.87 and cos A = 0.50. Step four: A = 80°, the ladder almost upright, sin A = 0.98 and cos A = 0.17.

As the foot slides away from the wall, the ladder lies flatter and $A$ shrinks towards $0^{\circ}$ . The height shrinks towards 0, while the base grows towards the full length of the ladder. So $\sin A$ gets closer and closer to 0, and $\cos A$ to 1. That is why mathematicians define

\begin{aligned} \sin 0^{\circ} & = 0 \\ \cos 0^{\circ} & = 1 \end{aligned}

The other ratios of $0^{\circ}$ follow: $\tan 0^{\circ} = \frac{0}{1} = 0$ and $\sec 0^{\circ} = \frac{1}{1} = 1$ . But $cosec 0^{\circ}$ and $\cot 0^{\circ}$ would need a division by $\sin 0^{\circ} = 0$ , which is impossible. We say they are not defined.

Now push the foot of the ladder towards the wall. As $A$ grows towards $90^{\circ}$ , the ladder stands almost upright: the height approaches the full ladder and the base shrinks towards 0. So we define

\begin{aligned} \sin 90^{\circ} & = 1 \\ \cos 90^{\circ} & = 0 \end{aligned}

Then $cosec 90^{\circ} = 1$ and $\cot 90^{\circ} = \frac{0}{1} = 0$ , while $\tan 90^{\circ}$ and $\sec 90^{\circ}$ would need a division by $\cos 90^{\circ} = 0$ , so they are not defined.

All the special values

Here are all the values together, for ready reference. The reciprocal ratios are in a second table.

Table 8.3Sine, cosine and tangent of the special anglesNCERT Table 8.1 · p. 125

$A$	$\sin A$	$\cos A$	$\tan A$
$0^{\circ}$	$0$	$1$	$0$
$30^{\circ}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$
$45^{\circ}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}$	$1$
$60^{\circ}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$
$90^{\circ}$	$1$	$0$	not defined

Table 8.4Cosecant, secant and cotangent of the special anglesNCERT Table 8.1 · p. 125

$A$	$cosec A$	$\sec A$	$\cot A$
$0^{\circ}$	not defined	$1$	not defined
$30^{\circ}$	$2$	$\frac{2}{\sqrt{3}}$	$\sqrt{3}$
$45^{\circ}$	$\sqrt{2}$	$\sqrt{2}$	$1$
$60^{\circ}$	$\frac{2}{\sqrt{3}}$	$2$	$\frac{1}{\sqrt{3}}$
$90^{\circ}$	$1$	not defined	$0$

Two trends stand out in the first table (Fig. 8.8). Moving from $0^{\circ}$ towards $90^{\circ}$ , the sine climbs steadily from 0 up to 1, while the cosine slides from 1 down to 0; the two meet only at $45^{\circ}$ . The tangent climbs as well, and near $90^{\circ}$ it grows past any number you name.

3/3The two bars match only at $45^{\circ}$ , where both are $\frac{1}{\sqrt{2}} \approx 0.71$ .

Fig. 8.8As

A

grows,

\sin A

rises and

\cos A

falls.NCERT p. 125

A bar chart with the angles 0°, 30°, 45°, 60° and 90° along the bottom and values from 0 to 1 up the side. Step one shows solid blue bars for sin A, rising through 0, 0.5, 0.71, 0.87 and 1. Step two adds hatched orange bars for cos A, falling through 1, 0.87, 0.71, 0.5 and 0, while the sine bars fade. Step three shows both sets of bars at full strength and marks 45°, where the two bars have the same height.

Using the special values

Once you know these values, you can find unknown sides and angles of a right triangle without measuring them.

3/3 $\tan 30^{\circ} = \frac{1.8}{d}$ gives $d = 1.8 \sqrt{3} \approx 3.12 m$ .

