Introduction to Trigonometry Class 10 Mathematics Notes

Trigonometry is a branch of mathematics that studies the relationships between the sides and angles of triangles. The word "trigonometry" itself comes from Greek words meaning "three" (trigonon) and "measure" (metron). This chapter introduces the fundamental concepts of trigonometry, focusing on its application to right-angled triangles.

Right-Angled Triangles and Trigonometric Ratios

In a right-angled triangle, we define specific terms for the sides relative to an acute angle. Let's consider a right-angled triangle ABC, with the right angle at B, and an acute angle A.

Hypotenuse: The side opposite the right angle (AC). It is always the longest side.
Side Opposite to Angle A: The side facing angle A (BC).
Side Adjacent to Angle A: The side that forms angle A, excluding the hypotenuse (AB).

These sides form the basis for defining six trigonometric ratios for an acute angle in a right-angled triangle. These ratios express the relationship between an angle and the lengths of its sides.

The six trigonometric ratios are:

Sine (sin): The ratio of the length of the side opposite the angle to the length of the hypotenuse.

\sin A = \frac{Side Opposite to Angle A}{Hypotenuse} = \frac{B C}{A C}

Cosine (cos): The ratio of the length of the side adjacent to the angle to the length of the hypotenuse.

\cos A = \frac{Side Adjacent to Angle A}{Hypotenuse} = \frac{A B}{A C}

Tangent (tan): The ratio of the length of the side opposite the angle to the length of the side adjacent to the angle.

\tan A = \frac{Side Opposite to Angle A}{Side Adjacent to Angle A} = \frac{B C}{A B}

Cosecant (cosec): The reciprocal of the sine ratio.

cosec A = \frac{1}{\sin A} = \frac{Hypotenuse}{Side Opposite to Angle A} = \frac{A C}{B C}

Secant (sec): The reciprocal of the cosine ratio.

\sec A = \frac{1}{\cos A} = \frac{Hypotenuse}{Side Adjacent to Angle A} = \frac{A C}{A B}

Cotangent (cot): The reciprocal of the tangent ratio.

\cot A = \frac{1}{\tan A} = \frac{Side Adjacent to Angle A}{Side Opposite to Angle A} = \frac{A B}{B C}

It's important to note that the symbols $\sin A$ , $\cos A$ , etc., represent the trigonometric ratio of angle A and are not the product of 'sin' and 'A' or 'cos' and 'A'.

Relationship Between Tangent and Cotangent

The tangent and cotangent ratios can also be expressed in terms of sine and cosine:

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

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

Invariance of Trigonometric Ratios

The values of trigonometric ratios for a given angle depend only on the angle itself, not on the size of the right-angled triangle. If we have two similar right-angled triangles with a common acute angle, their corresponding sides are proportional, leading to the same trigonometric ratios.

Notation for Powers of Trigonometric Ratios

For convenience, we often write $\sin^{2} A$ to mean $(\sin A)^{2}$ , $\cos^{2} A$ for $(\cos A)^{2}$ , and so on. However, this notation is different for cosecant: $cosec A$ is the reciprocal of $\sin A$ , and $\sin^{- 1} A$ (read as "sine inverse A") represents a different concept (inverse trigonometric function) which will be studied in higher classes.

Historical Context

The concept of sine originated from the Indian mathematician Aryabhata around 500 AD. He used the term "ardha-jya" (half-chord), which was later shortened to "jya" or "jiva". When translated into Arabic, "jiva" was retained, and then in Latin, it was translated to "sinus," meaning "curve." The abbreviated notation 'sin' was first used by English astronomer Edmund Gunter. The terms 'cosine' and 'tangent' have later origins.

Finding Other Trigonometric Ratios Given One Ratio

If we know the value of one trigonometric ratio for an acute angle, we can find the lengths of two sides of a right-angled triangle (in terms of a constant 'k') and then use the Pythagorean theorem to find the length of the third side. This allows us to calculate all other trigonometric ratios.

Pythagorean Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

(Hypotenuse)^{2} = (Side Opposite)^{2} + (Side Adjacent)^{2}

Example 1

Given: $\tan A = \frac{3}{4}$ in a right-angled triangle ABC, right-angled at B. To find: All other trigonometric ratios of angle A.

Solution: We know that $\tan A = \frac{Opposite}{Adjacent}$ . So, we can assume: Given: Side opposite to A (BC) = $3 k$ Given: Side adjacent to A (AB) = $4 k$ , where $k$ is a positive constant.

Now, using the Pythagorean theorem to find the hypotenuse (AC):

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

A C^{2} = (4 k)^{2} + (3 k)^{2}

A C^{2} = 16 k^{2} + 9 k^{2}

A C^{2} = 25 k^{2}

A C = \sqrt{25 k^{2}} = 5 k

Now we can find the other trigonometric ratios:

\sin A = \frac{B C}{A C} = \frac{3 k}{5 k} = \frac{3}{5}

\cos A = \frac{A B}{A C} = \frac{4 k}{5 k} = \frac{4}{5}

cosec A = \frac{1}{\sin A} = \frac{5}{3}

\sec A = \frac{1}{\cos A} = \frac{5}{4}

\cot A = \frac{1}{\tan A} = \frac{4}{3}

Answer: $\sin A = \frac{3}{5}$ , $\cos A = \frac{4}{5}$ , $cosec A = \frac{5}{3}$ , $\sec A = \frac{5}{4}$ , $\cot A = \frac{4}{3}$ .

Example 2

Given: If $\sin B = \sin Q$ , where $∠ B$ and $∠ Q$ are acute angles in right-angled triangles. To prove: $∠ B = ∠ Q$ .

Solution: Let's consider two right-angled triangles: $△ A B C$ (right-angled at C) and $△ P Q R$ (right-angled at R). We are given:

\sin B = \sin Q

\frac{A C}{A B} = \frac{P R}{P Q}

Let $\frac{A C}{A B} = \frac{P R}{P Q} = k$ (where $k$ is a positive constant). So, $A C = k \cdot A B$ and $P R = k \cdot P Q$ .

Now, using the Pythagorean theorem for $△ A B C$ :

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

B C^{2} = A B^{2} - (k \cdot A B)^{2}

B C^{2} = A B^{2} (1 - k^{2})

B C = A B \sqrt{1 - k^{2}}

And for $△ P Q R$ :

Q R^{2} = P Q^{2} - P R^{2}

Q R^{2} = P Q^{2} - (k \cdot P Q)^{2}

Q R^{2} = P Q^{2} (1 - k^{2})

Q R = P Q \sqrt{1 - k^{2}}

Now let's find $\cos B$ and $\cos Q$ :

\cos B = \frac{B C}{A B} = \frac{A B \sqrt{1 - k^{2}}}{A B} = \sqrt{1 - k^{2}}

\cos Q = \frac{Q R}{P Q} = \frac{P Q \sqrt{1 - k^{2}}}{P Q} = \sqrt{1 - k^{2}}

So, $\cos B = \cos Q$ .

