There is perhaps nothing which so occupies the middle position of mathematics as trigonometry.

J.F. Herbart (1890)

8.1 Introduction

You have already studied about triangles, and in particular, right triangles, in your earlier classes. Let us take some examples from our surroundings where right triangles can be imagined to be formed. For instance :

1. Suppose the students of a school are visiting Qutub Minar. Now, if a student is looking at the top of the Minar, a right triangle can be imagined to be made, as shown in Fig 8.1. Can the student find out the height of the Minar, without actually measuring it?

Fig. 8.1

2. Suppose a girl is sitting on the balcony of her house located on the bank of a river. She is looking down at a flower pot placed on a stair of a temple situated nearby on the other bank of the river. A right triangle is imagined to be made in this situation as shown in Fig.8.2. If you know the height at which the person is sitting, can you find the width of the river?

Fig. 8.2

3. Suppose a hot air balloon is flying in the air. A girl happens to spot the balloon in the sky and runs to her mother to tell her about it. Her mother rushes out of the house to look at the balloon.Now when the girl had spotted the balloon intially it was at point A. When both the mother and daughter came out to see it, it had already travelled to another point B. Can you find the altitude of $B$ from the ground?

Fig. 8.3

In all the situations given above, the distances or heights can be found by using some mathematical techniques, which come under a branch of mathematics called ’trigonometry’. The word ’trigonometry’ is derived from the Greek words ’tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). In fact, trigonometry is the study of relationships between the sides and angles of a triangle. The earliest known work on trigonometry was recorded in Egypt and Babylon. Early astronomers used it to find out the distances of the stars and planets from the Earth. Even today, most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts.

In this chapter, we will study some ratios of the sides of a right triangle with respect to its acute angles, called trigonometric ratios of the angle. We will restrict our discussion to acute angles only. However, these ratios can be extended to other angles also. We will also define the trigonometric ratios for angles of measure $0^{\circ}$ and $90^{\circ}$. We will calculate trigonometric ratios for some specific angles and establish some identities involving these ratios, called trigonometric identities.

8.2 Trigonometric Ratios

In Section 8.1, you have seen some right triangles imagined to be formed in different situations.

Let us take a right triangle $ABC$ as shown in Fig. 8.4.

Fig. 8.4

Here, $\angle CAB$ (or, in brief, angle $A$ ) is an acute angle. Note the position of the side $BC$ with respect to angle $A$. It faces $\angle A$. We call it the side opposite to angle $A$. $AC$ is the hypotenuse of the right triangle and the side $AB$ is a part of $\angle A$. So, we call it the side adjacent to angle $A$.

Note that the position of sides change when you consider angle $C$ in place of $A$ (see Fig. 8.5).

Fig. 8.5

You have studied the concept of ‘ratio’ in your earlier classes. We now define certain ratios involving the sides of a right triangle, and call them trigonometric ratios.

The trigonometric ratios of the angle $A$ in right triangle $A B C$ (see Fig. 8.4) are defined as follows :

$$ \begin{aligned} \text{sine of }\angle A=\dfrac{\text{ side opposite to angle } A}{\text{ hypotenuse }}=\dfrac{BC}{AC}\\ \text{cosine of } \angle A=\dfrac{\text{ side adjacent to angle } A}{\text{ hypotenuse }}=\dfrac{AB}{AC}\end{aligned} $$

$$\text{tangent of }\angle A=\dfrac{\text{ side opposite to angle } A}{\text{ side adjacent to angle } A}=\dfrac{BC}{AB}$$

$$\text{cosecant of }\angle A=\dfrac{1}{\text{ sine of } \angle A}=\dfrac{\text{ hypotenuse }}{\text{ side opposite to angle } A}=\dfrac{AC}{BC}$$

$$\text{secant of }\angle A=\dfrac{1}{\text{ cosine of } \angle A}=\dfrac{\text{ hypotenuse }}{\text{ side adjacent to angle } A}=\dfrac{AC}{AB}$$

$$\text{ cotangent of }\angle A=\dfrac{1}{\text{ tangent of } \angle A}=\dfrac{\text{ side adjacent to angle } A}{\text{ side opposite to angle } A}=\dfrac{AB}{BC}$$

The ratios defined above are abbreviated as $\sin A, \cos A, \tan A, cosec A, \sec A$ and $\cot$ A respectively. Note that the ratios $cosec \mathbf{A}, \sec \mathbf{A}$ and $\cot \mathbf{A}$ are respectively, the reciprocals of the ratios $\sin A, \cos A$ and $\tan A$.

Also, observe that $\tan A=\dfrac{B C}{A B}=\dfrac{\dfrac{B C}{A C}}{\dfrac{A B}{A C}}=\dfrac{\sin A}{\cos A}$ and $\cot A=\dfrac{\cos A}{\sin A}$.

So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.

Why don’t you try to define the trigonometric ratios for angle C in the right triangle? (See Fig. 8.5)

The first use of the idea of ‘sine’ in the way we use it today was in the work Aryabhatiyam by Aryabhata, in A.D. 500. Aryabhata used the word ardha-jya for the half-chord, which was shortened to jya or jiva in due course. When the Aryabhatiyam was translated into Arabic, the word jiva was retained as it is. The word jiva was translated into sinus, which means curve, when the Arabic version was translated into Latin. Soon the word sinus, also used as sine, became common in mathematical texts throughout Europe. An English Professor of astronomy Edmund Gunter (1581-1626), first used the abbreviated notation ‘sin’.

Aryabhata C.E. $476-550$

The origin of the terms ‘cosine’ and ’tangent’ was much later. The cosine function arose from the need to compute the sine of the complementary angle. Aryabhatta called it kotijya. The name cosinus originated with Edmund Gunter. In 1674, the English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’.

Remark : Note that the symbol sin A is used as an abbreviation for ’the sine of the angle A’. sin A is not the product of ‘sin’ and A. ‘sin’ separated from A has no meaning. Similarly, cos A is not the product of ‘cos’ and A. Similar interpretations follow for other trigonometric ratios also.

Now, if we take a point $P$ on the hypotenuse $AC$ or a point $Q$ on $AC$ extended, of the right triangle $ABC$ and draw $PM$ perpendicular to $AB$ and $QN$ perpendicular to $A B$ extended (see Fig. 8.6), how will the trigonometric ratios of $\angle A$ in $\triangle PAM$ differ from those of $\angle A$ in $\triangle CAB$ or from those of $\angle A$ in $\triangle$ QAN?

Fig. 8.6

To answer this, first look at these triangles. Is $\triangle$ PAM similar to $\triangle CAB$ ? From Chapter 6, recall the AA similarity criterion. Using the criterion, you will see that the triangles PAM and CAB are similar. Therefore, by the property of similar triangles, the corresponding sides of the triangles are proportional.

So, we have $\dfrac{AM}{AB}=\dfrac{AP}{AC}=\dfrac{MP}{BC}$

From this, we find

$ \begin{aligned} \dfrac{MP}{AP} & =\dfrac{BC}{AC}=\sin A . \end{aligned} $

Similarly, $ \begin{aligned} \dfrac{AM}{AP}=\dfrac{AB}{AC} & =\cos A, \dfrac{MP}{AM}=\dfrac{BC}{AB}=\tan A \text{ and so on. } \end{aligned} $

This shows that the trigonometric ratios of angle $A$ in $\triangle$ PAM not differ from those of angle $A$ in $\triangle CAB$.