Fig. 8.9One side and one angle are enough to find the rest of a right triangle.NCERT Fig. 8.19 · p. 125

A playground slide with its platform 1.8 m above the ground and a straight slide meeting the ground at 30°, so that the slide, the platform post and the ground form a right triangle. In the first step the slide length and the ground distance are marked with question marks. In the second step the slide is labelled 3.6 m, found from sin 30°. In the third step the ground distance is labelled about 3.12 m, found from tan 30°.

Example 8.6NCERT Example 6 · p. 125

How long is the slide?

The platform of a playground slide is 1.8 m above the ground, and the straight slide makes an angle of $30^{\circ}$ with the ground (Fig. 8.9). Find the length of the slide and the distance of its foot from the platform post.

Try it yourself first. Then reveal the solution one step at a time.

Example 8.7NCERT Example 7 · p. 126

Finding angles from two sides

A rope runs straight from the top of a 5 m pole to a peg in level ground, $5 \sqrt{3}$ m from the foot of the pole. Find the angle that the rope makes with the ground and the angle it makes with the pole.

Try it yourself first. Then reveal the solution one step at a time.

Notice what Example 8.6 and Example 8.7 needed: one length, plus one more piece of information, an angle in the first and a second length in the other. Since the right angle is already known, that is always enough to pin a right triangle down completely; every other side and angle follows.

Example 8.8NCERT Example 8 · p. 126

Two angles from two equations

Find the angles $A$ and $B$ , with $A > B$ and $0^{\circ} < A + B \leq 90^{\circ}$ , for which $\cos (A + B) = 0$ and $\tan (A - B) = \frac{1}{\sqrt{3}}$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

What is

\tan 60^{\circ}

\frac{1}{\sqrt{3}}

\sqrt{3}

\frac{\sqrt{3}}{2}

\frac{1}{2}

Show answer

Answer (B) $\sqrt{3}$

In half an equilateral triangle, the side opposite

60^{\circ}

\sqrt{3} a

and the adjacent side is

a

, so

\tan 60^{\circ} = \sqrt{3}

. The value

\frac{1}{\sqrt{3}}

\tan 30^{\circ}

: the same two sides the other way up.

8.4 Trigonometric Identities

In algebra, an identity is an equation that holds for every value of its variables, such as $(x + 1)^{2} = x^{2} + 2 x + 1$ . An ordinary equation such as $x + 1 = 5$ holds for just one value.

Definition

Trigonometric identity

An equation in the trigonometric ratios that holds for every angle at which both of its sides are defined.

Example $\tan A = \frac{\sin A}{\cos A}$ holds for every $A$ with $0^{\circ} \leq A < 90^{\circ}$ .

You have already met two identities, (8.1) and (8.2). The most important one comes straight from the Pythagoras theorem, and two more follow from it.

Three identities from one theorem

Take $△ A B C$ , right-angled at $B$ . The Pythagoras theorem says

A B^{2} + B C^{2} = A C^{2}

(8.3)

Divide (8.3) by $A C^{2}$ . Each fraction is now the square of a ratio of the angle $A$ :

\frac{A B^{2}}{A C^{2}} + \frac{B C^{2}}{A C^{2}} = 1

\sin^{2} A + \cos^{2} A = 1

(8.4)

At $A = 0^{\circ}$ this reads $0 + 1 = 1$ , and at $A = 90^{\circ}$ it reads $1 + 0 = 1$ . So (8.4) holds for all $A$ with $0^{\circ} \leq A \leq 90^{\circ}$ .

Divide (8.3) by $A B^{2}$ :

1 + \frac{B C^{2}}{A B^{2}} = \frac{A C^{2}}{A B^{2}}

1 + \tan^{2} A = \sec^{2} A

(8.5)

Check the ends: at $0^{\circ}$ both sides equal 1, but at $90^{\circ}$ neither $\tan A$ nor $\sec A$ exists. So (8.5) is valid for $0^{\circ} \leq A < 90^{\circ}$ .