Since $\cos B = \cos Q$ and $B$ and $Q$ are acute angles, it implies that $∠ B = ∠ Q$ .

Answer: $∠ B = ∠ Q$ .

Example 3

Given: In a right-angled triangle ACB, $∠ C = 90^{\circ}$ , $A B = 29$ units, $B C = 21$ units. $∠ A B C = θ$ . To find: (i) $\cos^{2} θ + \sin^{2} θ$ (ii) $\cos^{2} θ - \sin^{2} θ$ .

Solution: First, find the length of side AC using the Pythagorean theorem:

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

A C^{2} + (21)^{2} = (29)^{2}

A C^{2} + 441 = 841

A C^{2} = 841 - 441 = 400

A C = \sqrt{400} = 20 units

Now, find $\sin θ$ and $\cos θ$ :

\sin θ = \frac{Opposite}{Hypotenuse} = \frac{A C}{A B} = \frac{20}{29}

\cos θ = \frac{Adjacent}{Hypotenuse} = \frac{B C}{A B} = \frac{21}{29}

(i) Calculate $\cos^{2} θ + \sin^{2} θ$ :

\cos^{2} θ + \sin^{2} θ = {(\frac{21}{29})}^{2} + {(\frac{20}{29})}^{2}

= \frac{21^{2}}{29^{2}} + \frac{20^{2}}{29^{2}}

= \frac{441}{841} + \frac{400}{841}

= \frac{441 + 400}{841} = \frac{841}{841} = 1

(ii) Calculate $\cos^{2} θ - \sin^{2} θ$ :

\cos^{2} θ - \sin^{2} θ = {(\frac{21}{29})}^{2} - {(\frac{20}{29})}^{2}

= \frac{441}{841} - \frac{400}{841}

= \frac{441 - 400}{841} = \frac{41}{841}

Answer: (i) $1$ (ii) $\frac{41}{841}$ .

Example 4

Given: In a right-angled triangle ABC, $∠ B = 90^{\circ}$ , and $\tan A = 1$ . To verify: $2 \sin A \cos A = 1$ .

Solution: We are given $\tan A = 1$ .

\tan A = \frac{Opposite}{Adjacent} = \frac{B C}{A B}

So, $\frac{B C}{A B} = 1$ , which means $B C = A B$ . Let $B C = A B = k$ , where $k$ is a positive constant.

Now, find the hypotenuse AC using the Pythagorean theorem:

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

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

A C^{2} = 2 k^{2}

A C = \sqrt{2 k^{2}} = k \sqrt{2}

Now calculate $\sin A$ and $\cos A$ :

\sin A = \frac{B C}{A C} = \frac{k}{k \sqrt{2}} = \frac{1}{\sqrt{2}}

\cos A = \frac{A B}{A C} = \frac{k}{k \sqrt{2}} = \frac{1}{\sqrt{2}}

Finally, verify $2 \sin A \cos A$ :

2 \sin A \cos A = 2 \times (\frac{1}{\sqrt{2}}) \times (\frac{1}{\sqrt{2}})

= 2 \times \frac{1}{2} = 1

Thus, $2 \sin A \cos A = 1$ is verified.

Answer: Verified.

Example 5

Given: In $△ O P Q$ , $∠ P = 90^{\circ}$ , $O P = 7$ cm, and $O Q - P Q = 1$ cm. To find: $\sin Q$ and $\cos Q$ .

Solution: Let $P Q = x$ cm. Then, $O Q = 1 + P Q = (1 + x)$ cm.

Now, apply the Pythagorean theorem to $△ O P Q$ :

O Q^{2} = O P^{2} + P Q^{2}

(1 + x)^{2} = 7^{2} + x^{2}

1 + 2 x + x^{2} = 49 + x^{2}

Subtract $x^{2}$ from both sides:

1 + 2 x = 49

2 x = 49 - 1

2 x = 48

x = \frac{48}{2} = 24

So, $P Q = 24$ cm.

And $O Q = 1 + x = 1 + 24 = 25$ cm.

Now we can find $\sin Q$ and $\cos Q$ :

\sin Q = \frac{Opposite}{Hypotenuse} = \frac{O P}{O Q} = \frac{7}{25}

\cos Q = \frac{Adjacent}{Hypotenuse} = \frac{P Q}{O Q} = \frac{24}{25}

Answer: $\sin Q = \frac{7}{25}$ , $\cos Q = \frac{24}{25}$ .

Remember
The trigonometric ratios are defined for acute angles in a right-angled triangle. The hypotenuse is always the longest side, so the values of $\sin θ$ and $\cos θ$ are always between 0 and 1 (inclusive).

Exam-smart
Case/source questions might give you a diagram of a right-angled triangle with some sides labelled, or a word problem describing a scenario (e.g., a ladder leaning against a wall) that can be represented by a right-angled triangle. You'll need to identify the sides opposite, adjacent, and hypotenuse with respect to a given angle and then calculate trigonometric ratios or use them to find missing lengths.

Trigonometric Ratios of Some Specific Angles

This section focuses on determining the precise numerical values of trigonometric ratios for key angles: 0°, 30°, 45°, 60°, and 90°. These specific values are fundamental for solving many trigonometric problems and simplify calculations where precise angle measurements are known.

Trigonometric Ratios of 45°

Consider a right-angled triangle ABC, where $∠ B = 90^{\circ}$ . If $∠ A = 45^{\circ}$ , then $∠ C$ must also be $45^{\circ}$ (since the sum of angles in a triangle is $180^{\circ}$ , and $180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ}$ ). In such a triangle, the sides opposite equal angles are equal. Thus, $B C = A B$ . Let's assume $A B = a$ . Then $B C = a$ . Using the Pythagorean theorem, $A C^{2} = A B^{2} + B C^{2} = a^{2} + a^{2} = 2 a^{2}$ . So, the hypotenuse $A C = \sqrt{2 a^{2}} = a \sqrt{2}$ .

Now we can find the trigonometric ratios for 45°:

sin 45°: (Opposite side) / (Hypotenuse) = $B C / A C = a / (a \sqrt{2}) = 1 / \sqrt{2}$ .
cos 45°: (Adjacent side) / (Hypotenuse) = $A B / A C = a / (a \sqrt{2}) = 1 / \sqrt{2}$ .
tan 45°: (Opposite side) / (Adjacent side) = $B C / A B = a / a = 1$ .

And their reciprocals:

cosec 45° = $1 / \sin 45^{\circ} = \sqrt{2}$ .
sec 45° = $1 / \cos 45^{\circ} = \sqrt{2}$ .
cot 45° = $1 / \tan 45^{\circ} = 1$ .