In the same way, you should check that the value of sin A (and also of other trigonometric ratios) remains the same in $\triangle QAN$ also.

From our observations, it is now clear that 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.

Note : For the sake of convenience, we may write $\sin ^{2} A, \cos ^{2} A$, etc., in place of $(\sin A)^{2},(\cos A)^{2}$, etc., respectively. But $cosec A=(\sin A)^{-1} \neq \sin ^{-1} A$ (it is called sine inverse A). $\sin ^{-1} A$ has a different meaning, which will be discussed in higher classes. Similar conventions hold for the other trigonometric ratios as well. Sometimes, the Greek letter $\theta$ (theta) is also used to denote an angle.

We have defined six trigonometric ratios of an acute angle. If we know any one of the ratios, can we obtain the other ratios? Let us see.

If in a right triangle $ABC, \sin A=\dfrac{1}{3}$, then this means that $\dfrac{B C}{A C}=\dfrac{1}{3}$, i.e., the lengths of the sides $BC$ and $AC$ of the triangle $A B C$ are in the ratio $1: 3$ (see Fig. 8.7). So if $BC$ is equal to $k$, then $AC$ will be $3 k$, where $k$ is any positive number. To determine other trigonometric ratios for the angle $A$, we need to find the length of the third side $A B$. Do you remember the Pythagoras theorem? Let us use it to determine the required length $AB$.

Fig. 8.7

$ AB^{2}=AC^{2}-BC^{2}=(3 k)^{2}-(k)^{2}=8 k^{2}=(2 \sqrt{2} k)^{2} $

Therefore,$\quad\quad\quad AB= \pm 2 \sqrt{2} k$

So, we get $ AB=2 \sqrt{2} k \quad(\text{ Why is } AB \text{ not }-2 \sqrt{2} k ?) $

Now, $ \quad\quad\quad \quad \cos A=\dfrac{AB}{AC}=\dfrac{2 \sqrt{2} k}{3 k}=\dfrac{2 \sqrt{2}}{3} $

Similarly, you can obtain the other trigonometric ratios of the angle A.

Remark : Since the hypotenuse is the longest side in a right triangle, the value of $\sin A$ or $\cos A$ is always less than 1 (or, in particular, equal to 1 ).

Let us consider some examples.

Example 1 : Given $\tan A=\dfrac{4}{3}$, find the other trigonometric ratios of the angle $A$.

Solution : Let us first draw a right $\triangle ABC$ (see Fig 8.8).

Fig. 8.8

Now, we know that $\tan A=\dfrac{BC}{AB}=\dfrac{4}{3}$.

Therefore, if $BC=4 k$, then $AB=3 k$, where $k$ is a positive number.

Now, by using the Pythagoras Theorem, we have

$ AC^{2}=AB^{2}+BC^{2}=(4 k)^{2}+(3 k)^{2}=25 k^{2} $

So, $AC=5 k$

Now, we can write all the trigonometric ratios using their definitions.

$ \begin{matrix} \sin A=\dfrac{BC}{AC}=\dfrac{4 k}{5 k}=\dfrac{4}{5} \\ \cos A=\dfrac{AB}{AC}=\dfrac{3 k}{5 k}=\dfrac{3}{5} \end{matrix} $

Therefore, $\cot A=\dfrac{1}{\tan A}=\dfrac{3}{4}, cosec A=\dfrac{1}{\sin A}=\dfrac{5}{4}$ and $\sec A=\dfrac{1}{\cos A}=\dfrac{5}{3}$.

Example 2 : If $\angle B$ and $\angle Q$ are acute angles such that $\sin B=\sin Q$, then prove that $\angle B=\angle Q$.

Solution : Let us consider two right triangles $ABC$ and $PQR$ where $\sin B=\sin Q$ (see Fig. 8.9).

Fig. 8.9

We have $\sin B=\dfrac{A C}{A B}$

$ \text{ and } \quad \quad \quad \quad \quad \sin Q=\dfrac{P R}{P Q} $

Then $\quad \quad \quad \dfrac{AC}{AB}=\dfrac{PR}{PQ}$

Therefore, $$ \dfrac{AC}{PR}=\dfrac{AB}{PQ}=k \text{, say } \tag{1} $$

Now, using Pythagoras theorem,

$ \begin{aligned} BC & =\sqrt{AB^{2}-AC^{2}} \end{aligned} $

and $ \begin{aligned} QR & =\sqrt{PQ^{2}-PR^{2}} \end{aligned} $

$\text{So,} \quad \dfrac{BC}{QR}=\dfrac{\sqrt{AB^{2}-AC^{2}}}{\sqrt{PQ^{2}-PR^{2}}}=\dfrac{\sqrt{k^{2} PQ^{2}-k^{2} PR^{2}}}{\sqrt{PQ^{2}-PR^{2}}}=\dfrac{k \sqrt{PQ^{2}-PR^{2}}}{\sqrt{PQ^{2}-PR^{2}}}=k \tag{2}$

From (1) and (2), we have

$$ \dfrac{AC}{PR}=\dfrac{AB}{PQ}=\dfrac{BC}{QR} $$

Then, by using Theorem 6.4, $\Delta ACB \sim \Delta PRQ$ and therefore, $\angle B=\angle Q$.

Example 3 : Consider $\triangle ACB$, right-angled at $C$, in which $AB=29$ units, $BC=21$ units and $\angle ABC=\theta$ (see Fig. 8.10). Determine the values of

Fig. 8.10

(i) $\cos ^{2} \theta+\sin ^{2} \theta$,

(ii) $\cos ^{2} \theta-\sin ^{2} \theta$.

Solution : In $\triangle ACB$, we have

$ \begin{aligned} & A C=\sqrt{A B^{2}-B C^{2}}=\sqrt{(29)^{2}-(21)^{2}} \\ & =\sqrt{(29-21)(29+21)}=\sqrt{(8)(50)}=\sqrt{400}=20 \text{ units } \end{aligned} $

So, $\quad \sin \theta=\dfrac{A C}{A B}=\dfrac{20}{29}, \cos \theta=\dfrac{B C}{A B}=\dfrac{21}{29}$.

Now, (i) $\cos ^{2} \theta+\sin ^{2} \theta=(\dfrac{20}{29})^{2}+(\dfrac{21}{29})^{2}=\dfrac{20^{2}+21^{2}}{29^{2}}=\dfrac{400+441}{841}=1$,

and (ii) $\cos ^{2} \theta-\sin ^{2} \theta=(\dfrac{21}{29})^{2}-(\dfrac{20}{29})^{2}=\dfrac{(21+20)(21-20)}{29^{2}}=\dfrac{41}{841}$.

Example 4 : In a right triangle $A B C$, right-angled at $B$, if $\tan A=1$, then verify that

$2 \sin A \cos A=1$.

Solution : In $\triangle ABC, \tan A=\dfrac{BC}{AB}=1 \quad$ (see Fig 8.11)

Fig. 8.11

i.e., $ B C=A B $

Let $AB=BC=k$, where $k$ is a positive number.