Divide (8.3) by $B C^{2}$ :

\frac{A B^{2}}{B C^{2}} + 1 = \frac{A C^{2}}{B C^{2}}

\cot^{2} A + 1 = {cosec}^{2} A

(8.6)

This time the trouble is at the other end: $\cot A$ and $cosec A$ do not exist at $0^{\circ}$ , so (8.6) is valid for $0^{\circ} < A \leq 90^{\circ}$ .

There is a neat way to see all three at once. Scale the triangle so that its hypotenuse is 1; its shorter sides then become $\cos A$ and $\sin A$ . Scale it so that the adjacent side is 1, or so that the opposite side is 1, and the sides become the other ratios (Fig. 8.10). The Pythagoras theorem in each of these triangles is one of the identities.

1/3Hypotenuse 1: the legs are $\cos A$ and $\sin A$ , so $\cos^{2} A + \sin^{2} A = 1$ .

Fig. 8.10One triangle, three sizes, three identities.NCERT Fig. 8.21 · p. 128

One right triangle with the angle A at its left corner, drawn at three sizes. Step one: the hypotenuse is 1, so the base is cos A and the height is sin A, and the Pythagoras theorem reads cos²A + sin²A = 1. Step two: the triangle is enlarged until the base is 1, so the height is tan A and the hypotenuse is sec A, giving 1 + tan²A = sec²A. Step three: it is enlarged until the height is 1, so the base is cot A and the hypotenuse is cosec A, giving cot²A + 1 = cosec²A.

Using the identities

Identities give a second way to find all the ratios from one, without drawing a triangle. Suppose $\cot A = 2 \sqrt{2}$ for an acute angle $A$ . By (8.6),

\begin{aligned} {cosec}^{2} A & = 1 + \cot^{2} A \\ = 1 + 8 = 9 \end{aligned}

So $cosec A = 3$ (positive, since $A$ is acute) and $\sin A = \frac{1}{3}$ . By (8.4), $\cos A = \sqrt{1 - \frac{1}{9}} = \frac{2 \sqrt{2}}{3}$ . Finally $\tan A = \frac{1}{2 \sqrt{2}}$ and $\sec A = \frac{3}{2 \sqrt{2}}$ .

Example 8.9NCERT Example 9 · p. 129

Everything in terms of $\cos A$

For an angle $A$ between $0^{\circ}$ and $90^{\circ}$ , write $\sin A$ , $\tan A$ and $cosec A$ using only $\cos A$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 8.10NCERT Example 10 · p. 129

Using a difference of squares

Prove that $(1 + \cot^{2} A) (1 - \cos A) (1 + \cos A) = 1$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 8.11NCERT Example 11 · p. 130

Take out a common factor

Prove that $\frac{\cos A + \cos A \tan^{2} A}{\sin A + \sin A \cot^{2} A} = \tan A$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 8.12NCERT Example 12 · p. 130

Make an identity do the work

(a) Using $\sec^{2} θ - \tan^{2} θ = 1$ , show that $\sec θ - \tan θ = \frac{1}{\sec θ + \tan θ}$ . (b) If $\sec θ + \tan θ = 5$ , find $\sec θ$ , $\tan θ$ and $\sin θ$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

Which of these is not an identity (for the angles where both sides are defined)?

\sin^{2} A + \cos^{2} A = 1

\sec^{2} A - \tan^{2} A = 1

{cosec}^{2} A - \cot^{2} A = 1

\sin A + \cos A = 1

Show answer

Answer (D) $\sin A + \cos A = 1$

45^{\circ}

\sin A + \cos A = \frac{2}{\sqrt{2}} = \sqrt{2}

, not 1. It is the squares of

\sin A

and

\cos A

that add up to 1.

