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Trigonometry Identities, Applications - Introduction to Trigonometry, Class 10, Mathematics | Extra Documents, Videos & Tests for Class 10 PDF Download

T - IDENTITIES (Trigonometry Identities)

We know 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.
The three Fundamental Trigonometric Identities are –

(i) cos2 A + sin2 A = 1 ; 0� ≤ A ≤ 90�
(ii) 1 + tan2 A = sec2 A ; 0� ≤ A < 90�
(iii) 1+ cot2 A = cosec2 A; 0� < A ≤ 90�

Geometrical Proof : Consider a ΔABC, right angled at B. Then we have :
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
AB2 + BC2 = AC2 ...(i) By Pythagoras theorem

 

(i) cos2 A + sin2 A = 1 ; 0� ≤ A ≤ 90� Dividing each term of (i) by AC2, we get

Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
i.e.,Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m

i.e., (cos A)+ (sin A)2 = 1 i.e., cos2 A + sin2 A = 1 ...(ii)

This is true for all A such that 0� ≤ A ≤ 90�
So, this is a trigonometric identity.
(ii) 1 + tan2 A = sec2 A ; 0� ≤ A < 90� Let us now divide (i) by AB2. We get

Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
i.e., Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m

i.e., 1 + tan2 A = sec2A ...(iii)

This equation is true for A = 0�. Since tan A and sec A are not defined for A = 90�, so (iii) is true for all A such that 0� ≤ A < 90�

(iii) 1 + cotA = cosec2 A ; 0� < A ≤ 90�

Again, let us divide (i) by BC2, we get

Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m

⇒ 1 + cot2 A = cosec2 A ...(iv)

Since cosec A and cot A are not defined for A = 0�, therefore

(iv) is true for all A such that  0� < A ≤ 90� Using the above trigonometric identities, we can express each trigonometric ratio in terms of the other trigonometric ratios, i.e., if any one of the ratios is known, we can also determine the values of other trigonometric ratios.

Fundamental Identities (Results)

Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,mClass 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,mClass 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m

To prove Trigonometrical Identities The following methods are to be followed :

Method-I: Take the more complicated side of the identity (L.H.S. or R.H.S. as the case may be) and by using suitable trigonometric and algebraic formulae prove it equal to the other side. 

Method-II : When neither side of the identity is in a simple form, simplify the L.H.S. and R.H.S. separately by using suitable formulae (by expressing all the T-ratios occuring in the identity in terms of the sine and cosine and show that the results are equal).

Method-III : If the identity to be proved is true, transposing so as to get similar terms on the same side, or if cross-multiplication, and using suitable formulae, we get an identity which is true.

Ex.17 Express sin A, sec A and tan A in terms of cot A.

Sol. We know that

 sin A = Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m

sec A = Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m

...(Dividing num. and deno. by sin A)

 Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m and tan A = Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m

Ex.18 Prove

Sol. L H S = Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m

Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,mClass 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m = RHS

Hence, proved.

Ex.19 Prove (cosecA – sinA) (sec A – cos A) Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m

Sol. LHS = (cosecA – sinA) (sec A – cos A) = Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m

  Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m  [∵ sin2 A + cosA = 1]

= sin A cos A = Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m  [∵ sin2 A + cos2 A = 1]

Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m    [Dividing the numerator and denominator by sin A cos A.]

 Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m

Hence, proved.

APPLICATIONS OF TRIGONOMETRY

Many times, we have to find the height and distances of many objects in real life. We use trigonometry to solve problems, such as finding the height of a tower, height of a flagmast, distance between two objects, where measuring directly is trouble, some and some times impossible. In those cases, we adopt indirect methods which involves solution of right triangles.
Thus Trigonometry is very useful in geography, astronomy and navigation. It helps us to prepare maps, determine the position of a landmass in relation to the longitudes and latitudes. Surveyors have made use of this knowledge since ages.
Angle of Elevation The angle between the horizontal line drawn through the observer eye and line joining the eye to

any object is called the angle of elevation of the object, if the object is at a higher level than the eye i.e., If a horizontal line OX is drawn through O, the eye of the observer, and P is an object in the vertical plane through OX, then if P is above OX, as in fig. ∠XOP is called the angle of elevation or the altitude of P as seen from O.

 Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m

Angle of Depression The angle between the horizontal line drawn through the observer eye and the line joining the eye to any object is called the angle of depression of the object if the object is at a lower level than the eye. i.e., lf a horizontal line OX is drawn through O, the eye of an observer, and P is an object in the vertical plane through OX, then if P is below OX, as in fig. ∠XOP is called the angle of depression of P as seen from O.

 Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m

REMARK :

 Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m

1. The angle of elevation as well as angle of depression are measured with reference to horizontal line.
2. All objects such as towers, trees, mountains etc. shall be considered as linear for mathematical convenience,throughout this section.  
3. The height of the observer, is neglected, if it is not given in the problem.
4. Angle of depression of P as seen from O is equal to the angle of elevation of O, as seen from P.i.e., ∠AOP = ∠OPX.  
5. To find one side of a right angled triangle when another side and an acute angle are given, the hypotenusealso being regarded as a side.
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m a certain T-ratio of the given angle.  
6. The angle of elevation increases as the object moves towards the right of the line of sight.
7. The angle of depression increases as the object moves towards the left of the line of sight.

COMPETITION WINDOW

BEARING OF A POINT

The true bearing to a point is the angle measured in degrees in a clockwise direction from the north line.
We will refer to the true bearing simply as the bearing.
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
e.g. (i) the bearing of point P is 65� (ii) the bearing of point Q is 300� A bearing is used to represent the direction of one point relative to another point. e.g., the bearing of A from B is 60�. The bearing of B from A is 240�.
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
 

TRY OUT THE FOLLOWING

State the bearing of the point P in each of the following diagrams :

Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,mClass 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m


ANSWERS

(a) 48�  (b) 240� (c) 140� (d) 290� Ex.20  An observer 1.5 m tall, is 28.5 m away from a tower 30 m high. Determine the angle of elevation of the top of the tower from his eye.  

Sol. Let AB be the height of the tower, CD the height of the observer with his eye at the point D, AB = 30 m, CD = 1.5 m.
Through D, draw DE II CA then ∠BDE = q where q is the angle of elevation of the top of the tower from

his eye. AC = horizontal distance between the tower and the observer = 28.5 m BE = AB - AE = (30 - 1.5) m = 28.5 m BDE is right triangle at E,
then Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
 = tan θ ⇒ tan θ = 1
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
⇒ tan θ = 1 = tan 45� ⇒ θ = 45�.  Required angle of elevation of the tower = θ = 45�.

 Ex.23 A captain of an aeroplane flying at an altitude of 1000 metres sights two ships as shown in the figure. If the angle of depressions are 60� and 30�, find the distance between the ships.
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m

Sol. Let A be the position of the captain of an aeroplane flying at the altitude of 1000 metres from the ground.
AB = the altitude of the aerplane from the ground = 1000 m P and Q be the position of two ships.
Let PB = x metres, and BQ = y metres.
Required : PQ = Distance between the ships = (x + y) metres.
ABP is rt. Δ at B ABQ is rt. Δ at B 
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
 577.3 my = 1000 (1.732) = 1732 m

Required distance between the ships = (x + y) metres = (577.3 + 1732) m = 2309.3 m