Now, $$ \begin{aligned} AC & =\sqrt{AB^{2}+BC^{2}} \\ & =\sqrt{(k)^{2}+(k)^{2}}=k \sqrt{2} \end{aligned} $$

Therefore, $ \sin A=\dfrac{B C}{A C}=\dfrac{1}{\sqrt{2}} \text{ and } \cos A=\dfrac{A B}{A C}=\dfrac{1}{\sqrt{2}} $

So, $\quad 2 \sin A \cos A=2(\dfrac{1}{\sqrt{2}})(\dfrac{1}{\sqrt{2}})=1$, which is the required value.

Example 5 : In $\Delta OPQ$, right-angled at $P$, $OP=7 cm$ and $OQ-PQ=1 cm$ (see Fig. 8.12). Determine the values of $\sin Q$ and $\cos Q$.

Fig. 8.12

Solution : In $\triangle OPQ$, we have

$ OQ^{2}=OP^{2}+PQ^{2} $

i.e., $ (1+P Q)^{2}=OP^{2}+PQ^{2} $(why?)

i.e., $ \quad 1+PQ^{2}+2 PQ=OP^{2}+PQ^{2} $

i.e., $ \quad 1+2 P Q=7^{2} \quad \quad \quad \quad \text{(Why?)} $

i.e., $ PQ=24 ~cm \text{ and } OQ=1+PQ=25 ~cm $

So, $\quad \sin Q=\dfrac{7}{25}$ and $\cos Q=\dfrac{24}{25}$.

EXERCISE 8.1

1. In $\triangle ABC$, right-angled at $B, AB=24 cm, BC=7 cm$. Determine :

(i) $\sin A, \cos A$

(ii) $\sin C, \cos C$

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Solution

Applying Pythagoras theorem for $\triangle A B C$, we obtain

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

$=(24 cm)^{2}+(7 cm)^{2}$

$=(576+49) cm^{2}$

$=625 cm^{2}$

$\therefore AC=\sqrt{625} cm=25 cm$

(i) $\sin A=\dfrac{\text{ Side opposite to } \angle A}{\text{ Hypotenuse }}=\dfrac{BC}{AC}$

$=\dfrac{7}{25}$

$\cos A=$

$\dfrac{\text{ Side adjacent to } \angle A}{\text{ Hypotenuse }}=\dfrac{AB}{AC}=\dfrac{24}{25}$

(ii)

$ \begin{aligned} & \sin C=\dfrac{\text{ Side opposite to } \angle C}{\text{ Hypotenuse }}=\dfrac{AB}{AC} \\ & =\dfrac{24}{25} \end{aligned} $

$ \begin{aligned} & \dfrac{\text{ Side adjacent to } \angle C}{\text{ Hypotenuse }}=\dfrac{BC}{AC} \\ & \cos C= \\ & =\dfrac{7}{25} \end{aligned} $

2. In Fig. 8.13, find tan $P-\cot R$.

Fig. 8.13

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Solution

Applying Pythagoras theorem for $\triangle PQR$, we obtain

$PR^{2}=PQ^{2}+QR^{2}$

$(13 cm)^{2}=(12 cm)^{2}+QR^{2}$

$169 cm^{2}=144 cm^{2}+QR^{2}$

$25 cm^{2}=QR^{2}$

$QR=5 cm$

$ \begin{aligned} \tan P & =\dfrac{\text{ Side opposite to } \angle P}{\text{ Side adjacent to } \angle P}=\dfrac{QR}{PQ} \\ & =\dfrac{5}{12} \end{aligned} $

$ \begin{aligned} \cot R & =\dfrac{\text{ Side adjacent to } \angle R}{\text{ Side opposite to } \angle R}=\dfrac{QR}{PQ} \\ & =\dfrac{5}{12} \end{aligned} $

$\tan P-\cot R=\dfrac{5}{12}-\dfrac{5}{12}=0$

3. If $\sin A=\dfrac{3}{4}$, calculate $\cos A$ and $\tan A$.

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Solution

Let $\triangle A B C$ be a right-angled triangle, right-angled at point $B$.

Given that,

$ \begin{aligned} & \sin A=\dfrac{3}{4} \\ & \dfrac{BC}{AC}=\dfrac{3}{4} \end{aligned} $

Let $BC$ be $3 k$. Therefore, $AC$ will be $4 k$, where $k$ is a positive integer.

Applying Pythagoras theorem in $\triangle A B C$, we obtain

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

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

$16 k^{2}-9 k^{2}=AB^{2}$

$7 k^{2}=A B^{2}$

$AB=\sqrt{7} k$

$\cos A=\dfrac{\text{ Side adjacent to } \angle A}{\text{ Hypotenuse }}$

$=\dfrac{AB}{AC}=\dfrac{\sqrt{7 k}}{4 k}=\dfrac{\sqrt{7}}{4}$

$\tan A=\dfrac{\text{ Side opposite to } \angle A}{\text{ Side adjacent to } \angle A}$

$=\dfrac{BC}{AB}=\dfrac{3 k}{\sqrt{7} k}=\dfrac{3}{\sqrt{7}}$

4. Given $15 \cot A=8$, find $\sin A$ and $\sec A$.

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Solution

Consider a right-angled triangle, right-angled at $B$.

$\cot A=\dfrac{\text{ Side adjacent to } \angle A}{\text{ Side opposite to } \angle A}$

$=\dfrac{AB}{BC}$

It is given that,

$ \begin{aligned} \cot A & =\dfrac{8}{15} \\ \dfrac{AB}{BC} & =\dfrac{8}{15} \end{aligned} $

Let $A B$ be $8 k$.Therefore, $B C$ will be $15 k$, where $k$ is a positive integer.

Applying Pythagoras theorem in $\triangle A B C$, we obtain

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

$=(8 k)^{2}+(15 k)^{2}$

$=64 k^{2}+225 k^{2}$

$=289 k^{2}$

$AC=17 k$

$ \begin{aligned} \sin A & =\dfrac{\text{ Side opposite to } \angle A}{\text{ Hypotenuse }}=\dfrac{BC}{AC} \\ & =\dfrac{15 k}{17 k}=\dfrac{15}{17} \end{aligned} $

$\sec A=\dfrac{\text{ Hypotenuse }}{\text{ Side adjacent to } \angle A}$

$ =\dfrac{AC}{AB}=\dfrac{17}{8} $

5. Given $\sec \theta=\dfrac{13}{12}$, calculate all other trigonometric ratios.

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Solution

Consider a right-angle triangle $\triangle A B C$, right-angled at point $B$.

$\sec \theta=\dfrac{\text{ Hypotenuse }}{\text{ Side adjacent to } \angle \theta}$

$\dfrac{13}{12}=\dfrac{AC}{AB}$

If $AC$ is $13 k, AB$ will be $12 k$, where $k$ is a positive integer.