8.5 Summary

Recap

Introduction to Trigonometry: key points and summary

Name the sides from the angle you look at: opposite, adjacent, and the hypotenuse across from the right angle (Section 8.2).
$\sin A = \frac{opp}{hyp}$ , $\cos A = \frac{adj}{hyp}$ , $\tan A = \frac{opp}{adj}$ ; cosec, sec and cot are their reciprocals (Table 8.1).
$\tan A = \frac{\sin A}{\cos A}$ and $\cot A = \frac{\cos A}{\sin A}$ ((8.1)).
A ratio depends only on the angle, not on the size of the triangle (similar triangles).
From one ratio of an acute angle, the Pythagoras theorem gives the other five (one ratio to all six).
For acute $A$ , $\sin A$ and $\cos A$ lie between 0 and 1, $\sec A$ and $cosec A$ exceed 1, and $\tan A$ can be any positive number (ranges).
Half a square gives the ratios of $45^{\circ}$ ; half an equilateral triangle gives those of $30^{\circ}$ and $60^{\circ}$ (Table 8.3).
$\sin 0^{\circ} = 0$ , $\cos 0^{\circ} = 1$ , $\sin 90^{\circ} = 1$ , $\cos 90^{\circ} = 0$ ; $\tan 90^{\circ}$ , $\sec 90^{\circ}$ , $cosec 0^{\circ}$ and $\cot 0^{\circ}$ are not defined (0° and 90°).
From $0^{\circ}$ to $90^{\circ}$ , $\sin A$ rises from 0 to 1 and $\cos A$ falls from 1 to 0 (Fig. 8.8).
One side and one more part fix a right triangle (Example 8.6).
$\sin^{2} A + \cos^{2} A = 1$ ; $1 + \tan^{2} A = \sec^{2} A$ for $0^{\circ} \leq A < 90^{\circ}$ ; $\cot^{2} A + 1 = {cosec}^{2} A$ for $0^{\circ} < A \leq 90^{\circ}$ (identities).

1/3Opposite, adjacent and hypotenuse give $\sin A$ , $\cos A$ and $\tan A$ .

Fig. 8.11The whole chapter on one card.

A revision card in three steps. Step one shows a right triangle with angle A, its sides labelled adj (blue), opp (orange) and hyp (purple), and beside it sin A = opp/hyp, cos A = adj/hyp and tan A = opp/adj. Step two shows the two triangles that give the special values: half a square with legs 1 and 1, hypotenuse root 2 and angles 45°, and half an equilateral triangle with sides 1, root 3 and 2 and angles 30° and 60°. Step three fades the triangle and lists the identities sin²A + cos²A = 1, 1 + tan²A = sec²A and 1 + cot²A = cosec²A.

Flash cards · 10

Test yourself

1 / 10

Quick revision

1 of 3

Which of these is not defined at

A = 90^{\circ}

\sin A

\sec A

\cot A

cosec A

Show answer

Answer (B) $\sec A$

\sec 90^{\circ}

would be

\frac{1}{\cos 90^{\circ}} = \frac{1}{0}

. The others are fine:

\sin 90^{\circ} = 1

\cot 90^{\circ} = 0

and

cosec 90^{\circ} = 1

Board-exam practice

These questions follow the formats of the CBSE board paper: multiple choice, assertion–reason, short answers, a proof and a case study.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 8.1, Q1	Practice 8.1 · Q1
NCERT Exercise 8.1, Q2	Practice 8.1 · Q2
NCERT Exercise 8.1, Q3	Practice 8.1 · Q3
NCERT Exercise 8.1, Q4	Practice 8.1 · Q4
NCERT Exercise 8.1, Q5	Practice 8.1 · Q5
NCERT Exercise 8.1, Q6	Practice 8.1 · Q6
NCERT Exercise 8.1, Q7	Practice 8.1 · Q7
NCERT Exercise 8.1, Q8	Practice 8.1 · Q8
NCERT Exercise 8.1, Q9	Practice 8.1 · Q9
NCERT Exercise 8.1, Q10	Practice 8.1 · Q10
NCERT Exercise 8.1, Q11	Practice 8.1 · Q11
NCERT Exercise 8.1, Q11 (iii)	Practice 8.1 · Q12
NCERT Exercise 8.2, Q1 (i)–(iv)	Practice 8.2 · Q1
NCERT Exercise 8.2, Q1 (v)	Practice 8.2 · Q2
NCERT Exercise 8.2, Q2	Practice 8.2 · Q3
NCERT Exercise 8.2, Q3	Practice 8.2 · Q4
NCERT Exercise 8.2, Q4	Practice 8.2 · Q5
NCERT Exercise 8.3, Q1	Practice 8.3 · Q1
NCERT Exercise 8.3, Q2	Practice 8.3 · Q2
NCERT Exercise 8.3, Q3	Practice 8.3 · Q3
NCERT Exercise 8.3, Q4 (i)–(iv)	Practice 8.3 · Q4
NCERT Exercise 8.3, Q4 (vi)–(viii)	Practice 8.3 · Q5
NCERT Exercise 8.3, Q4 (ix)–(x)	Practice 8.3 · Q6
NCERT Exercise 8.3, Q4 (v)	Practice 8.3 · Q7

Quick check: MCQs on Introduction to Trigonometry

8.1 Introduction2 questions
Ratios depend on the angle, not on the size4 questions
From one ratio to all six3 questions
Using the special values4 questions
Using the identities4 questions
Quick revision3 questions

Glossary · 10

Key terms (English–Hindi)

adjacent sideसंलग्न भुजाsanlagn bhuja: For an acute angle of a right triangle, the side other than the hypotenuse that forms the angle.First seen in the chapter →

cosecant, secant, cotangent: The reciprocals of sine, cosine and tangent: $cosec A = \frac{1}{\sin A}$ , $\sec A = \frac{1}{\cos A}$ , $\cot A = \frac{1}{\tan A}$ .First seen in the chapter →
cosineकोज्याkojya: $\cos A$ = side adjacent to $A$ ÷ hypotenuse.First seen in the chapter →

hypotenuseकर्णkarn: The side opposite the right angle in a right triangle; it is the longest side.First seen in the chapter →

opposite sideसम्मुख भुजाsammukh bhuja: For an acute angle of a right triangle, the side that does not touch the angle.First seen in the chapter →

sineज्याjya: $\sin A$ = side opposite $A$ ÷ hypotenuse.First seen in the chapter →

tangent (ratio)स्पर्शज्याsparshjya: $\tan A$ = side opposite $A$ ÷ side adjacent to $A$ ; also $\frac{\sin A}{\cos A}$ .First seen in the chapter →
trigonometric identityत्रिकोणमितीय सर्वसमिकाtrikonmitiya sarvasamika: An equation in trigonometric ratios that is true for every angle for which both sides are defined.First seen in the chapter →
trigonometric ratioत्रिकोणमितीय अनुपातtrikonmitiya anupaat: A ratio of two sides of a right triangle taken for one of its acute angles, such as $\sin A$ or $\cos A$ .First seen in the chapter →
trigonometryत्रिकोणमितिtrikonmiti: The branch of mathematics that studies how the sides and angles of triangles are related.First seen in the chapter →

Introduction to Trigonometry: Class 10 Maths · Chapter 8

What you will learn in Introduction to Trigonometry

8.1 Introduction

8.2 Trigonometric Ratios

The six ratios

Ratios depend on the angle, not on the size

From one ratio to all six

Practice 8.1: questions with answers

8.3 Trigonometric Ratios of Some Specific Angles

Ratios of 45°

Ratios of 30° and 60°

Ratios of 0° and 90°

All the special values

Using the special values

Practice 8.2: questions with answers

8.4 Trigonometric Identities

Three identities from one theorem

Using the identities

Practice 8.3: questions with answers

8.5 Summary

Introduction to Trigonometry: key points and summary

Board-exam practice

Practice questions with answers (NCERT “Board-exam practice”)

Find your NCERT exercise question

Quick check: MCQs on Introduction to Trigonometry

Key terms (English–Hindi)

You reached the end