8 Trigonometry

Introduction to Trigonometry & Measurement of Angle:

The word 'trigonometry' is derived from the Greek words 'trigon' and 'metron' and it means 'measuring
the sides of a triangle'. The subject was originally developed to solve geometric problems involving
triangles. It was studied by sea captains for navigation, surveyor to map out the new lands, by engineers
and others. Currently, trigonometry is used in many areas such as the science of seismology, designing
electric circuits, describing the state of an atom, predicting the heights of tides in the ocean, analysing
a musical tone and in many other areas.
(a) Measurement of angles : Commonly two systems of measurement of angles are used.
(i) Sexagesimal or English System : Here 1 right angle = 90° (degrees)]
1° = 60' (minutes)
1' = 60" (seconds)
(ii) Circular system : Here an angle is measured in radians. One radian corresponds to the angle
subtended by an arc of length 'r ' at the centre of the circle of radius r. It is a constant quantity
and does not depend upon the radius of the circle.
D R
(b) Relation between the these systems : =
90  / 2

(c) If  is the angle subtended at the centre of a circle of radius 'r', by an arc of length '' then = .
r
Note that here , r are in the same units and q is always in radians.

Solved Example

Example 1: If the arcs of same length in two circles subtend angles of 60° and 75° at their centres.
Find the ratio of their radii.
Solution: Let r1 and r2 be the radii of the given circles and let their arcs of same length 's' subtend
angles of 60° and 75° at their centres.
c c c c
         5 
Now, 60 =  60   =  and 75 =  75   = 
 180  3  180   12 
 s 5 s  5
 = and =  r1 = s and r =s
3 r1 12 r2 3 12 2

 5
 r1 = r  4r1 = 5r2  r1 : r2 = 5 : 4
3 12 2

Trigonometry 1
Concept Builders - 1

(i) The radius of a circle is 30 cm. Find the length of an arc of this circle if the length of the chord of the arc is
30 cm.

T-Ratios (or Trigonometric Functions)

In a right angle triangle
p b p
sin = ;cos  = ; tan = ;
h h b
h h b
cosec  = sec  = and cot  =
p b p
'p' is perpendicular ; 'b' is base and 'h' is hypotenuse.
Note : The quantity by which the cosine falls short of unity i.e. 1 – cos, is called the versed sine  of
 and also by which the sine falls short of unity i.e. 1– sin is called the coversed sine of .

Basic Trigonometric Identities:

(1) sin. cosec = 1 (2) cos . sec = 1
sin cos 
(3) tan. cot = 1 (4) tan = & cot  =
cos  sin
(5) sin2 + cos2 = 1 or sin2 = 1 – cos2 or cos2 = 1 – sin2 
(6) sec2 – tan2 = 1 or sec2 = 1 + tan2 or tan2 = sec2 – 1
1
(7) sec + tan  =
sec  − tan 
(8) cosec  – cot  = 1 or cosec2 = 1 + cot2 or cot2 = cosec2 – 1
2 2

1
(9) cosec + cot  =
cosec  − cot 
(10) Expressing trigonometrical ratio in terms of each other :

sin  cos  tan  cot  sec  cosec 

tan 1 sec 2  − 1 1
sin  sin  1 − cos  2
1 + tan  2
1 + cot  2
sec  cos ec
1 cot  1 cos ec 2  − 1
cos  1 − sin 2
cos 
1 + tan 2  1 + cot 2  sec  cos ec
sin  1 − cos 2  1 1
tan  tan  sec2  −1
1 − sin 2
cos  cot  cos ec2  − 1

1 − sin 2  cos  1 1
cot  cot  cos ec2 −1
sin  1 − cos  2
tan  sec2  − 1
1 1 1 + cot 2  cos ec
sec  1 + tan  2 sec 
1 − sin 2
cos  cot  cos ec2  − 1
1 1 1 + tan 2  sec 
cosec  1 + cot 2  cosec 
sin  1 − cos  2
tan  sec2  − 1