Example 1

Given: A right-angled triangle ABC, right-angled at B, with $∠ A = 45^{\circ}$ and $A B = 10$ cm. To find: The lengths of sides BC and AC. Solution: Since $∠ A = 45^{\circ}$ and $∠ B = 90^{\circ}$ , $∠ C = 45^{\circ}$ . This means $△ A B C$ is an isosceles right-angled triangle with $A B = B C$ . So, $B C = 10$ cm. Using the Pythagorean theorem:

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

A C^{2} = (10 cm)^{2} + (10 cm)^{2}

A C^{2} = 100 {cm}^{2} + 100 {cm}^{2}

A C^{2} = 200 {cm}^{2}

A C = \sqrt{200} cm = 10 \sqrt{2} cm

Answer: $B C = 10$ cm, $A C = 10 \sqrt{2}$ cm.

Trigonometric Ratios of 30° and 60°

Consider an equilateral triangle ABC where $A B = B C = A C$ . Each angle is $60^{\circ}$ . Draw a perpendicular AD from A to BC. This perpendicular bisects BC and also bisects $∠ A$ . So, $B D = D C = B C / 2$ , and $∠ B A D = ∠ C A D = 60^{\circ} / 2 = 30^{\circ}$ . Consider the right-angled triangle ABD, with $∠ A D B = 90^{\circ}$ , $∠ A B D = 60^{\circ}$ , and $∠ B A D = 30^{\circ}$ . Let $A B = 2 a$ . Then $B D = a$ . Using the Pythagorean theorem in $△ A B D$ : $A D^{2} = A B^{2} - B D^{2} = (2 a)^{2} - (a)^{2} = 4 a^{2} - a^{2} = 3 a^{2}$ . So, $A D = \sqrt{3 a^{2}} = a \sqrt{3}$ .

Now we can find the trigonometric ratios for 30° and 60°:

For 30° (using $△ A B D$ ):

sin 30°: (Opposite side to 30°) / (Hypotenuse) = $B D / A B = a / (2 a) = 1 / 2$ .
cos 30°: (Adjacent side to 30°) / (Hypotenuse) = $A D / A B = (a \sqrt{3}) / (2 a) = \sqrt{3} / 2$ .
tan 30°: (Opposite side to 30°) / (Adjacent side to 30°) = $B D / A D = a / (a \sqrt{3}) = 1 / \sqrt{3}$ .

For 60° (using $△ A B D$ ):

sin 60°: (Opposite side to 60°) / (Hypotenuse) = $A D / A B = (a \sqrt{3}) / (2 a) = \sqrt{3} / 2$ .
cos 60°: (Adjacent side to 60°) / (Hypotenuse) = $B D / A B = a / (2 a) = 1 / 2$ .
tan 60°: (Opposite side to 60°) / (Adjacent side to 60°) = $A D / B D = (a \sqrt{3}) / a = \sqrt{3}$ .

And their reciprocals:

cosec 30° = $1 / \sin 30^{\circ} = 2$ .
sec 30° = $1 / \cos 30^{\circ} = 2 / \sqrt{3}$ .
cot 30° = $1 / \tan 30^{\circ} = \sqrt{3}$ .
cosec 60° = $1 / \sin 60^{\circ} = 2 / \sqrt{3}$ .
sec 60° = $1 / \cos 60^{\circ} = 2$ .
cot 60° = $1 / \tan 60^{\circ} = 1 / \sqrt{3}$ .

Example 2

Given: A right-angled triangle PQR, right-angled at Q, with $P Q = 5$ cm and $∠ P R Q = 30^{\circ}$ . To find: The lengths of sides QR and PR. Solution: In $△ P Q R$ : We are given $P Q = 5$ cm (opposite to $∠ R$ ) and $∠ R = 30^{\circ}$ .

To find QR (adjacent to $∠ R$ ): Use tan.

\tan R = P Q / Q R

\tan 30^{\circ} = 5 cm / Q R

1 / \sqrt{3} = 5 cm / Q R

$Q R = 5 \sqrt{3}$ cm.

To find PR (hypotenuse): Use sin.

\sin R = P Q / P R

\sin 30^{\circ} = 5 cm / P R

1 / 2 = 5 cm / P R

$P R = 10$ cm.

Answer: $Q R = 5 \sqrt{3}$ cm, $P R = 10$ cm.

Trigonometric Ratios of 0° and 90°

We can understand the values for 0° and 90° by considering a right-angled triangle ABC where $∠ B = 90^{\circ}$ .

As $∠ A$ approaches 0°:

Side BC (opposite to A) becomes very small, approaching 0.
Hypotenuse AC approaches the length of side AB (adjacent to A).
$\sin A = B C / A C \to 0 / A B = 0$ .
$\cos A = A B / A C \to A B / A B = 1$ .
$\tan A = B C / A B \to 0 / A B = 0$ .

Therefore, we define:

sin 0° = 0
cos 0° = 1
tan 0° = 0
cot 0° = $1 / \tan 0^{\circ}$ , which is undefined.
sec 0° = $1 / \cos 0^{\circ} = 1 / 1 = 1$ .
cosec 0° = $1 / \sin 0^{\circ}$ , which is undefined.

As $∠ A$ approaches 90°:

Side AB (adjacent to A) becomes very small, approaching 0.
Hypotenuse AC approaches the length of side BC (opposite to A).
$\sin A = B C / A C \to B C / B C = 1$ .
$\cos A = A B / A C \to 0 / B C = 0$ .
$\tan A = B C / A B \to B C / 0$ , which is undefined.

Therefore, we define:

sin 90° = 1
cos 90° = 0
tan 90° = $1 / \tan 90^{\circ}$ , which is undefined.
cot 90° = $1 / \tan 90^{\circ} = 0$ .
sec 90° = $1 / \cos 90^{\circ}$ , which is undefined.
cosec 90° = $1 / \sin 90^{\circ} = 1 / 1 = 1$ .

Table of Trigonometric Ratios for Specific Angles

Here is a summary of the values:

Angle ( $θ$ )	sin $θ$	cos $θ$	tan $θ$	cosec $θ$	sec $θ$	cot $θ$
0°	0	1	0	Undefined	1	Undefined
30°	1/2	$\sqrt{3} / 2$	$1 / \sqrt{3}$	2	$2 / \sqrt{3}$	$\sqrt{3}$
45°	$1 / \sqrt{2}$	$1 / \sqrt{2}$	1	$\sqrt{2}$	$\sqrt{2}$	1
60°	$\sqrt{3} / 2$	1/2	$\sqrt{3}$	$2 / \sqrt{3}$	2	$1 / \sqrt{3}$
90°	1	0	Undefined	1	Undefined	0

Key Observations from the Table:

As the angle $θ$ increases from 0° to 90°, sin $θ$ increases from 0 to 1.
As the angle $θ$ increases from 0° to 90°, cos $θ$ decreases from 1 to 0.

