CBSE Class 10 Chapter 8 Introduction of Trigonometry Notes

Aryabhatta, an Indian mathematician, first introduced Trigonometry in his book Aryabhattam.

This mathematical concept shows the connections between angles and the sides of a right angle triangle.

Trigonometric Ratios

• As given figure, Angle A is Acute angle. 
• In respect of Angle A side BC called opposite side to Angle A and side AB called Adjacent side to angle A because it is a part of angle A.

• As given figure, Angle C is Acute angle. 
• In respect of Angle C side AB called opposite side to Angle C and side BC called Adjacent side to angle C because it is a part of angle C.

So, the trigonometric ratios of the angle A and C in right angle as shown in figures can be defined as follow:

• Sine $(\sin)$: The sine of an angle is the ratio of the length of the side opposite the angle to the length of the hypotenuse in a right-angled triangle.

$\sin \text {(A)}=\frac {\text {Opposite side to Angle A}}{\text {Hypotenuse}}=\frac {BC}{AC}$

$\sin \text {(C)}=\frac {\text {Opposite side to Angle C}}{\text {Hypotenuse}}=\frac {AB}{AC}$

• Cosine $(\cos)$: The cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse in a right-angled triangle.

$\cos \text {(A)}=\frac {\text {Adjacent side to angle A}}{\text {Hypotenuse}}=\frac {AB}{AC}$

$\cos \text {(C)}=\frac {\text {Adjacent side to angle C}}{\text {Hypotenuse}}=\frac {BC}{AC}$

• Tangent $(\tan)$: The tangent of an angle is the ratio of the length of the side opposite the angle to the length of the adjacent side in a right-angled triangle.

$\tan \text {(A)}=\frac {\text {Opposite side to angle A}}{\text {Adjacent side to angle A}}=\frac {BC}{AB}$

$\tan \text {(C)}=\frac {\text {Opposite side to angle C}}{\text {Adjacent side to angle C}}=\frac {AB}{BC}$

• Tangent relation with sin and cos is $\tan \theta = \frac {\sin \theta}{cos \theta}$

Lets check it for angle A and C :

We know that - 

$\sin \theta = \frac {\text {Opposite side to Angle}}{\text {Hypotenuse}}$ 

 $\cos \theta =\frac {\text {Adjacent side to angle }}{\text {Hypotenuse}}$

$\tan \theta =\frac {\text {Opposite side to angle}}{\text {Adjacent side to angle}}$ 

Putting value from above given figure for Angle A  We get,

$\tan \text {(A)}=\frac {BC}{AB}=\frac {\frac {BC}{AC}}{\frac {AB}{AC}}=\frac {\sin \text { A}}{cos \text { A}}$

Putting value from above given figure for Angle C  We get, 

 $\tan \text {(C)}=\frac {AB}{BC}=\frac {\frac {AB}{AC}}{\frac {BC}{AC}}=\frac {\sin \text { C}}{cos \text { C}}$ 

These ratios can be remembered using the acronym SOH-CAH-TOA.

In addition to these three primary ratios, there are three reciprocal trigonometric ratios:

• Cosecant (cosec): It is the reciprocal of the sine. The cosecant represents the ratio of the hypotenuse to the side opposite the given angle in a right-angled triangle.

$\text {cosec (A)}=\frac {1}{\sin (A)}=\frac {\text {Hypotenuse}}{\text {opposite side to angle A}}=\frac {AC}{BC}$

$\text {cosec (C)}=\frac {1}{\sin (C)}=\frac {\text {Hypotenuse}}{\text {Opposite side to angle C}}=\frac {AC}{AB}$

• Secant $(\sec)$: It is the reciprocal of the cosine. The secant represents the ratio of the hypotenuse to the side adjacent to the given angle in a right-angled triangle.

$\sec \text {(A)}=\frac {1}{\cos (A)}=\frac {\text {Hypotenuse}}{\text {Adjacent side to angle A}}=\frac {AC}{AB}$

$\sec \text {(C)}=\frac {1}{\cos (C)}=\frac {\text {Hypotenuse}}{\text {Adjacent side to angle C}}=\frac {AC}{BC}$

• Cotangent $(\cot)$: It is the reciprocal of the tangent. The cotangent represents the ratio of the side adjacent to the angle to the side opposite the angle in a right-angled triangle.