Applying Pythagoras theorem in $\triangle ABC$, we obtain

$(A C)^{2}=(A B)^{2}+(B C)^{2}$

$(13 k)^{2}=(12 k)^{2}+(B C)^{2}$

$169 k^{2}=144 k^{2}+BC^{2}$

$25 k^{2}=BC^{2}$

$BC=5 k$

$\sin \theta=\dfrac{\text{ Side opposite to } \angle \theta}{\text{ Hypotenuse }}=\dfrac{B C}{A C}=\dfrac{5 k}{13 k}=\dfrac{5}{13}$

$\cos \theta=\dfrac{\text{ Side adjacent to } \angle \theta}{\text{ Hypotenuse }}=\dfrac{AB}{AC}=\dfrac{12 k}{13 k}=\dfrac{12}{13}$

$\tan \theta=\dfrac{\text{ Side opposite to } \angle \theta}{\text{ Side adjacent to } \angle \theta}=\dfrac{BC}{AB}=\dfrac{5 k}{12 k}=\dfrac{5}{12}$

$\cot \theta=\dfrac{\text{ Side adjacent to } \angle \theta}{\text{ Side opposite to } \angle \theta}=\dfrac{AB}{BC}=\dfrac{12 k}{5 k}=\dfrac{12}{5}$

$cosec \theta=\dfrac{\text{ Hypotenuse }}{\text{ Side opposite to } \angle \theta}=\dfrac{AC}{BC}=\dfrac{13 k}{5 k}=\dfrac{13}{5}$

6. If $\angle A$ and $\angle B$ are acute angles such that $\cos A=\cos B$, then show that $\angle A=\angle B$.

Show Answer

Solution

Let us consider a triangle $A B C$ in which $C D \perp A B$.

It is given that

$\cos A=\cos B$ $\Rightarrow \dfrac{AD}{AC}=\dfrac{BD}{BC}$

We have to prove $\angle A=\angle B$. To prove this, let us extend $A C$ to $P$ such that $B C=C P$.

From equation (1), we obtain

$\dfrac{AD}{BD}=\dfrac{AC}{BC}$

$\Rightarrow \dfrac{AD}{BD}=\dfrac{AC}{CP}$

(By construction, we have $BC=CP$ )

By using the converse of B.P.T,

$CD \mid BP$

$\Rightarrow \angle ACD=\angle CPB$ (Corresponding angles) ..

And, $\angle BCD=\angle CBP$ (Alternate interior angles)

By construction, we have $BC=CP$.

$\therefore \angle CBP=\angle CPB$ (Angle opposite to equal sides of a triangle) $\ldots$ (5)

From equations (3), (4), and (5), we obtain

$\angle ACD=\angle BCD$.

In $\triangle CAD$ and $\triangle CBD$,

$\angle ACD=\angle BCD$ [Using equation (6)]

$\angle CDA=\angle CDB[.$ Both $90^{\circ}$ ]

Therefore, the remaining angles should be equal.

$\therefore \angle CAD=\angle CBD$

$\Rightarrow \angle A=\angle B$

Alternatively,

Let us consider a triangle $A B C$ in which $C D \perp A B$.

It is given that,

$\cos A=\cos B$

$\Rightarrow \dfrac{AD}{AC}=\dfrac{BD}{BC}$

$\Rightarrow \dfrac{AD}{BD}=\dfrac{AC}{BC}$

Let $\dfrac{AD}{BD}=\dfrac{AC}{BC}=k$

$\Rightarrow AD=k BD$.

And, $AC=k BC$

Using Pythagoras theorem for triangles CAD and CBD, we obtain

$C D^{2}=A C^{2}-A D^{2}$..

And, $CD^{2}=BC^{2}-BD^{2}$.

From equations (3) and (4), we obtain

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

$\Rightarrow(k B C)^{2}-(k B D)^{2}=BC^{2}-BD^{2}$

$\Rightarrow k^{2}(BC^{2}-BD^{2})=BC^{2}-BD^{2}$

$\Rightarrow k^{2}=1$

$\Rightarrow k=1$

Putting this value in equation (2), we obtain

$AC=BC$

$\Rightarrow \angle A=\angle B$ (Angles opposite to equal sides of a triangle)

7. If $\cot \theta=\dfrac{7}{8}$, evaluate :

(i) $\dfrac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$,

(ii) $\cot ^{2} \theta$

Show Answer

Solution

Let us consider a right triangle $A B C$, right-angled at point $B$.

$ \begin{aligned} \cot \theta & =\dfrac{\text{ Side adjacent to } \angle \theta}{\text{ Side opposite to } \angle \theta}=\dfrac{BC}{AB} \\ & =\dfrac{7}{8} \end{aligned} $

If $BC$ is $7 k$, then $AB$ will be $8 k$, where $k$ is a positive integer.

Applying Pythagoras theorem in $\triangle A B C$, we obtain

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

$=(8 k)^{2}+(7 k)^{2}$

$=64 k^{2}+49 k^{2}$

$=113 k^{2}$

$AC=\sqrt{113} k$

$\sin \theta=\dfrac{\text{ Side opposite to } \angle \theta}{\text{ Hypotenuse }}=\dfrac{AB}{AC}$

$=\dfrac{8 k}{\sqrt{113} k}=\dfrac{8}{\sqrt{113}}$

$\cos \theta=\dfrac{\text{ Side adjacent to } \angle \theta}{\text{ Hypotenuse }}=\dfrac{BC}{AC}$

$=\dfrac{7 k}{\sqrt{113} k}=\dfrac{7}{\sqrt{113}}$

(i) $\dfrac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\dfrac{(1-\sin ^{2} \theta)}{(1-\cos ^{2} \theta)}$ $=\dfrac{1-(\dfrac{8}{\sqrt{113}})^{2}}{1-(\dfrac{7}{\sqrt{113}})^{2}}=\dfrac{1-\dfrac{64}{113}}{1-\dfrac{49}{113}}$

$=\dfrac{\dfrac{49}{113}}{64}=\dfrac{49}{64}$

113

(ii) $\cot ^{2} \theta=(\cot \theta)^{2}=(\dfrac{7}{8})^{2}=\dfrac{49}{64}$

8. If $3 \cot A=4$, check whether $\dfrac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A$ or not.

Show Answer

Solution

It is given that $3 \cot A=4$

Or, $\cot A=\dfrac{4}{3}$

Consider a right triangle $A B C$, right-angled at point $B$.

$\cot A=\dfrac{\text{ Side adjacent to } \angle A}{\text{ Side opposite to } \angle A}$

$\dfrac{AB}{BC}=\dfrac{4}{3}$

If $AB$ is $4 k$, then $BC$ will be $3 k$, where $k$ is a positive integer.