Example 3

Given: $\sin (A - B) = 1 / 2$ and $\cos (A + B) = 1 / 2$ , where $0^{\circ} < A + B \leq 90^{\circ}$ and $A > B$ . To find: Values of angles A and B. Solution: From the table: $\sin (A - B) = 1 / 2$ implies $A - B = 30^{\circ}$ (Equation 1). $\cos (A + B) = 1 / 2$ implies $A + B = 60^{\circ}$ (Equation 2).

Add Equation 1 and Equation 2:

(A - B) + (A + B) = 30^{\circ} + 60^{\circ}

2 A = 90^{\circ}

$A = 45^{\circ}$ .

Substitute $A = 45^{\circ}$ into Equation 2:

45^{\circ} + B = 60^{\circ}

B = 60^{\circ} - 45^{\circ}

$B = 15^{\circ}$ .

Check conditions: $0^{\circ} < A + B \leq 90^{\circ}$ ( $0^{\circ} < 60^{\circ} \leq 90^{\circ}$ , true) and $A > B$ ( $45^{\circ} > 15^{\circ}$ , true).

Answer: $A = 45^{\circ}$ , $B = 15^{\circ}$ .

Example 4

Given: Evaluate $\sin 60^{\circ} \cos 30^{\circ} + \sin 30^{\circ} \cos 60^{\circ}$ . To find: The value of the expression. Solution: Using the values from the table:

\sin 60^{\circ} = \sqrt{3} / 2

\cos 30^{\circ} = \sqrt{3} / 2

\sin 30^{\circ} = 1 / 2

\cos 60^{\circ} = 1 / 2

Substitute these values into the expression:

(\sqrt{3} / 2) \times (\sqrt{3} / 2) + (1 / 2) \times (1 / 2)

= (\sqrt{3} \times \sqrt{3}) / (2 \times 2) + (1 \times 1) / (2 \times 2)

= 3 / 4 + 1 / 4

= 4 / 4

$= 1$ .

Answer: 1.

Example 5

Given: Evaluate $\frac{5 \cos^{2} 60^{\circ} + 4 \sec^{2} 30^{\circ} - \tan^{2} 45^{\circ}}{\sin^{2} 30^{\circ} + \cos^{2} 30^{\circ}}$ . To find: The value of the expression. Solution: From the table: $\cos 60^{\circ} = 1 / 2$ , so $\cos^{2} 60^{\circ} = (1 / 2)^{2} = 1 / 4$ . $\sec 30^{\circ} = 2 / \sqrt{3}$ , so $\sec^{2} 30^{\circ} = (2 / \sqrt{3})^{2} = 4 / 3$ . $\tan 45^{\circ} = 1$ , so $\tan^{2} 45^{\circ} = 1^{2} = 1$ . $\sin 30^{\circ} = 1 / 2$ , so $\sin^{2} 30^{\circ} = (1 / 2)^{2} = 1 / 4$ . $\cos 30^{\circ} = \sqrt{3} / 2$ , so $\cos^{2} 30^{\circ} = (\sqrt{3} / 2)^{2} = 3 / 4$ .

Substitute these values into the expression: Numerator: $5 (1 / 4) + 4 (4 / 3) - 1 = 5 / 4 + 16 / 3 - 1$ To sum these, find a common denominator (12):

= (5 \times 3) / 12 + (16 \times 4) / 12 - 12 / 12

= 15 / 12 + 64 / 12 - 12 / 12

$= (15 + 64 - 12) / 12 = 67 / 12$ .

Denominator: $\sin^{2} 30^{\circ} + \cos^{2} 30^{\circ} = 1 / 4 + 3 / 4 = 4 / 4 = 1$ . (Note: This denominator is an example of the trigonometric identity $\sin^{2} θ + \cos^{2} θ = 1$ , which will be covered in the next section).

The expression becomes: $(67 / 12) / 1 = 67 / 12$ .

Answer: $67 / 12$ .

True/False Statements about Trigonometric Ratios

Let's analyze the given statements.

The value of tan A is always less than 1.

False. For example, $\tan 60^{\circ} = \sqrt{3}$ which is approximately 1.732, and $\tan 90^{\circ}$ is undefined (approaches infinity). Tan values can be greater than 1.

sec A = $\frac{12}{5}$ for some value of angle A.

True. We know that $\sec A = hypotenuse / adjacent side$ . If we consider a right-angled triangle where the adjacent side is 5 units and the hypotenuse is 12 units, then $\sec A = 12 / 5$ . Since the hypotenuse is always the longest side in a right-angled triangle, $12 / 5 > 1$ . The range of $\sec A$ for acute angles is $[1, \infty)$ . $12 / 5$ falls within this range.

cos A is the abbreviation used for the cosecant of angle A.

False. cos A is the abbreviation for cosine of angle A. Cosecant is abbreviated as cosec A.

cot A is the product of cot and A.

False. cot A represents the cotangent of the angle A. It is a ratio of sides and not a product of 'cot' and 'A'.

sin $θ$ = $\frac{5}{4}$ for some angle $θ$ .

False. In a right-angled triangle, the sine of an angle is the ratio of the opposite side to the hypotenuse. The hypotenuse is always the longest side. Therefore, $\sin θ$ can never be greater than 1. Since $5 / 4 > 1$ , this statement is false.

Remember
The values of trigonometric ratios for 0°, 30°, 45°, 60°, and 90° are specific and must be memorized. They are derived from geometric constructions of special triangles.

Exam-smart
Questions often involve substituting these specific values into expressions to evaluate them, or solving equations where angles correspond to these specific values. You might also get questions that ask you to justify why a given value is possible or impossible for a trigonometric ratio of an angle.

Trigonometric Ratios of Some Specific Angles

This section focuses on finding the precise trigonometric ratios for specific standard angles: $0^{\circ}$ , $30^{\circ}$ , $45^{\circ}$ , $60^{\circ}$ , and $90^{\circ}$ . These values are fundamental and frequently used in trigonometry and its applications.

Trigonometric Ratios of $45^{\circ}$

To determine the ratios for $45^{\circ}$ , consider an isosceles right-angled triangle ABC, where $∠ B = 90^{\circ}$ and $∠ A = ∠ C = 45^{\circ}$ . Since the angles are equal, the sides opposite to them are also equal. Let $A B = B C = a$ .