$\cot \text {(A)}=\frac {1}{\tan (A)}=\frac {\text {Adjacent side to angle A}}{\text {Opposite side to angle A}}=\frac {AB}{BC}$

$\cot \text {(C)}=\frac {1}{\tan (C)}=\frac {\text {Adjacent side to angle C}}{\text {Opposite side to angle C}}=\frac {BC}{AB}$

• Cotangent relation with sin and cos is $\cot \theta = \frac {\cos \theta}{sin \theta}$

Lets check it for angle A and C :

We know that - 

$\sin \theta = \frac {\text {Opposite side to Angle}}{\text {Hypotenuse}}$ 

$\cos \theta =\frac {\text {Adjacent side to angle }}{\text {Hypotenuse}}$

$\cot \theta =\frac {1}{\tan (A)}$

Putting value from above given figure for Angle A  We get,

$\cot \text {(A)}=\frac {1}{\tan (A)}=\frac {1}{\frac {\sin (A)}{\cos (A)}}=\frac {\cos (A)}{\sin (A)}$

Putting value from above given figure for Angle C  We get, 

$\cot \text {(C)}=\frac {1}{\tan (C)}=\frac {1}{\frac {\sin (C)}{\cos (C)}}=\frac {\cos (C)}{\sin (C)}$

Note: The values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.

Trigonometric Ratios of Some Specific Angles

There are some specific angles that are defined for trigonometric ratios are $0^{0}, 30^{0}, 45^{0}, 60^{0} \text { and } 90^{0}$

Trigonometric Ratios of $45^\circ$:

If one of the angles of a right-angled triangle is $45^\circ$, then another angle will also be equal to $45^\circ$.

Consider a right-angled triangle ABC at B, such that:

$\angle A = \angle C = 45^\circ$

Thus, $BC = AB = a$ (let's denote it as $a$.

By using the Pythagorean theorem, we have:

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

$AC^2 = a^2 + a^2$

$AC^2 = 2a^2$

$AC = a\sqrt{2}$

Now, applying trigonometric ratios:

$\sin 45^\circ = \frac{\text{Opposite side to angle } 45^\circ}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}}$

$\cos 45^\circ = \frac{\text{Adjacent side to angle } 45^\circ}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}}$

$\tan 45^\circ = \frac{\text{Opposite side to angle } 45^\circ}{\text{Adjacent side}} = \frac{BC}{AB} = \frac{a}{a} = 1$

Similarly, by applying trigonometric ratios we can find reciprocal of sin, cos and tan as fallow: 

$\text {cosec } 45^\circ = \frac{1}{\sin 45^\circ} = \frac{1}{\frac{1}{\sqrt{2}}}= \sqrt{2}$

$\sec 45^\circ = \frac{1}{\cos 45^\circ} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}$

$\cot 45^\circ = \frac{1}{\tan 45^\circ} = \frac{1}{1}= 1$

Trigonometric Ratios of $30^\circ$ and $60^\circ$:

Lets take an equilateral triangle ABC, such that;

$AB = BC = AC = 2a$

$\angle A = \angle B = \angle C = 60^\circ$

Draw a perpendicular AD from vertex A that meets BC at D

By congruency of the triangle, we can write;

$\Delta ABD \cong \Delta ACD$

Hence, $BD = DC$

$\angle BAD = \angle CAD$ (By CPCT)

Now, in triangle ABD, $\angle BAD = 30^\circ$ and $\angle ABD = 60^\circ$

By using Pythagoras theorem,

$AD^2 = AB^2 - BD^2$

$AD^2= (2a)^2 - (a)^2$

$AD^2= 3a^2$

$AD = a\sqrt{3}$

Now, applying trigonometric ratios for  $30^\circ$ angle will be;

$\sin 30^\circ = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2}$

$\cos 30^\circ = \frac{AD}{AB} = \frac{a\sqrt{3}}{2a} = \frac{\sqrt{3}}{2}$

$\tan 30^\circ = \frac{BD}{AD} = \frac{a}{a\sqrt{3}} = \frac{1}{\sqrt{3}}$

Similarly, by applying trigonometric ratios we can find reciprocal of sin, cos and tan as fallow: 

$\text {cosec } 30^\circ = \frac{1}{\sin 30} = 2$

$\sec 30^\circ = \frac{1}{\cos 30} = \frac{2}{\sqrt{3}}$

$\cot 30^\circ = \frac{1}{\tan 30} = \sqrt{3}$

Similarly, we can derive the values of trigonometric ratios for $60^\circ$.