In $\triangle ABC$,

$(A C)^{2}=(A B)^{2}+(B C)^{2}$

$=(4 k)^{2}+(3 k)^{2}$

$ \begin{aligned} & =16 k^{2}+9 k^{2} \\ & =25 k^{2} \\ & A C=5 k \\ & \cos A=\dfrac{\text{ Side adjacent to } \angle A}{\text{ Hypotenuse }}=\dfrac{AB}{AC} \\ & =\dfrac{4 k}{5 k}=\dfrac{4}{5} \\ & \sin A=\dfrac{\text{ Side opposite to } \angle A}{\text{ Hypotenuse }}=\dfrac{BC}{AC} \\ & =\dfrac{3 k}{5 k}=\dfrac{3}{5} \\ & \tan A=\dfrac{\text{ Side opposite to } \angle A}{\text{ Hypotenuse }}=\dfrac{BC}{AB} \\ & =\dfrac{3 k}{4 k}=\dfrac{3}{4} \\ & \dfrac{1-\tan ^{2} A}{1+\tan ^{2} A}=\dfrac{1-(\dfrac{3}{4})^{2}}{1+(\dfrac{3}{4})^{2}}=\dfrac{1-\dfrac{9}{16}}{1+\dfrac{9}{16}} \\ & =\dfrac{\dfrac{7}{16}}{\dfrac{25}{16}}=\dfrac{7}{25} \\ & \cos ^{2} A-\sin ^{2} A=(\dfrac{4}{5})^{2}-(\dfrac{3}{5})^{2} \\ & =\dfrac{16}{25}-\dfrac{9}{25}=\dfrac{7}{25} \\ & \therefore \dfrac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A \end{aligned} $

9. In triangle $A B C$, right-angled at $B$, if $\tan A=\dfrac{1}{\sqrt{3}}$, find the value of:

(i) $\sin A \cos C+\cos A \sin C$

(ii) $\cos A \cos C-\sin A \sin C$

Show Answer

Solution

$\tan A=\dfrac{1}{\sqrt{3}}$

$\dfrac{BC}{AB}=\dfrac{1}{\sqrt{3}}$

If $BC$ is $k$, then $AB$ will be $\sqrt{3} k$, where $k$ is a positive integer.

In $\triangle ABC$,

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

$=(\sqrt{3} k)^{2}+(k)^{2}$

$=3 k^{2}+k^{2}=4 k^{2}$

$\therefore AC=2 k$

$\sin A=\dfrac{\text{ Side opposite to } \angle A}{\text{ Hypotenuse }}=\dfrac{BC}{AC}=\dfrac{k}{2 k}=\dfrac{1}{2}$

$\cos A=\dfrac{\text{ Side adjacent to } \angle A}{\text{ Hypotenuse }}=\dfrac{AB}{AC}=\dfrac{\sqrt{3} k}{2 k}=\dfrac{\sqrt{3}}{2}$

$\sin C=\dfrac{\text{ Side opposite to } \angle C}{\text{ Hypotenuse }}=\dfrac{AB}{AC}=\dfrac{\sqrt{3} k}{2 k}=\dfrac{\sqrt{3}}{2}$

$\cos C=\dfrac{\text{ Side adjacent to } \angle C}{\text{ Hypotenuse }}=\dfrac{BC}{AC}=\dfrac{k}{2 k}=\dfrac{1}{2}$

(i) $\sin A \cos C+\cos A \sin C$

$ \begin{aligned} & =(\dfrac{1}{2})(\dfrac{1}{2})+(\dfrac{\sqrt{3}}{2})(\dfrac{\sqrt{3}}{2})=\dfrac{1}{4}+\dfrac{3}{4} \\ & =\dfrac{4}{4}=1 \end{aligned} $

(ii) $\cos A \cos C-\sin A \sin C$ $=(\dfrac{\sqrt{3}}{2})(\dfrac{1}{2})-(\dfrac{1}{2})(\dfrac{\sqrt{3}}{2})=\dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4}=0$

10. In $\Delta PQR$, right- angled at $Q, PR+QR=25 cm$ and $PQ=5 cm$. Determine the values of sin $P, \cos P$ and $\tan P$.

Show Answer

Solution

Given that, $PR+QR=25$

$PQ=5$

Let PR be $x$.

Therefore, $QR=25-x$

Applying Pythagoras theorem in $\triangle PQR$, we obtain

$PR^{2}=PQ^{2}+QR^{2}$

$x^{2}=(5)^{2}+(25-x)^{2}$

$x^{2}=25+625+x^{2}-50 x$

$50 x=650$

$x=13$

Therefore, $PR=13 cm$

$QR=(25-13) cm=12 cm$

$\sin P=\dfrac{\text{ Side opposite to } \angle P}{\text{ Hypotenuse }}=\dfrac{QR}{PR}=\dfrac{12}{13}$

$\cos P=\dfrac{\text{ Side adjacent to } \angle P}{\text{ Hypotenuse }}=\dfrac{PQ}{PR}=\dfrac{5}{13}$

$\tan P=\dfrac{\text{ Side opposite to } \angle P}{\text{ Side adjacent to } \angle P}=\dfrac{QR}{PQ}=\dfrac{12}{5}$

11. State whether the following are true or false. Justify your answer.

(i) The value of $\tan A$ is always less than 1.

(ii) sec $A=\dfrac{12}{5}$ for some value of angle $A$.

(iii) $\cos A$ is the abbreviation used for the cosecant of angle $A$.

(iv) $\cot A$ is the product of cot and $A$.

(v) $\sin \theta=\dfrac{4}{3}$ for some angle $\theta$.

Show Answer

Solution

(i) Consider a $\triangle A B C$, right-angled at $B$.

$\tan A=\dfrac{\text{ Side opposite to } \angle A}{\text{ Side adjacent to } \angle A}$

$=\dfrac{12}{5}$

But $\dfrac{12}{5}>1$

$\therefore \tan A>1$

So, $\tan A<1$ is not always true.

Hence, the given statement is false.

(ii)

$ \sec A=\dfrac{12}{5} $

$\dfrac{\text{ Hypotenuse }}{\text{ Side adjacent to } \angle A}=\dfrac{12}{5}$

$\dfrac{AC}{AB}=\dfrac{12}{5}$

Let $A C$ be $12 k, A B$ will be $5 k$, where $k$ is a positive integer.

Applying Pythagoras theorem in $\triangle ABC$, we obtain

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

$(12 k)^{2}=(5 k)^{2}+BC^{2}$

$144 k^{2}=25 k^{2}+BC^{2}$

$BC^{2}=119 k^{2}$

$BC=10.9 k$

It can be observed that for given two sides $AC=12 k$ and $AB=5 k$,

$B C$ should be such that,

$A C-A B<B C<A C+A B$

$12 k-5 k<BC<12 k+5 k$

$7 k<BC<17 k$

However, $BC=10.9 k$. Clearly, such a triangle is possible and hence, such value of sec $A$ is possible.

Hence, the given statement is true.

(iii) Abbreviation used for cosecant of angle $A$ is $cosec A$. And $\cos A$ is the abbreviation used for cosine of angle $A$.

Hence, the given statement is false.

(iv) $\cot A$ is not the product of $\cot$ and $A$. It is the cotangent of $\angle A$.

Hence, the given statement is false.

(v) $\sin \theta=\dfrac{4}{3}$

We know that in a right-angled triangle,

$ \sin \theta=\dfrac{\text{ Side opposite to } \angle \theta}{\text{ Hypotenuse }} $

In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of $\sin \theta$ is not possible.

Hence, the given statement is false

8.3 Trigonometric Ratios of Some Specific Angles

From geometry, you are already familiar with the construction of angles of $30^{\circ}, 45^{\circ}$, $60^{\circ}$ and $90^{\circ}$. In this section, we will find the values of the trigonometric ratios for these angles and, of course, for $0^{\circ}$.