2 Trigonometry
Solved Example

Example 2: If sin  + sin2 = 1, then prove that

cos12 + 3cos10 + 3cos8 + cos6–1 = 0
Solution: Given that sin = 1 – sin2 = cos2
L.H.S. = cos6(cos2 + 1)3 – 1= sin3(1 + sin)3 – 1= (sin + sin2)3 – 1 = 1 – 1 = 0

Example 3: 4(sin6 + cos6) – 6 ( sin4 + cos4) is equal to

(A) 0 (B) 1 (C) –2 (D) none of these
Solution: 4[(sin  + cos ) – 3 sin  cos  (sin  + cos )] – 6[ (sin  + cos2)2 – 2sin2 cos2]
2 2 3 2 2 2 2 2

= 4[1 – 3 sin2 cos2] – 6[1 –2 sin2 cos2]

= 4 – 12 sin2 cos2 – 6 + 12 sin2 cos2 = –2
Ans. (C)

Concept Builders - 2

4
(i) If cot  = , then find the value of sin, cos and cosec in first quadrant.
3
(II) sin + cosec = 2, then find the value of sin8 + cosec8

New Definition of T-Ratios:

By using rectangular coordinates the definitions of trigonometric functions can be extended to angles
of any size in the following way(see diagram). A point P is taken with coordinates (x, y). The radius
vector OP has length r and the angle is taken as the directed angle measured anticlockwise from the
x-axis. The three main trigonometric functions are then defined in terms of r and the coordinates x
and y

sin = y/r,
cos = x/r
tan = y/x,(The other function are reciprocals of these)
This can give negative values of the trigonometric functions.

Trigonometry 3
Signs of Trigonometric Functions In Different Quadrants:

Trigonometric Functions of Allied Angles:

(a) sin (2n + ) = sin , cos (2n + ) = cos , where n  I
(b)
sin (–) = – sin  cos (–) = cos 
sin(90° – ) = cos cos(90° – ) = sin
sin(90° + ) = cos cos(90° + ) = –sin
sin(180° – ) = sin cos(180° – ) = –cos
sin(180° + ) = –sin cos(180° + ) = –cos
sin(270° – ) = –cos cos(270° – ) = –sin
sin(270° + ) = –cos cos(270° + ) = sin
sin (360° – ) = –sin cos(360° – ) = cos
sin (360° + ) = sin cos(360° + ) = cos

Values of T-Ratios of Some Standard Angles:

Angles 0º 30º 45º 60º 90º 180º 270º

T-ratio 0 /6 /4 /3 /2  3/2
0
sin  1/2 1/ 2 3/2 1 0 –1

1
cos  3/2 1/ 2 1/2 0 –1 0

tan  0 1/ 3 1 3 N.D. 0 N.D.

cot  N.D. 3 1 1/ 3 0 N.D. 0

sec  1 2/ 3 2 2 N.D. –1 N.D.

cosec  N.D. 2 2 2/ 3 1 N.D. –1

4 Trigonometry
N.D. → Not Defined
(a) sin n = 0 ; cos n =(–1)n; tan n = 0 where n  I
 
(b) sin(2n+1) = (−1)n ;cos(2n + 1) = 0 where n  I
2 2

1 1
Example 4: and tan =
If sin  = − then  is equal to -
2 3
(A) 30° (B) 150° (C) 210° (D) none of these
Solution: Let us first find out  lying between 0 and 360°.
1
Since sin = −
2
1
  = 210° or 330° and tan  =   = 30 or 210
3
7
Hence,  = 210 or is the value satisfying both. Ans. (C)
6

Concept Builders - 3

1 3
(i) If cos = − and     , then find the value of 4tan2 – 3cosec2.
2 2
(ii) Prove that :
(a) cos570° sin510° + sin(–330°) cos(–390°) = 0
11 9 3  17 3−2 3
(b) tan − 2 sin − cosec2 + 4 cos2 =
3 3 4 4 6 2

Trigonometric Ratios of The Sum & Difference of Two Angles :

(i) sin (A + B) = sin A cos B + cos A sin B. (ii) sin (A – B) = sin A cos B – cos A sin B.
(iii) cos (A + B) = cos A cos B – sin A sin B (iv) cos (A – B) = cos A cos B + sin A sin B
tan A + tan B tanA − tanB
(V) tan (A + B) = (iv) tan(A − B) =
1 − tan A tan B 1 + tanA tanB
cotBcot A − 1 cotBcot A + 1
(vii) cot(A + B) = (viii) cot(A − B) =
cotB + cot A cotB − cot A
Some more results :
(i) sin2 A – sin2 B = sin (A + B). sin(A – B) = cos2 B – cos2 A.
(ii) cos2 A – sin2 B = cos (A+B). cos (A – B).