$\sin 60^\circ = \frac{AD}{AB} = \frac{a\sqrt{3}}{2a} = \frac{\sqrt{3}}{2}$

$\cos 60^\circ = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2}$

$\tan 60^\circ = \frac{AD}{BD} = \frac{a\sqrt{3}}{a} = \sqrt{3}$

Similarly, by applying trigonometric ratios we can find reciprocal of sin, cos and tan as fallow: 

$\text {cosec } 60^\circ = \frac{1}{\sin 60} = \frac{1}{\frac{\sqrt{3}}{2}}= \frac{2}{\sqrt{3}}$

$\sec 60^\circ = \frac{1}{\cos 60} = \frac{1}{\frac{1}{2}}= 2$

$\cot 60^\circ = \frac{1}{\tan 60} =  \frac {1}{\sqrt{3}}$

Trigonometric Ratios of $0^\circ$ and $0^\circ$:

Let's take a ABC right angle triangle at B, $\angle {A}$ is a acute angle.

If  we reduced $\angle {A}$ then side AC will come near to side AB. So, if

$\angle {A}=0^{0}$ then almost $AC = AB$ and $BC = 0$

Hence, trignometric ratios for $0^\circ$

$\sin {A} = \frac {BC}{AC} = 0$

$\cos {A} = \frac {AB}{AC} = 1$

$\tan {A} = \frac {\sin {A}}{\cos {A}} = \frac {0}{1} = 0$

also their reciprocal are

$\text {cosec } {A} = \frac {1}{\sin {A}} = \frac {1}{0}= \text {not defined}$

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

$\cot {A} = \frac {1}{\tan {A}} = \frac {1}{0}= \text {not defined}$

Trignometric ratios for $90^\circ$

Let's again take a ABC right angle triangle at B, $\angle {C}$ is a acute angle.

If  we reduced $\angle {C}$ then side AC will come near to side BC. So, if

$\angle {C}=0^{0}$ then almost $AC = BC$ and $AB = 0$

Hence, trignometric ratios for $90^\circ$

$\sin {C} = \frac {AB}{AC} = 1$

$\cos {C} = \frac {AB}{AC} = 0$

$\tan {C} = \frac {\sin {C}}{\cos {C}} = \frac {1}{0} = \text {Not defined}$

also their reciprocal are

$\text {cosec } {C} = \frac {1}{\sin {C}} = \frac {1}{1}= 1$

$\sec {C} = \frac {1}{\cos {C}} = \frac {1}{0}= \text {Not defined}$

$\cot {C} = \frac {1}{\tan {C}} = \frac {1}{\ text {Not defined}}= 0$

Trigonometric Ratio table for specific angle

$\angle {\theta}$	$0^{0}$	$30^{0}$	$45^{0}$	$60^{0}$	$90^{0}$
$\sin {\theta}$	$0$	$\frac {1}{2}$	$\frac {1}{\sqrt 2}$	$\frac {\sqrt 3}{2}$	$1$
$\cos {\theta}$	$1$	$\frac {\sqrt 3}{2}$	$\frac {1}{\sqrt 2}$	$\frac {1}{2}$	$0$
$\tan {\theta}$	$0$	$\frac {1}{\sqrt 3}$	$1$	$\sqrt 3$	$\text {Not defined}$
$\text { cosec} {\theta}$	$\text {Not defined}$	$2$	$\sqrt 2$	$\frac {2}{\sqrt 3}$	$1$
$\sec {\theta}$	$1$	$\frac {2}{\sqrt 3}$	$\sqrt 2$	$2$	$\text {Not defined}$
$\cot {\theta}$	$\text {Not defined}$	$\sqrt 3$	$1$	$\frac {1}{\sqrt 3}$	$0$

Trigonometric Identities

• Trigonometric identities are an equations that involve trigonometric ratios of an angle and are true for all values of the angles involved.
• These identities are fundamental in trigonometry and are used to simplify expressions, solve equations and establish relationships between different trigonometric functions.
• Trigonometric identities are applicable specifically to right-angled triangles.
• The foundational trigonometric identities are derived from the six fundamental trigonometric ratios: sine, cosine, tangent, cosecant, secant, and cotangent.