Using the Pythagorean theorem ( $A C^{2} = A B^{2} + B C^{2}$ ):

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

A C = \sqrt{2 a^{2}} = a \sqrt{2}

Now, we can find the trigonometric ratios:

sin $45^{\circ}$ : (side opposite to $45^{\circ}$ ) / hypotenuse = $B C / A C = a / (a \sqrt{2}) = 1 / \sqrt{2}$
cos $45^{\circ}$ : (side adjacent to $45^{\circ}$ ) / hypotenuse = $A B / A C = a / (a \sqrt{2}) = 1 / \sqrt{2}$
tan $45^{\circ}$ : (side opposite to $45^{\circ}$ ) / (side adjacent to $45^{\circ}$ ) = $B C / A B = a / a = 1$

The reciprocal ratios are:

cosec $45^{\circ}$ = $1 / \sin 45^{\circ} = \sqrt{2}$
sec $45^{\circ}$ = $1 / \cos 45^{\circ} = \sqrt{2}$
cot $45^{\circ}$ = $1 / \tan 45^{\circ} = 1$

Example 1

Given: A right-angled triangle ABC, $∠ B = 90^{\circ}$ , $∠ C = 45^{\circ}$ . To find: sin $45^{\circ}$ , cos $45^{\circ}$ , tan $45^{\circ}$ .

Solution: Since $∠ C = 45^{\circ}$ and $∠ B = 90^{\circ}$ , then $∠ A = 180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ}$ . Thus, $△ A B C$ is an isosceles right-angled triangle with $A B = B C$ . Let $A B = B C = k$ . By Pythagoras theorem, $A C^{2} = A B^{2} + B C^{2} = k^{2} + k^{2} = 2 k^{2}$ . So, $A C = k \sqrt{2}$ .

sin $45^{\circ}$ = Opposite side / Hypotenuse = $B C / A C = k / (k \sqrt{2}) = 1 / \sqrt{2}$ cos $45^{\circ}$ = Adjacent side / Hypotenuse = $A B / A C = k / (k \sqrt{2}) = 1 / \sqrt{2}$ tan $45^{\circ}$ = Opposite side / Adjacent side = $B C / A B = k / k = 1$

Answer: sin $45^{\circ} = 1 / \sqrt{2}$ , cos $45^{\circ} = 1 / \sqrt{2}$ , tan $45^{\circ} = 1$ .

Trigonometric Ratios of $30^{\circ}$ and $60^{\circ}$

To find the ratios for $30^{\circ}$ and $60^{\circ}$ , consider an equilateral triangle ABC where each angle is $60^{\circ}$ . Draw a perpendicular AD from vertex A to the base BC.

Since $△ A B C$ is equilateral, AD bisects BC and $∠ B A C$ . So, $B D = D C = B C / 2$ and $∠ B A D = ∠ C A D = 60^{\circ} / 2 = 30^{\circ}$ . $△ A B D$ is a right-angled triangle with $∠ D = 90^{\circ}$ , $∠ A B D = 60^{\circ}$ , and $∠ B A D = 30^{\circ}$ .

Let the side length of the equilateral triangle be $2 a$ . So, $A B = 2 a$ . Then $B D = B C / 2 = (2 a) / 2 = a$ . Using the Pythagorean theorem in $△ A B D$ :

A D^{2} = A B^{2} - B D^{2} = (2 a)^{2} - a^{2} = 4 a^{2} - a^{2} = 3 a^{2}

A D = \sqrt{3 a^{2}} = a \sqrt{3}

Now we can find the trigonometric ratios for $30^{\circ}$ and $60^{\circ}$ :

For $30^{\circ}$ (using $∠ B A D$ ):

sin $30^{\circ}$ : Opposite / Hypotenuse = $B D / A B = a / (2 a) = 1 / 2$
cos $30^{\circ}$ : Adjacent / Hypotenuse = $A D / A B = (a \sqrt{3}) / (2 a) = \sqrt{3} / 2$
tan $30^{\circ}$ : Opposite / Adjacent = $B D / A D = a / (a \sqrt{3}) = 1 / \sqrt{3}$

The reciprocal ratios are:

cosec $30^{\circ}$ = $1 / \sin 30^{\circ} = 2$
sec $30^{\circ}$ = $1 / \cos 30^{\circ} = 2 / \sqrt{3}$
cot $30^{\circ}$ = $1 / \tan 30^{\circ} = \sqrt{3}$

For $60^{\circ}$ (using $∠ A B D$ ):

sin $60^{\circ}$ : Opposite / Hypotenuse = $A D / A B = (a \sqrt{3}) / (2 a) = \sqrt{3} / 2$
cos $60^{\circ}$ : Adjacent / Hypotenuse = $B D / A B = a / (2 a) = 1 / 2$
tan $60^{\circ}$ : Opposite / Adjacent = $A D / B D = (a \sqrt{3}) / a = \sqrt{3}$

The reciprocal ratios are:

cosec $60^{\circ}$ = $1 / \sin 60^{\circ} = 2 / \sqrt{3}$
sec $60^{\circ}$ = $1 / \cos 60^{\circ} = 2$
cot $60^{\circ}$ = $1 / \tan 60^{\circ} = 1 / \sqrt{3}$

Example 2

Given: An equilateral triangle ABC with side length 10 cm. AD is the altitude from A to BC. To find: sin $60^{\circ}$ , cos $30^{\circ}$ .

Solution: Side length $A B = B C = A C = 10$ cm. Since AD is an altitude, D is the midpoint of BC. So, $B D = B C / 2 = 10 / 2 = 5$ cm. In right-angled $△ A B D$ , $∠ A D B = 90^{\circ}$ , $∠ B = 60^{\circ}$ , $∠ B A D = 30^{\circ}$ . $A B = 10$ cm (hypotenuse). $B D = 5$ cm (adjacent to $60^{\circ}$ , opposite to $30^{\circ}$ ). Using Pythagoras theorem for AD: $A D^{2} = A B^{2} - B D^{2} = 10^{2} - 5^{2} = 100 - 25 = 75$ . $A D = \sqrt{75} = \sqrt{25 \times 3} = 5 \sqrt{3}$ cm (opposite to $60^{\circ}$ , adjacent to $30^{\circ}$ ).

sin $60^{\circ}$ = Opposite / Hypotenuse = $A D / A B = (5 \sqrt{3}) / 10 = \sqrt{3} / 2$ . cos $30^{\circ}$ = Adjacent / Hypotenuse = $A D / A B = (5 \sqrt{3}) / 10 = \sqrt{3} / 2$ .

Answer: sin $60^{\circ} = \sqrt{3} / 2$ , cos $30^{\circ} = \sqrt{3} / 2$ .

Trigonometric Ratios of $0^{\circ}$ and $90^{\circ}$

To understand the ratios for $0^{\circ}$ and $90^{\circ}$ , consider a right-angled triangle ABC with $∠ B = 90^{\circ}$ .

As $∠ A$ approaches $0^{\circ}$ :

Side BC (opposite to A) approaches 0.
Side AB (adjacent to A) approaches the hypotenuse AC.
sin A = Opposite/Hypotenuse = $B C / A C$ . As BC $\to$ 0 and AC is finite, sin A $\to$ 0.
cos A = Adjacent/Hypotenuse = $A B / A C$ . As AB $\to$ AC, cos A $\to$ 1.
tan A = sin A / cos A $\to$ 0 / 1 = 0.