Example 5: Prove that 3 cosec20° – sec20° = 4.

3 1 3 cos20 − sin20
Solution: L.H.S. = − =
sin20 cos20 sin20  cos20
 3 1 
cos 20 − sin20 
4
 2
= 
2 = (
 4 sin60  cos 20 − cos60  sin20 )
 
2sin20 cos 20 sin40

= 4.
(
sin 60 − 20 ) = 4  sin40
= 4 = R.H.S.
sin40 sin40

Trigonometry 5
Example 6: Prove that tan70° = cot70° + 2cot40° .
tan20 + tan50
Solution : L.H.S. = tan 70° = tan(20° + 50°) =
1 − tan20 tan50
or tan70° – tan20° tan50° tan70° = tan20° + tan50°
or tan70° = tan70° tan50° tan20° + tan20° + tan50° = 2 tan 50° + tan20°
= cot70° + 2cot40° = R.H.S

Concept Builders - 4

3 9 
(i) If sinA = and cosB = ,0  A & B  , then find the value of the following:
5 41 2
(a) sin(A + B) (b) sin(A – B) (c) cos(A + B) (d) cos(A – B)
(II) If x + y = 45°, then prove that :
(a) (1 + tanx)(1 + tany) = 2 (b) (cotx – 1)(coty – 1) = 2
(Remember these results)

Sine Formulae:
In any triangle ABC
a b c abc
= = == = 2R
sinA sinB sinC 2
where R is circumradius and  is area of triangle.

Example 7: Angles of a triangle are in 4 : 1 : 1 ratio. The ratio between its greatest side and perimeter is
3 3 3 1
(A) (B) (C) (D)
2+ 3 2+ 3 2− 3 2+ 3
Solution : Angles are in ratio 4 : 1 : 1.
 angles are 120°, 30°, 30°.
If sides opposite to these angles are a, b, c respectively, then a will be the greatest
side.
a b c
Now from sine formula = =
 
sin 120 sin30 sin30
a b c
 = =
3 / 2 1/ 2 1/ 2
a b c
 = = = k (say)
3 1 1
then a = 3k, perimeter = (2 + 3)k
3k 3
 required ratio = = Ans. (B)
(2 + 3)k 2+ 3

6 Trigonometry
Example 8: In triangle ABC, if b = 3, c = 4 and B = /3, then number of such triangles is –
(A) 1 (B) 2 (C) 0 (D) infinite
sin B sin C
Solution : Using sine formulae =
b c
sin  / 3 sin C 3 sin C 2
 =  =  sin C =  1 which is not possible.
3 4 6 4 3
Hence there exist no triangle with given elements. Ans.(C)

Example 9: The sides of a triangle are three consecutive natural numbers and its largest angle is
twice the smallest one. Determine the sides of the triangle.

Solution : Let the sides be n, n + 1, n + 2 cms.

i.e. AC = n, AB = n + 1, BC = n + 2
Smallest angle is B and largest one is A.
Here, A = 2B
Also, A + B + C = 180°
 3B + C = 180°  C = 180° – 3B
We have, sine law as,
sin A sin B sin C sin2B sin B sin(180 − 3B)
= =  = =
n+2 n n+1 n+2 n n+1
sin2B sin B sin3B
 = =
n+2 n n+1
(ii) (iii)
from (i) and (ii);
2sin B cos B sin B n+2
=  cos B = ..........(iv)
n+2 n 2n
and from (ii) and (iii);
sin B 3 sin B − 4 sin3 B
=
n n+1