Fundamental Trigonometric Identities and their Proofs:

Lets Take a right angle triangle ABC which is right angled at B as per figure.

Here,

Adjacent side to $\angle {A}$ = $\text {Base}$

Opposite side to $\angle {A}$ = $\text {Perpendicular}$

Now, by applying Pythagoras Theorem, we have

$$\text {(Hypotenuse)}^{2} = \text {(Base)}^{2}+\text {(Perpendicular)}^{2}$$

$$AC^{2} = AB^{2}+BC^{2}     ………………………..(i)$$

Trigonometric identities -1:

By dividing each term of equation (i) by $AC^{2}$ we get

$$\frac{AB^{2}}{AC^{2}}+\frac{BC^{2}}{AC^{2}}=\frac{AC^{2}}{AC^{2}}$$

$$\left(\frac{AB}{AC}\right)^{2}+\left(\frac{BC}{AC}\right)^{2}=\left(\frac{AC}{AC}\right)^{2}$$

As we know that $\left(\frac{AB}{AC}\right)^{2}=\cos A$ $\left(\frac{BC}{AC}\right)^{2}=\sin A$ So,

$$(\cos A)^{2}+(\sin A)^{2}=1$$

$$\bf \cos^{2} A+\sin^{2} A=1 .............(ii)$$

Here, Identity 1 is valid for $0^{0}\leq A \leq 90^{0}$

Trigonometric identities -2:

By dividing each term of equation (i) by $AB^{2}$ we get

$$\frac{AB^{2}}{AB^{2}}+\frac{BC^{2}}{AB^{2}}=\frac{AC^{2}}{AB^{2}}$$

$$\left(\frac{AB}{AB}\right)^{2}+\left(\frac{BC}{AB}\right)^{2}=\left(\frac{AC}{AB}\right)^{2}$$

As we know that $\left(\frac{AC}{AB}\right)^{2}=\sec A$ $\left(\frac{BC}{AB}\right)^{2}=\tan A$ So,

$$\bf 1+\tan^{2} A=\sec^{2} A .............(iii)$$

This equation true, $A=0^{0}$ and $A=90^{0}$ is not defined for $\sec$ and $\tan$. so this Identity  is valid for all $0^{0}\leq A < 90^{0}$

Trigonometric identities -3:

By dividing each term of equation (i) by $BC^{2}$ we get

$$\frac{AB^{2}}{BC^{2}}+\frac{BC^{2}}{BC^{2}}=\frac{AC^{2}}{BC^{2}}$$

$$\left(\frac{AB}{BC}\right)^{2}+\left(\frac{BC}{BC}\right)^{2}=\left(\frac{AC}{BC}\right)^{2}$$

As we know that $\left(\frac{AC}{BC}\right)^{2}=\text {cosec } A$ $\left(\frac{AB}{BC}\right)^{2}=\cot A$ So,

$$\bf \cot^{2}A+1=\text {Cosec}^{2} A .............(iii)$$

This equation is not defined  if$A=0^{0}$  for $\text {cosec}$ and $\cot$. so this Identity  is valid for all $0^{0}< A \leq 90^{0}$

These three identities are known as Pythagorean Trigonometric Identities.