Therefore, we define:

sin $0^{\circ} = 0$
cos $0^{\circ} = 1$
tan $0^{\circ} = 0$
cosec $0^{\circ}$ is undefined (since it's $1 / \sin 0^{\circ}$ )
sec $0^{\circ}$ = $1 / \cos 0^{\circ} = 1 / 1 = 1$
cot $0^{\circ}$ is undefined (since it's $\cos 0^{\circ} / \sin 0^{\circ}$ )

As $∠ A$ approaches $90^{\circ}$ :

Side AB (adjacent to A) approaches 0.
Side BC (opposite to A) approaches the hypotenuse AC.
sin A = Opposite/Hypotenuse = $B C / A C$ . As BC $\to$ AC, sin A $\to$ 1.
cos A = Adjacent/Hypotenuse = $A B / A C$ . As AB $\to$ 0 and AC is finite, cos A $\to$ 0.
tan A = sin A / cos A $\to$ 1 / 0, which is undefined.

Therefore, we define:

sin $90^{\circ} = 1$
cos $90^{\circ} = 0$
tan $90^{\circ}$ is undefined.
cosec $90^{\circ}$ = $1 / \sin 90^{\circ} = 1 / 1 = 1$
sec $90^{\circ}$ is undefined (since it's $1 / \cos 90^{\circ}$ )
cot $90^{\circ}$ = $\cos 90^{\circ} / \sin 90^{\circ} = 0 / 1 = 0$

Table of Trigonometric Ratios for Specific Angles

The values derived above are summarized in the following table:

Angle ( $θ$ )	$\sin θ$	$\cos θ$	$\tan θ$	$cosec θ$	$\sec θ$	$\cot θ$
$0^{\circ}$	0	1	0	Undefined	1	Undefined
$30^{\circ}$	$1 / 2$	$\sqrt{3} / 2$	$1 / \sqrt{3}$	2	$2 / \sqrt{3}$	$\sqrt{3}$
$45^{\circ}$	$1 / \sqrt{2}$	$1 / \sqrt{2}$	1	$\sqrt{2}$	$\sqrt{2}$	1
$60^{\circ}$	$\sqrt{3} / 2$	$1 / 2$	$\sqrt{3}$	$2 / \sqrt{3}$	2	$1 / \sqrt{3}$
$90^{\circ}$	1	0	Undefined	1	Undefined	0

Observation: As the angle $θ$ increases from $0^{\circ}$ to $90^{\circ}$ , $\sin θ$ increases from 0 to 1, and $\cos θ$ decreases from 1 to 0.

Example 3

Given: A right-angled triangle ABC, $∠ B = 90^{\circ}$ , $A B = 5$ cm, $∠ C = 30^{\circ}$ . To find: Lengths of BC and AC.

Solution: We have the opposite side (AB) and the angle C. To find BC (adjacent to C):

\tan C = Opposite / Adjacent

\tan 30^{\circ} = A B / B C

1 / \sqrt{3} = 5 / B C

$B C = 5 \sqrt{3}$ cm.

To find AC (hypotenuse):

\sin C = Opposite / Hypotenuse

\sin 30^{\circ} = A B / A C

1 / 2 = 5 / A C

$A C = 10$ cm.

Answer: $B C = 5 \sqrt{3}$ cm, $A C = 10$ cm.

Example 4

Given: A right-angled triangle PQR, $∠ Q = 90^{\circ}$ , $P Q = 3$ cm, $P R = 6$ cm. To find: $∠ Q P R$ and $∠ P R Q$ .

Solution: In $△ P Q R$ : $\sin R = Opposite / Hypotenuse = P Q / P R = 3 / 6 = 1 / 2$ . Since $\sin R = 1 / 2$ , the angle R is $30^{\circ}$ . So, $∠ P R Q = 30^{\circ}$ .

Now, $∠ Q P R = 180^{\circ} - 90^{\circ} - ∠ P R Q = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ}$ .

Answer: $∠ P R Q = 30^{\circ}$ , $∠ Q P R = 60^{\circ}$ .

Example 5

Given: $\sin (A - B) = 1 / 2$ and $\cos (A + B) = 1 / 2$ , with $0^{\circ} < A + B \leq 90^{\circ}$ and $A > B$ . To find: Values of A and B.

Solution: From $\sin (A - B) = 1 / 2$ , we know that $A - B = 30^{\circ}$ (since $A > B$ , $A - B$ is positive and less than $90^{\circ}$ ). From $\cos (A + B) = 1 / 2$ , we know that $A + B = 60^{\circ}$ (since $0^{\circ} < A + B \leq 90^{\circ}$ ).

We have a system of linear equations:

$A - B = 30^{\circ}$
$A + B = 60^{\circ}$

Adding (1) and (2):

(A - B) + (A + B) = 30^{\circ} + 60^{\circ}

2 A = 90^{\circ}

A = 45^{\circ}

Substituting $A = 45^{\circ}$ into equation (2):

45^{\circ} + B = 60^{\circ}

$B = 60^{\circ} - 45^{\circ} = 15^{\circ}$ .

Answer: $A = 45^{\circ}$ , $B = 15^{\circ}$ .

Remember
The values for $0^{\circ}$ , $30^{\circ}$ , $45^{\circ}$ , $60^{\circ}$ , and $90^{\circ}$ are crucial. Memorizing them, or the pattern of their generation, is key. Notice how $\sin θ$ values increase and $\cos θ$ values decrease in this range.

Exam-smart
Questions will often involve substituting these values into expressions to simplify them, or using them to find unknown angles or side lengths in right-angled triangles. Problems might also present equations like sin(X) = 1/2 and ask for angle X, requiring you to recall the specific angle value.

Try this: Evaluate: $\sin 30^{\circ} \cos 60^{\circ} + \cos 30^{\circ} \sin 60^{\circ}$ . Answer: 1

Trigonometric Identities

A trigonometric identity is an equation involving trigonometric ratios that holds true for all possible values of the angle(s) for which the ratios are defined. These identities are fundamental in simplifying trigonometric expressions and proving other relationships.

In a right-angled triangle $△ A B C$ , with the right angle at $B$ , the Pythagorean theorem states:

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

Dividing this equation by $A C^{2}$ :

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

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

Since $\frac{A B}{A C} = \cos A$ and $\frac{B C}{A C} = \sin A$ , we get the first fundamental identity:

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

This identity is true for all angles $A$ between $0^{\circ}$ and $90^{\circ}$ .

Dividing the Pythagorean theorem by $A B^{2}$ :

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

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

Since $\frac{B C}{A B} = \tan A$ and $\frac{A C}{A B} = \sec A$ , we get the second identity:

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

This identity is true for all angles $A$ such that $0^{\circ} \leq A < 90^{\circ}$ , as $\tan A$ and $\sec A$ are undefined at $A = 90^{\circ}$ .