Ex.24 Two poles of equal heights are standing opposite to each other on either side of a road, which is 80 metres wide. From a point between them on the road, the angles of elevation of their top are 30� and 60�. Find the position of the point and also the height of the poles.
Sol. Let AB and CD be two poles of equal height standing opposite to each other on either side of the road BD. ⇒ AB = CD = h metres.
Let P be the observation point on the road BD. The angles of elevation of their top are 30� and 60�. ∠APB = 30�, ∠CPD = 60�
The width of the road = BD = 80 m, let PD = x metres Then BP = (80 – x) metres consider right DCDP, we have :
h II I h 30� 60�
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m           ...(i)
In right ΔABP, we have : AB
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,mClass 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,mClass 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m          ...(ii)
From (i) and (ii), we get :
Class 10 Maths,CBSE Class 10,Maths,Class 10,Introduction to Trigonometry,m
⇒ (80 – x) = 3x ⇒ 4x = 80 ⇒ x = 20
Height of each pole = AB = CD = Trigonometry Identities, Applications - Introduction to Trigonometry, Class 10, Mathematics | Extra Documents, Videos & Tests for Class 10 = 20 (1.732) = 34.64 metres.
Position of point P is 20 m from the first and 60 m from the second pole. i.e., the position of the point P is 20 m from either of the poles.

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FAQs on Trigonometry Identities, Applications - Introduction to Trigonometry, Class 10, Mathematics - Extra Documents, Videos & Tests for Class 10

1. What are the Pythagorean identities in trigonometry?
Ans. The Pythagorean identities in trigonometry are three equations that are derived from the Pythagorean theorem. They are: 1. sin²θ + cos²θ = 1 2. tan²θ + 1 = sec²θ 3. 1 + cot²θ = csc²θ These identities are fundamental in trigonometry and are used to simplify and solve trigonometric equations.
2. How do trigonometry identities help in solving problems?
Ans. Trigonometry identities are useful in solving problems involving angles and sides of triangles. They provide relationships between trigonometric functions that can be used to simplify expressions, prove other identities, and solve equations. By using these identities, complex trigonometric equations can be simplified into simpler forms, making it easier to find solutions.
3. What are the applications of trigonometry in real life?
Ans. Trigonometry has various applications in real life, some of which include: 1. Architecture and Construction: Trigonometry is used in designing and constructing buildings, bridges, and other structures. It helps in calculating angles, distances, and heights, ensuring accurate and stable constructions. 2. Navigation: Trigonometry is crucial in navigation and GPS systems. It helps determine positions, distances, and directions between different locations, making it easier to navigate accurately. 3. Astronomy: Trigonometry plays a significant role in astronomy. It helps calculate distances between celestial bodies, determine their relative positions, and predict astronomical events. 4. Engineering: Trigonometry is widely used in various engineering fields. It helps in designing electrical circuits, analyzing forces and motion, and solving complex mathematical problems related to engineering. 5. Physics: Trigonometry is essential in physics for analyzing and describing various physical phenomena, such as wave motion, vibrations, and oscillations.
4. How can I prove trigonometric identities?
Ans. To prove trigonometric identities, you can use various methods, such as: 1. Algebraic Manipulation: Rearrange the given equation using algebraic properties and trigonometric identities to match the form of the identity you want to prove. 2. Pythagorean Identities: Use the Pythagorean identities to simplify expressions involving trigonometric functions and replace them with equivalent forms. 3. Double-Angle Identities: Utilize the double-angle identities to express trigonometric functions in terms of other trigonometric functions. 4. Sum and Difference Identities: Apply the sum and difference identities to express trigonometric functions in terms of other trigonometric functions. 5. Simplification Techniques: Simplify complex expressions by factoring, canceling common terms, or using algebraic identities. By applying these techniques and manipulating the equations step by step, you can eventually reach the desired identity.
5. How are trigonometry identities useful in calculus?
Ans. Trigonometry identities are highly relevant in calculus as they simplify trigonometric functions and make integration and differentiation easier. By using trigonometric identities, complex expressions involving trigonometric functions can be simplified into simpler forms, allowing for more efficient calculations. For example, the integration of trigonometric functions can be simplified by applying identities such as the double-angle or half-angle identities. These identities help to transform the integral into a more manageable form, making it easier to solve. Similarly, when differentiating trigonometric functions, identities can be used to simplify the expressions and obtain the derivative more easily. Trigonometry identities also play a crucial role in solving differential equations involving trigonometric functions, providing a foundation for various applications of calculus in physics, engineering, and other fields.
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