(
sin B sin B 3 − 4 sin B
=
2
)
n n+1

n+ 1
n
(
= 3 − 4 1 − cos2 B) ............(v)
from (iv) and (v), we get
2
n+ 1 n + 2
= −1 + 4  
n  2n 
n+ 1  n2 + 4n + 4 
 +1= 
n  n2 
 
2n + 1 n2 + 4n + 4
 =  2n2 + n = n2 + 4n + 4
n n2

 n2 – 3n – 4 = 0  (n – 4)(n + 1) = 0
n = 4 or – 1
where n  –1
 n = 4. Hence the sides are 4, 5, 6

Trigonometry 7
Concept Builders - 5


(i) If in a ABC, A = and b : c = 2 : 3 , find B .
6
(II) Show that, in any ABC : a sin(B – C) + b sin(C – A) + c sin(A – B) = 0.
sin A sin(A − B)
(iii) If in a ABC, = , show that a2, b2, c2 are in A.P.
sinC sin(B − C)

Cosine Formulae:
b2 + c2 − a 2 c2 + a2 − b2 a 2 + b2 − c2
(a) cosA = (b) cosB = (c) cosC =
2bc 2ca 2ab
or a2 = b2 + c2 – 2bc cosA

Example 10: In a triangle ABC, if B = 30° and c = 3 b, then A can be equal to –

(A) 45° (B) 60° (C) 90° (D) 120°
c2 + a2 − b2 3 3b2 + a2 − b2
Solution : We have cos B =  =
2ca 2 2  3b  a
 a2 – 3ab + 2b2 = 0  (a – 2b) (a – b) = 0
 Either a = b  A = 30°
or a = 2b
 a2 = 4b2 = b2 + c2  A = 90°. Ans. (C)

Example 11: In a triangle ABC, (a2 –b2 – c2) tan A + (a2 – b2 +c2) tan B is equal to –
(A) (a2 + b2 –c2) tan C (B) (a2 + b2 + c2) tan C
(C) (b + c –a ) tan C
2 2 2
(D) none of these
Solution : Using cosine law :
The given expression is equal to –2 bc cos A tan A + 2 ac cos B tan B
 sin A sin B 
= 2abc  − + =0 Ans. (D)
 a b 

Concept Builders - 6

(i) If a : b : c = 4 : 5 : 6, then show that C = 2A.

(II) In any ABC, prove that
cos A cosB cos C a 2 + b2 + c2
(a) + + =
a b c 2abc
b2 c2 a2 a4 + b4 + c4
(b) cos A + cosB + cos C =
a b c 2abc

Angle of elevation
In order to see an object which is at a higher level compared to the ground level we are to
look up. the line joining the object and the eye of the observer is known as the line of sight
and the angle which this line of sight makes with the horizontal drawn through the eye of
the observer is known as the angle of elevation.
Therefore, the angle of elevation of an object helps in finding out its height (Figure).

8 Trigonometry
Angle of Depression
When the object is at a lower level than the observer's eyes, he has to look downwards to
have a view of the object. In that case, the angle which the line of sight makes with the
horizontal through the observer's eye is known as the angle of depression (Figure).

Example 12: A man is standing on the deck of a ship, which is 8 m above water level. He observes
the angle of elevation of the top of a hill as 60º and the angle of depression of the ase
of the hill as 30º. Calculate the distance of the hill from the ship and the height of the
hill.
Solution : Let x be distance of hill from man and h + 8 be height of hill which is required.
In right triangle ACB,

AC h h
 tan60 = =  3=
BC x x
In right triangle BCD,
CD 8 1 8
tan30 = =  = x =8 3
BC x 3 x
 Height of hill = h + 8 = 3  x + 8( 3)(8 3) + 8 = 32m.
Distance of ship from hill = x = 8 3m

Trigonometry 9
Example 13: A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff
of height 5 meters. At a point on the plane, the angle of elevation of the bottom and the
top of the flag staff are respectively 30° and 60°. Find the height of tower.
Solution : Let AB be the tower of height h metre and BC be the height of flag staff surmounted on
the tower.
Let the point on the plane be D at a distance x meter from the foot of the tower
In ABD
AB 1 h
tan 30 =  =  x = 3h ...... (i)
AD 3 x