Dividing the Pythagorean theorem by $B C^{2}$ :

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

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

Since $\frac{A B}{B C} = \cot A$ and $\frac{A C}{B C} = \csc A$ , we get the third identity:

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

This identity is true for all angles $A$ such that $0^{\circ} < A \leq 90^{\circ}$ , as $\cot A$ and $\csc A$ are undefined at $A = 0^{\circ}$ .

These three identities are known as the fundamental trigonometric identities:

Identity	Description	Application
$\sin^{2} A + \cos^{2} A = 1$	Relates the sine and cosine of an angle.	Expressing one in terms of the other, solving equations
$1 + \tan^{2} A = \sec^{2} A$	Relates the tangent and secant of an angle.	Simplifying expressions, solving equations
$1 + \cot^{2} A = \csc^{2} A$	Relates the cotangent and cosecant of an angle.	Simplifying expressions, solving equations

These identities allow us to express any trigonometric ratio in terms of another, which is useful when only one ratio is known.

Example 9 : Express the ratios $\cos A$ , $\tan A$ , and $\sec A$ in terms of $\sin A$ .

Given: The need to express $\cos A$ , $\tan A$ , and $\sec A$ in terms of $\sin A$ . To find: Expressions for $\cos A$ , $\tan A$ , and $\sec A$ solely using $\sin A$ .

Solution: We use the fundamental identity $\sin^{2} A + \cos^{2} A = 1$ .

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

Taking the square root of both sides:

\cos A = \pm \sqrt{1 - \sin^{2} A}

For acute angles $A$ ( $0^{\circ} \leq A < 90^{\circ}$ ), $\cos A$ is positive, so we write:

\cos A = \sqrt{1 - \sin^{2} A}

Now we find $\tan A$ :

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

Substituting the expression for $\cos A$ :

\tan A = \frac{\sin A}{\sqrt{1 - \sin^{2} A}}

Finally, we find $\sec A$ :

\sec A = \frac{1}{\cos A}

Substituting the expression for $\cos A$ :

\sec A = \frac{1}{\sqrt{1 - \sin^{2} A}}

Answer:

\cos A = \sqrt{1 - \sin^{2} A}

\tan A = \frac{\sin A}{\sqrt{1 - \sin^{2} A}}

\sec A = \frac{1}{\sqrt{1 - \sin^{2} A}}

Example 10 : Prove that $\sec A (1 – \sin A) (\sec A + \tan A) = 1$ .

Given: The expression $\sec A (1 – \sin A) (\sec A + \tan A)$ . To find: Prove that the expression equals 1.

Solution: Let's simplify the Left Hand Side (LHS):

LHS = \sec A (1 – \sin A) (\sec A + \tan A)

Rewrite $\sec A$ and $\tan A$ in terms of $\sin A$ and $\cos A$ :

\sec A = \frac{1}{\cos A}, \tan A = \frac{\sin A}{\cos A}

Substitute these into the LHS:

LHS = (\frac{1}{\cos A}) (1 – \sin A) (\frac{1}{\cos A} + \frac{\sin A}{\cos A})

Combine the terms in the third parenthesis:

LHS = (\frac{1}{\cos A}) (1 – \sin A) (\frac{1 + \sin A}{\cos A})

Multiply the terms:

LHS = \frac{(1 – \sin A) (1 + \sin A)}{\cos A \cdot \cos A}

Use the difference of squares formula $(a - b) (a + b) = a^{2} - b^{2}$ in the numerator:

LHS = \frac{1^{2} – \sin^{2} A}{\cos^{2} A}

LHS = \frac{1 – \sin^{2} A}{\cos^{2} A}

From the identity $\sin^{2} A + \cos^{2} A = 1$ , we know that $1 – \sin^{2} A = \cos^{2} A$ . Substitute this into the numerator:

LHS = \frac{\cos^{2} A}{\cos^{2} A}

LHS = 1

This is equal to the Right Hand Side (RHS).

Answer: LHS = RHS, hence the identity is proven.

Example 11 : Prove that $\frac{\cot A – \cos A}{\cot A + \cos A} = \frac{\csc A – 1}{\csc A + 1}$ .

Given: The expression $\frac{\cot A – \cos A}{\cot A + \cos A}$ . To find: Prove that it equals $\frac{\csc A – 1}{\csc A + 1}$ .

Solution: Let's simplify the Left Hand Side (LHS):

LHS = \frac{\cot A – \cos A}{\cot A + \cos A}

Rewrite $\cot A$ in terms of $\sin A$ and $\cos A$ :

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

Substitute this into the LHS:

LHS = \frac{\frac{\cos A}{\sin A} – \cos A}{\frac{\cos A}{\sin A} + \cos A}

Factor out $\cos A$ from the numerator and denominator:

LHS = \frac{\cos A (\frac{1}{\sin A} – 1)}{\cos A (\frac{1}{\sin A} + 1)}

Cancel out $\cos A$ (assuming $\cos A \neq 0$ ):

LHS = \frac{\frac{1}{\sin A} – 1}{\frac{1}{\sin A} + 1}

Now, substitute $\csc A = \frac{1}{\sin A}$ :

LHS = \frac{\csc A – 1}{\csc A + 1}

This is the Right Hand Side (RHS).

Answer: LHS = RHS, hence the identity is proven.

Example 12 : Prove that $\frac{\sin θ – \cos θ + 1}{\sin θ + \cos θ – 1} = \frac{1 + \sin θ}{\cos θ}$ , using the identity $\sec^{2} θ = 1 + \tan^{2} θ$ .

Given: The expression $\frac{\sin θ – \cos θ + 1}{\sin θ + \cos θ – 1}$ and the identity $\sec^{2} θ = 1 + \tan^{2} θ$ . To find: Prove that the expression equals $\frac{1 + \sin θ}{\cos θ}$ .

Solution: The hint suggests using the identity $\sec^{2} θ = 1 + \tan^{2} θ$ . To involve secant and tangent, we divide the numerator and denominator of the LHS by $\cos θ$ :

LHS = \frac{\frac{\sin θ}{\cos θ} – \frac{\cos θ}{\cos θ} + \frac{1}{\cos θ}}{\frac{\sin θ}{\cos θ} + \frac{\cos θ}{\cos θ} – \frac{1}{\cos θ}}

LHS = \frac{\tan θ – 1 + \sec θ}{\tan θ + 1 – \sec θ}

Rearrange the terms in the numerator and denominator:

LHS = \frac{(\tan θ + \sec θ) – 1}{(\tan θ – \sec θ) + 1}

We know that $\sec^{2} θ - \tan^{2} θ = 1$ . So, we can write $1$ as $\sec^{2} θ - \tan^{2} θ$ .