Reciprocal Trigonometric Identities

There are three Reciprocal trigonometric identites :

• $\sin \theta = \frac{1}{\text {cosec } \theta}$ or $\text {cosec} \theta = \frac{1}{\sin \theta}$
• Proof:

As per figure in $\triangle ABC$ , let

$\text {Opposite side to} \angle A = BC = a$ 

$\text {Adjacent side to} \angle A = AB = b$ 

$\text {Hypotenuse}= AC = c$

Then by definition of sine in a right angle triangle:

$\sin A = \frac {BC}{AC}=\frac {a}{c}$

Taking reciprocal of both side:

$\frac {1}{\sin A} = \frac {AC}{BC}=\frac {c}{a}$

As we know $\frac {AC}{BC}=\text {cosec }\theta$

By putting value

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

Therefore, $\text {cosec }\theta = \frac {1}{\sin \theta}$

• $\cos \theta = \frac{1}{\sec \theta}$ or $\sec \theta = \frac{1}{\cos \theta}$
• Proof:



As per figure in $\triangle ABC$ , let

$\text {Opposite side to} \angle A = BC = a$ 

$\text {Adjacent side to} \angle A = AB = b$ 

$\text {Hypotenuse}= AC = c$

Then by definition of cosine in a right angle triangle:

$\cos A = \frac {AB}{AC}=\frac {b}{c}$

Taking reciprocal of both side:

$\frac {1}{\cos A} = \frac {AC}{AB}=\frac {c}{b}$

As we know $\frac {AC}{AB}=\sec\theta$

By putting value

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

Therefore, $\sec \theta = \frac {1}{\cos \theta}$

• $\tan \theta = \frac{1}{\cot \theta}$ or $\cot \theta = \frac{1}{\tan \theta}$
• Proof:



As per figure in $\triangle ABC$ , let

$\text {Opposite side to} \angle A = BC = a$ 

$\text {Adjacent side to} \angle A = AB = b$ 

$\text {Hypotenuse}= AC = c$

Then by definition of tangent in a right angle triangle:

$\tan A = \frac {BC}{AB}=\frac {a}{b}$

Taking reciprocal of both side:

$\frac {1}{\tan A} = \frac {1}{\frac {BC}{AB}}=\frac {1}{\frac {a}{b}}$

$\frac {1}{\tan A} = \frac {AB}{BC}=\frac {b}{a}$

As we know $\frac {AB}{BC}=\cot\theta$

By putting value

$\frac {1}{\tan A} =\cot\theta$

Therefore, $\cot \theta = \frac {1}{\tan \theta}$

Ratio Trigonometric Identities

There are two trigonometric ratio identities:

• $\tan \theta = \frac{\sin \theta}{\cos \theta}$
• Proof:

As per figure in $\triangle ABC$ , let

$\text {Opposite side to} \angle A = BC$ 

$\text {Adjacent side to} \angle A = AB$ 

$\text {Hypotenuse}= AC$

Using the definitions of sine, cosine, and tangent in the context of this triangle:

$\sin\theta = \frac {\text {Opposite side to angle}}{\text {Hypotenuse}}=\frac {BC}{AC}$

$\cos\theta = \frac {\text {Adjacent side to angle}}{\text {Hypotenuse}}=\frac {AB}{AC}$

So,

$\tan\theta = \frac {\sin\theta}{\cos\theta}=\frac {\frac {BC}{AC}}{\frac{AB}{AC}}$

$\tan\theta = \frac {\sin\theta}{\cos\theta}=\frac {BC}{AB}$

Therefore, $\tan\theta = \frac {\sin\theta}{\cos\theta}$ proven.

• $\cot \theta = \frac{\cos \theta}{\sin \theta}$
• Proof:

As per figure in $\triangle ABC$ , let

$\text {Opposite side to} \angle A = BC$ 

$\text {Adjacent side to} \angle A = AB$ 

$\text {Hypotenuse}= AC$

Using the definitions of sine and cosine in the context of this triangle:

$\sin\theta = \frac {\text {Opposite side to angle}}{\text {Hypotenuse}}=\frac {BC}{AC}$

$\cos\theta = \frac {\text {Adjacent side to angle}}{\text {Hypotenuse}}=\frac {AB}{AC}$

So,

$\cot\theta = \frac {\cos\theta}{\sin\theta}=\frac {\frac {AB}{AC}}{\frac{BC}{AC}}$

$\cot\theta = \frac {\cos\theta}{\sin\theta}=\frac {AB}{BC}$

Therefore, $\cot\theta = \frac {\cos\theta}{\sin\theta}$ proven.