Trigonometric Ratios of $45^{\circ}$

In $\triangle ABC$, right-angled at $B$, if one angle is $45^{\circ}$, then the other angle is also $45^{\circ}$, i.e., $\angle A=\angle C=45^{\circ}$ (see Fig. 8.14).

Fig. 8.14

So,

$ BC=AB \quad(\text{ Why? }) $

Now, Suppose $BC=AB=a$.

Then by Pythagoras Theorem, $AC^{2}=AB^{2}+BC^{2}=a^{2}+a^{2}=2 a^{2}$,

and, therefore, $\quad AC=a \sqrt{2}$.

Using the definitions of the trigonometric ratios, we have :

$ \begin{aligned} & \sin 45^{\circ}=\dfrac{\text{ side opposite to angle } 45^{\circ}}{\text{ hypotenuse }}=\dfrac{BC}{AC}=\dfrac{a}{a \sqrt{2}}=\dfrac{1}{\sqrt{2}} \\ & \cos 45^{\circ}=\dfrac{\text{ side adjacent to angle } 45^{\circ}}{\text{ hypotenuse }}=\dfrac{AB}{AC}=\dfrac{a}{a \sqrt{2}}=\dfrac{1}{\sqrt{2}} \\ & \tan 45^{\circ}=\dfrac{\text{ side opposite to angle } 45^{\circ}}{\text{ side adjacent to angle } 45^{\circ}}=\dfrac{BC}{AB}=\dfrac{a}{a}=1 \end{aligned} $

Also, $cosec 45^{\circ}=\dfrac{1}{\sin 45^{\circ}}=\sqrt{2}, \sec 45^{\circ}=\dfrac{1}{\cos 45^{\circ}}=\sqrt{2}, \cot 45^{\circ}=\dfrac{1}{\tan 45^{\circ}}=1$.

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

Let us now calculate the trigonometric ratios of $30^{\circ}$ and $60^{\circ}$. Consider an equilateral triangle $ABC$. Since each angle in an equilateral triangle is $60^{\circ}$, therefore, $\angle A=\angle B=\angle C=60^{\circ}$.

Draw the perpendicular $AD$ from $A$ to the side $BC$ (see Fig. 8.15).

Fig. 8.15

$ \begin{aligned} \text{Now } \quad \quad \quad & \Delta ABD \cong \Delta ACD \quad(\text{ Why? }) \\ \text{Therefore, } \quad \quad \quad & BD =DC \\ \text{ and } \quad \quad \quad & \angle BAD =\angle CAD \quad(CPCT) \end{aligned} $

Now observe that:

$\triangle ABD$ is a right triangle, right- angled at $D$ with $\angle BAD=30^{\circ}$ and $\angle ABD=60^{\circ}$ (see Fig. 8.15).

As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the triangle. So, let us suppose that $AB=2 a$.

Then, $ BD=\dfrac{1}{2} BC=a $

and $ AD^{2}=AB^{2}-BD^{2}=(2 a)^{2}-(a)^{2}=3 a^{2}, $

Therefore, $ AD=a \sqrt{3} $

Now, we have : $ \begin{aligned} & \sin 30^{\circ}=\dfrac{BD}{AB}=\dfrac{a}{2 a}=\dfrac{1}{2}, \cos 30^{\circ}=\dfrac{AD}{AB}=\dfrac{a \sqrt{3}}{2 a}=\dfrac{\sqrt{3}}{2} \\ & \tan 30^{\circ}=\dfrac{BD}{AD}=\dfrac{a}{a \sqrt{3}}=\dfrac{1}{\sqrt{3}} . \end{aligned} $

Also, $\quad cosec 30^{\circ}=\dfrac{1}{\sin 30^{\circ}}=2, \quad \sec 30^{\circ}=\dfrac{1}{\cos 30^{\circ}}=\dfrac{2}{\sqrt{3}}$

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

Similarly, $ \begin{aligned} \sin 60^{\circ} & =\dfrac{AD}{AB}=\dfrac{a \sqrt{3}}{2 a}=\dfrac{\sqrt{3}}{2}, \cos 60^{\circ}=\dfrac{1}{2}, \tan 60^{\circ}=\sqrt{3}, \\ \text{cosec} 60^{\circ} & =\dfrac{2}{\sqrt{3}}, \sec 60^{\circ}=2 \text{ and } \cot 60^{\circ}=\dfrac{1}{\sqrt{3}} . \end{aligned} $

Trigonometric Ratios of $0^{\circ}$ and $\mathbf{9 0}$

Let us see what happens to the trigonometric ratios of angle $A$, if it is made smaller and smaller in the right triangle $ABC$ (see Fig. 8.16), till it becomes zero. As $\angle$ A gets smaller and smaller, the length of the side $BC$ decreases. The point $C$ gets closer to point $B$, and finally when $\angle A$ becomes very close to $0^{\circ}, AC$ becomes almost the same as $AB$ (see Fig. 8.17).

Fig. 8.16

Fig. 8.17

When $\angle A$ is very close to $0^{\circ}, BC$ gets very close to 0 and so the value of $\sin A=\dfrac{BC}{AC}$ is very close to 0 . Also, when $\angle A$ is very close to $0^{\circ}, AC$ is nearly the same as $A B$ and so the value of $\cos A=\dfrac{A B}{A C}$ is very close to 1 .

This helps us to see how we can define the values of sin A and $\cos A$ when $A=0^{\circ}$. We define : $\boldsymbol{{}\operatorname { s i n }} \mathbf{0}^{\circ}=\mathbf{0}$ and $\cos \mathbf{0}^{\circ}=\mathbf{1}$.

Using these, we have :

$\tan 0^{\circ}=\dfrac{\sin 0^{\circ}}{\cos 0^{\circ}}=0, \cot 0^{\circ}=\dfrac{1}{\tan 0^{\circ}}$, which is not defined. (Why?)

sec $0^{\circ}=\dfrac{1}{\cos 0^{\circ}}=1$ and $cosec 0^{\circ}=\dfrac{1}{\sin 0^{\circ}}$, which is again not defined.(Why?)

Now, let us see what happens to the trigonometric ratios of $\angle A$, when it is made larger and larger in $\triangle ABC$ till it becomes $90^{\circ}$. As $\angle A$ gets larger and larger, $\angle C$ gets smaller and smaller. Therefore, as in the case above, the length of the side $A B$ goes on decreasing. The point $A$ gets closer to point $B$. Finally when $\angle A$ is very close to $90^{\circ}$, $\angle C$ becomes very close to $0^{\circ}$ and the side $AC$ almost coincides with side $BC$ (see Fig. 8.18).

Fig. 8.18

When $\angle C$ is very close to $0^{\circ}, \angle A$ is very close to $90^{\circ}$, side $AC$ is nearly the same as side $BC$, and so sin $A$ is very close to 1 . Also when $\angle A$ is very close to $90^{\circ}$, $\angle C$ is very close to $0^{\circ}$, and the side $AB$ is nearly zero, so cos $A$ is very close to 0 .

So, we define :

$ \sin 90^{\circ}=1 \text{ and } \cos 90^{\circ}=0 . $

Now, why don’t you find the other trigonometric ratios of $90^{\circ}$ ?

We shall now give the values of all the trigonometric ratios of $0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}$ and $90^{\circ}$ in Table 8.1, for ready reference.