LHS = \frac{(\tan θ + \sec θ) – (\sec^{2} θ – \tan^{2} θ)}{(\tan θ – \sec θ) + 1}

Factor the difference of squares in the numerator:

LHS = \frac{(\tan θ + \sec θ) – (\sec θ – \tan θ) (\sec θ + \tan θ)}{(\tan θ – \sec θ) + 1}

Factor out $(\tan θ + \sec θ)$ from the numerator:

LHS = \frac{(\tan θ + \sec θ) [1 – (\sec θ – \tan θ)]}{(\tan θ – \sec θ) + 1}

LHS = \frac{(\tan θ + \sec θ) (1 – \sec θ + \tan θ)}{(\tan θ – \sec θ) + 1}

Notice that the term $(1 – \sec θ + \tan θ)$ in the numerator is the same as $(\tan θ – \sec θ + 1)$ in the denominator.

LHS = \frac{(\tan θ + \sec θ) (\tan θ – \sec θ + 1)}{(\tan θ – \sec θ + 1)}

Cancel out the common term $(\tan θ – \sec θ + 1)$ :

LHS = \tan θ + \sec θ

Now, express this in terms of $\sin θ$ and $\cos θ$ :

LHS = \frac{\sin θ}{\cos θ} + \frac{1}{\cos θ}

LHS = \frac{\sin θ + 1}{\cos θ}

This is the Right Hand Side (RHS).

Answer: LHS = RHS, hence the identity is proven.

Remember
When proving identities, it is often helpful to express all trigonometric ratios in terms of $\sin θ$ and $\cos θ$ . Also, remember to use the fundamental identities to substitute and simplify.

Exam-smart
Questions often ask to prove an identity. You'll be given an equation and need to manipulate one side (or both) using the fundamental identities and algebraic rules to show they are equal. Case studies might involve scenarios where a complex trigonometric expression needs to be simplified to a single value or a simpler ratio, using these identities.

Try this: Prove the identity $\frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \sin A + \cos A$ . Answer: LHS = $\frac{\cos A}{1 - \frac{\sin A}{\cos A}} + \frac{\sin A}{1 - \frac{\cos A}{\sin A}}$ = $\frac{\cos A}{\frac{\cos A - \sin A}{\cos A}} + \frac{\sin A}{\frac{\sin A - \cos A}{\sin A}}$ = $\frac{\cos^{2} A}{\cos A - \sin A} + \frac{\sin^{2} A}{\sin A - \cos A}$ = $\frac{\cos^{2} A}{\cos A - \sin A} - \frac{\sin^{2} A}{\cos A - \sin A}$ = $\frac{\cos^{2} A - \sin^{2} A}{\cos A - \sin A}$ = $\frac{(\cos A - \sin A) (\cos A + \sin A)}{\cos A - \sin A}$ = $\cos A + \sin A$ = RHS.

Introduction to Trigonometry — Notes

Read the notes

Right-Angled Triangles and Trigonometric Ratios

Relationship Between Tangent and Cotangent

Invariance of Trigonometric Ratios

Notation for Powers of Trigonometric Ratios

Historical Context

Finding Other Trigonometric Ratios Given One Ratio

Example 1

Example 2

Example 3

Example 4

Example 5

Trigonometric Ratios of Some Specific Angles

Trigonometric Ratios of 45°

Example 1

Trigonometric Ratios of 30° and 60°

Example 2

Trigonometric Ratios of 0° and 90°

Table of Trigonometric Ratios for Specific Angles

Example 3

Example 4

Example 5

True/False Statements about Trigonometric Ratios

Trigonometric Ratios of Some Specific Angles

Trigonometric Ratios of $45^{\circ}$

Example 1

Trigonometric Ratios of $30^{\circ}$ and $60^{\circ}$

Example 2

Trigonometric Ratios of $0^{\circ}$ and $90^{\circ}$

Table of Trigonometric Ratios for Specific Angles

Example 3

Example 4

Example 5

Trigonometric Identities

Example 9 : Express the ratios $\cos A$ , $\tan A$ , and $\sec A$ in terms of $\sin A$ .

Example 10 : Prove that $\sec A (1 – \sin A) (\sec A + \tan A) = 1$ .

Example 11 : Prove that $\frac{\cot A – \cos A}{\cot A + \cos A} = \frac{\csc A – 1}{\csc A + 1}$ .

Example 12 : Prove that $\frac{\sin θ – \cos θ + 1}{\sin θ + \cos θ – 1} = \frac{1 + \sin θ}{\cos θ}$ , using the identity $\sec^{2} θ = 1 + \tan^{2} θ$ .

All Class 10 Mathematics chapters

Introduction to Trigonometry — Notes

Read the notes

Right-Angled Triangles and Trigonometric Ratios

Relationship Between Tangent and Cotangent

Invariance of Trigonometric Ratios

Notation for Powers of Trigonometric Ratios

Historical Context

Finding Other Trigonometric Ratios Given One Ratio

Example 1

Example 2

Example 3

Example 4

Example 5

Trigonometric Ratios of Some Specific Angles

Trigonometric Ratios of 45°

Example 1

Trigonometric Ratios of 30° and 60°

Example 2

Trigonometric Ratios of 0° and 90°

Table of Trigonometric Ratios for Specific Angles

Example 3

Example 4

Example 5

True/False Statements about Trigonometric Ratios

Trigonometric Ratios of Some Specific Angles

Trigonometric Ratios of 45∘

Example 1

Trigonometric Ratios of 30∘ and 60∘

Example 2

Trigonometric Ratios of 0∘ and 90∘

Table of Trigonometric Ratios for Specific Angles

Example 3

Example 4

Example 5

Trigonometric Identities

Example 9 : Express the ratios cos⁡A, tan⁡A, and sec⁡A in terms of sin⁡A.

Example 10 : Prove that sec⁡A(1–sin⁡A)(sec⁡A+tan⁡A)=1.

Example 11 : Prove that cot⁡A–cos⁡Acot⁡A+cos⁡A=csc⁡A–1csc⁡A+1.

Example 12 : Prove that sin⁡θ–cos⁡θ+1sin⁡θ+cos⁡θ–1=1+sin⁡θcos⁡θ, using the identity sec⁡2θ=1+tan⁡2θ.

All Class 10 Mathematics chapters

Trigonometric Ratios of $45^{\circ}$

Trigonometric Ratios of $30^{\circ}$ and $60^{\circ}$

Trigonometric Ratios of $0^{\circ}$ and $90^{\circ}$

Example 9 : Express the ratios $\cos A$ , $\tan A$ , and $\sec A$ in terms of $\sin A$ .

Example 10 : Prove that $\sec A (1 – \sin A) (\sec A + \tan A) = 1$ .

Example 11 : Prove that $\frac{\cot A – \cos A}{\cot A + \cos A} = \frac{\csc A – 1}{\csc A + 1}$ .

Example 12 : Prove that $\frac{\sin θ – \cos θ + 1}{\sin θ + \cos θ – 1} = \frac{1 + \sin θ}{\cos θ}$ , using the identity $\sec^{2} θ = 1 + \tan^{2} θ$ .