Table 8.1

Remark : From the table above you can observe that as $\angle A$ increases from $0^{\circ}$ to $90^{\circ}$, sin A increases from 0 to 1 and cos A decreases from 1 to 0 .

Let us illustrate the use of the values in the table above through some examples.

Example 6 : In $\triangle ABC$, right-angled at $B$, $AB=5 ~cm$ and $\angle ACB=30^{\circ}$ (see Fig. 8.19). Determine the lengths of the sides $BC$ and $AC$.

Fig. 8.19

Solution : To find the length of the side $BC$, we will choose the trigonometric ratio involving $BC$ and the given side $A B$. Since $B C$ is the side adjacent to angle $C$ and $AB$ is the side opposite to angle $C$, therefore

$ \dfrac{AB}{BC}=\tan C $

i.e., $ \dfrac{5}{BC}=\tan 30^{\circ}=\dfrac{1}{\sqrt{3}} $

which gives $ BC=5 \sqrt{3} cm $

To find the length of the side AC, we consider

$ \begin{aligned} & \sin 30^{\circ} =\dfrac{AB}{AC} \quad \quad \quad \text{(Why?)} \\ \text{ i.e., }\quad \quad \quad & \dfrac{1}{2} =\dfrac{5}{AC} \\ \text{ i.e., }\quad \quad \quad & AC =10 ~cm \end{aligned} $

Note that alternatively we could have used Pythagoras theorem to determine the third side in the example above,

i.e., $ A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{5^{2}+(5 \sqrt{3})^{2}} cm=10 cm . $

Example 7 : In $\Delta PQR$, right-angled at $Q$ (see Fig. 8.20), PQ=3 cm and PR=6 cm. Determine $\angle QPR$ and $\angle PRQ$.

Fig. 8.20

Solution : Given PQ=3 cm and PR=6 cm.

Therefore, $ \dfrac{P Q}{P R}=\sin R $

or $ \sin R=\dfrac{3}{6}=\dfrac{1}{2} $

So, $ \angle PRQ=30^{\circ} $

and therefore, $ \angle QPR=60^{\circ} . \quad(\text{ Why? }) $

You may note that if one of the sides and any other part (either an acute angle or any side) of a right triangle is known, the remaining sides and angles of the triangle can be determined.

Example 8 : If $\sin (A-B)=\dfrac{1}{2}, \cos (A+B)=\dfrac{1}{2}, 0^{\circ}<A+B \leq 90^{\circ}, A>B$, find $A$ and B.

Solution : Since, $\sin (A-B)=\dfrac{1}{2}$, therefore, $A-B=30^{\circ} \quad($ Why? $) \quad \quad \quad \quad (1)$

Also, since $\cos (A+B)=\dfrac{1}{2}$, therefore, $A+B=60^{\circ} \quad$ (Why?) $\quad \quad \quad \quad (2)$

Solving (1) and (2), we get : $A=45^{\circ}$ and $B=15^{\circ}$.

EXERCISE 8.2

1. Evaluate the following:

(i) $\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}$

(ii) $2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$

(iii) $\dfrac{\cos 45^{\circ}}{\sec 30^{\circ}+cosec 30^{\circ}}$

(iv) $\dfrac{\sin 30^{\circ}+\tan 45^{\circ}-cosec 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$

(v) $\dfrac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}$

2. Choose the correct option and justify your choice :

(i) $\dfrac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=$

(A) $\sin 60^{\circ}$

(B) $\cos 60^{\circ}$

(C) $\tan 60^{\circ}$

(D) $\sin 30^{\circ}$

(ii) $\dfrac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$

(A) $\tan 90^{\circ}$

(B) 1

(C) $\sin 45^{\circ}$

(D) 0

(iii) $\sin 2 A=2 \sin A$ is true when $A=$

(A) $0^{\circ}$

(B) $30^{\circ}$

(C) $45^{\circ}$

(D) $60^{\circ}$

(iv) $\dfrac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=$

(A) $\cos 60^{\circ}$

(B) $\sin 60^{\circ}$

(C) $\tan 60^{\circ}$

(D) $\sin 30^{\circ}$

3. If $\tan (A+B)=\sqrt{3}$ and $\tan (A-B)=\dfrac{1}{\sqrt{3}} ; 0^{\circ}<A+B \leq 90^{\circ} ; A>B$, find $A$ and $B$.

4. State whether the following are true or false. Justify your answer.

(i) $\sin (A+B)=\sin A+\sin B$.

(ii) The value of $\sin \theta$ increases as $\theta$ increases.

(iii) The value of $\cos \theta$ increases as $\theta$ increases.

(iv) $\sin \theta=\cos \theta$ for all values of $\theta$.

(v) $\cot A$ is not defined for $A=0^{\circ}$.

8.4 Trigonometric Identities

You may recall that an equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved.

In this section, we will prove one trigonometric identity, and use it further to prove other useful trigonometric identities.

In $\triangle ABC$, right-angled at $B$ (see Fig. 8.21), we have:

$ AB^{2}+BC^{2}=AC^{2} \tag{1} $

Fig. 8.21

Dividing each term of (1) by $AC^{2}$, we get

$ \dfrac{AB^{2}}{AC^{2}}+\dfrac{BC^{2}}{AC}=\dfrac{AC^{2}}{A C^{2}} $

i.e., $ (\dfrac{AB}{AC})^{2}+(\dfrac{BC}{AC})^{2}=(\dfrac{AC}{AC})^{2} $

i.e., $ (\cos A)^{2}+(\sin A)^{2}=1 $

i.e., $ \cos ^{2} A+\sin ^{2} A=1 \tag{2} $

This is true for all A such that $0^{\circ} \leq A \leq 90^{\circ}$. So, this is a trigonometric identity.

Let us now divide (1) by $AB^{2}$. We get

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

or, $ (\dfrac{AB}{AB})^{2}+(\dfrac{BC}{AB})^{2}=(\dfrac{AC}{AB})^{2} $

$ \text{ i.e., } \quad 1+\tan ^{2} A=\sec ^{2} A \tag{3} $

Is this equation true for $A=0^{\circ}$ ? Yes, it is. What about $A=90^{\circ}$ ? Well, $\tan A$ and sec $A$ are not defined for $A=90^{\circ}$. So, (3) is true for all A such that $0^{\circ} \leq A<90^{\circ}$.

Let us see what we get on dividing (1) by $BC^{2}$. We get

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

$ \text{i.e.,} \quad \quad \quad(\dfrac{A B}{B C})^{2}+(\dfrac{B C}{B C})^{2}=(\dfrac{A C}{B C})^{2}$

$$\text{i.e.,} \quad \quad \quad \cot^{2} \mathbf{A}+1=cosec^{2} \mathbf{A} \tag{4}$$

Note that $\text{cosec} A$ and $\cot A$ are not defined for $A=0^{\circ}$. Therefore (4) is true for all A such that $0^{\circ}<A \leq 90^{\circ}$.

Using these identities, we can express each trigonometric ratio in terms of other trigonometric ratios, i.e., if any one of the ratios is known, we can also determine the values of other trigonometric ratios.

Let us see how we can do this using these identities. Suppose we know that $\tan A=\dfrac{1}{\sqrt{3}} \cdot$ Then, $\cot A=\sqrt{3}$.

Since, $\sec ^{2} A=1+\tan ^{2} A=1+\dfrac{1}{3}=\dfrac{4}{3}$, sec $A=\dfrac{2}{\sqrt{3}}$, and $\cos A=\dfrac{\sqrt{3}}{2}$.

Again, $\sin A=\sqrt{1-\cos ^{2} A}=\sqrt{1-\dfrac{3}{4}}=\dfrac{1}{2}$. Therefore, cosec $A=2$.

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

Solution : Since

$ \cos ^{2} A+\sin ^{2} A=1 \text{, therefore, } $

$ \cos ^{2} A=1-\sin ^{2} A \text{, i.e., } \cos A= \pm \sqrt{1-\sin ^{2} A} $

This gives

$ \cos A=\sqrt{1-\sin ^{2} A} \quad \text{ (Why?) } $

Hence, $\quad \tan A=\dfrac{\sin A}{\cos A}=\dfrac{\sin A}{\sqrt{1-\sin ^{2} A}}$ and sec $A=\dfrac{1}{\cos A}=\dfrac{1}{\sqrt{1-\sin ^{2} A}}$

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

Solution :

LHS $=\sec A(1-\sin A)(\sec A+\tan A)=(\dfrac{1}{\cos A})(1-\sin A)(\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A})$

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

Example 11 : Prove that $\dfrac{\cot A-\cos A}{\cot A+\cos A}=\dfrac{cosec A-1}{cosec A+1}$

Solution : LHS $=\dfrac{\cot A-\cos A}{\cot A+\cos A}=\dfrac{\dfrac{\cos A}{\sin A}-\cos A}{\dfrac{\cos A}{\sin A}+\cos A}$

$ =\dfrac{\cos A(\dfrac{1}{\sin A}-1)}{\cos A(\dfrac{1}{\sin A}+1)}=\dfrac{(\dfrac{1}{\sin A}-1)}{(\dfrac{1}{\sin A}+1)}=\dfrac{cosec A-1}{cosec A+1}=\text{ RHS } $

Example 12 : Prove that $\dfrac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\dfrac{1}{\sec \theta-\tan \theta}$, using the identity $\sec ^{2} \theta=1+\tan ^{2} \theta$.

Solution : Since we will apply the identity involving sec $\theta$ and $\tan \theta$, let us first convert the LHS (of the identity we need to prove) in terms of sec $\theta$ and $\tan \theta$ by dividing numerator and denominator by $\cos \theta$.

$$ \begin{aligned} \text{ LHS } & =\dfrac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\dfrac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta} \\ & =\dfrac{(\tan \theta+\sec \theta)-1}{(\tan \theta-\sec \theta)+1}=\dfrac{{(\tan \theta+\sec \theta)-1}(\tan \theta-\sec \theta)}{{(\tan \theta-\sec \theta)+1}(\tan \theta-\sec \theta)} \\ & =\dfrac{(\tan ^{2} \theta-\sec ^{2} \theta)-(\tan \theta-\sec \theta)}{{\tan \theta-\sec \theta+1}(\tan \theta-\sec \theta)} \\ & =\dfrac{-1-\tan \theta+\sec \theta}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)} \\ & =\dfrac{-1}{\tan \theta-\sec \theta}=\dfrac{1}{\sec \theta-\tan \theta} \end{aligned} $$

which is the RHS of the identity, we are required to prove.

EXERCISE 8.3

1. Express the trigonometric ratios $\sin A, \sec A$ and $\tan A$ in terms of $\cot A$.

2. Write all the other trigonometric ratios of $\angle A$ in terms of sec $A$.

3. Choose the correct option. Justify your choice.

(i) $9 \sec ^{2} A-9 \tan ^{2} A=$

(A) 1

(B) 9

(C) 8

(D) 0

(ii) $(1+\tan \theta+\sec \theta)(1+\cot \theta-cosec \theta)=$

(A) 0

(B) 1

(C) 2

(D) -1

(iii) $(\sec A+\tan A)(1-\sin A)=$

(A) $\sec A$

(B) $\sin A$

(C) $cosec A$

(D) $\cos A$

(iv) $\dfrac{1+\tan ^{2} A}{1+\cot ^{2} A}=$

(A) $\sec ^{2} A$

(B) -1

(C) $\cot ^{2} A$

(D) $\tan ^{2} A$

4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) $(cosec \theta-\cot \theta)^{2}=\dfrac{1-\cos \theta}{1+\cos \theta}$

(ii) $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2 \sec A$

(iii) $\dfrac{\tan \theta}{1-\cot \theta}+\dfrac{\cot \theta}{1-\tan \theta}=1+\sec \theta cosec \theta$

[Hint : Write the expression in terms of $\sin \theta$ and $\cos \theta$ ]

(iv) $\dfrac{1+\sec A}{\sec A}=\dfrac{\sin ^2 A}{1-\cos A}$

[Hint : Simplify LHS and RHS separately]

(v) $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec} A+\cot A$, wsing the identity $\operatorname{cosec}^2 A=1+\cot ^2 A$.

(vi) $\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$

(vii) $\dfrac{\sin \theta-2 \sin ^3 \theta}{2 \cos ^3 \theta-\cos \theta}=\tan \theta$

(viii) $(\sin \mathrm{A}+\operatorname{cosec} \mathrm{A})^2+(\cos \mathrm{A}+\sec \mathrm{A})^2=7+\tan ^2 \mathrm{~A}+\cot ^2 \mathrm{~A}$

(ix) $(\operatorname{cosec} A-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}$

[Hint : Simplify LHS and RHS separately]

(x) $\left(\dfrac{1+\tan ^2 A}{1+\cot ^2 A}\right)=\left(\dfrac{1-\tan A}{1-\cot A}\right)^2-\tan ^2 A$

8.5 Summary

In this chapter, you have studied the following points :

1. In a right triangle $A B C$, right-angled at $B$,

$ \begin{aligned} & \sin A=\dfrac{\text{ side opposite to angle } A}{\text{ hypotenuse }}, \cos A=\dfrac{\text{ side adjacent to angle } A}{\text{ hypotenuse }} \\ & \tan A=\dfrac{\text{ side opposite to angle } A}{\text{ side adjacent to angle } A} . \end{aligned} $

2. $\text{cosec} A=\dfrac{1}{\sin A} ; \sec A=\dfrac{1}{\cos A} ; \tan A=\dfrac{1}{\cot A}, \tan A=\dfrac{\sin A}{\cos A}$.

3. If one of the trigonometric ratios of an acute angle is known, the remaining trigonometric ratios of the angle can be easily determined.

4. The values of trigonometric ratios for angles $0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}$ and $90^{\circ}$.

5. The value of $\sin A$ or $\cos A$ never exceeds 1 , whereas the value of sec $A$ or $cosec A$ is always greater than or equal to 1 .

6. $\sin ^{2} A+\cos ^{2} A=1$,

$\sec ^{2} A-\tan ^{2} A=1$ for $0^{\circ} \leq A<90^{\circ}$,

$\text{cosec}^{2} A=1+\cot ^{2} A$ for $0^{\circ}<A \leq 90^{\circ}$.