In ADC
AC 5+h
tan 60° =  3=
AD x
5+h
 x= ...... (ii)
3
From (i) and (ii)
5+h
 3h=  3h = 5 + h
3
5
 2h = 5  h= = 2.5 m
2
So, the height of tower = 2.5 m

Example 14: The angle of elevation of an aeroplane from a point on the ground is 45º. After a flight
of 15 sec, the elevation changes to 30°. If the aeroplane is flying at a height of 3000
metres, find the speed of the aeroplane.
Solution : Let the point on the ground is E which is y metres from point B and let after 15 sec flight
it covers x metres distance.

In  AEB
AB 3000
tan 45° =  1=
EB y

10 Trigonometry
 y = 3000 m .......(i)
In CED
CD
 tan 30° = .
ED
1 3000
 = ( AB = CD)
3 x+y
 x + y = 3000 3 ...... (ii)
From equation (i) and (ii)
 x + 3000 = 3000 3  x = 3000 3 – 3000  x = 3000 ( 3 – 1)
 x + 3000 × (1.732 – 1)  x = 3000 × 0.732  x = 2196 m.
Dis tance cov ered 2196 2196 8
Speed of Aeroplane = = m\sec =  Km\hr = 527.04
Time taken 15 15 5
Km/hr.
Hence, the speed of aeroplane is 527.04 Km/hr/

Example 15: If the angle of elevation of a cloud from a point h metres above a lake is and the angle
of depression of its reflection in the lake is , prove that the distance of the cloud from
2h sec 
the point of observation is .
tan  − tan 
Solution : Let AB be the surface of the lake and let C be a point of observation such that AC = h
metres. Let D be the position of the cloud and D' be its reflection in the lake. Then BD
= BD.

In DCE
DE
tan  =
CE
H
 CE = ...... (i)
tan 
In CED'
ED '
tan  =
EC
h+H+h
 CE =
tan 
2h + H
 CE = ...... (ii)
tan 
From (i) & (ii)
H 2h + H
 =
tan  tan 
 H tan  = 2h tan  + H tan   H tan  – H tan  = 2h tan 
 H(tan  – tan ) = 2h tan 
2h tan 
 H= ....... (iii)
tan  − tan 
In DCE
DE DE H
sin a =  CD =  CD =
CD sin  sin 

Trigonometry 11
Substituting the value of H from (iii)
sin 
2h
2h tan  cos  2h sec 
CD =  CD = CD =
(tan  − tan  ) sin  (tan  − tan  ) sin  tan  − tan 
2h sec 
Hence, the distance of the cloud from the point of observation is .
tan  − tan 
Hence Proved.

12 Trigonometry
Exercise–I

Single Correct Type Questions

Trigonometry 13
Exercise–II

PART – I PREVIOUS YEAR NSEJS PROBLEMS

1. The expression (1 – tan A + sec A) (1 – cot A + cosec A) has value : [NSEJS-2008]

(A) -1 (B) 0 (C) +1 (D) +2

2. A person on the top of a tower observes a scooter moving with uniform velocity towards the
base of the tower. He finds that the angle of depression changes from 30° to 60° in 18 minutes.
The scooter will reach the base of the tower in next : [NSEJS-2008]
(A) 9 minutes (B) 18 / ( 3 − 1) minutes
(C) 6 3 minutes (D) the time depends upon the height of the tower

 sec  − 1  2  sin − 1 
3. The trigonometric equation is : cot2    + sec    has the value :
 1 + sin   1 + sec  
[NSEJS-2009]
(A) -1 (B) 0 (C) 1 (D) 2

4. If sin  + cosec  = 2, then [sin8  + cosec8 ] will have the value [NSEJS-2010]
(A) 2 (B) 24 (C) 26 (D) 28

5. An aeroplane is flying horizontally at a height of 3150 m above a horizontal plane ground. At a

particular instant it passes another aeroplane vertically below it. At this instant, the angles of
elevation of the planes form a point on the ground are 30° and 60°. Hence, the distance between
the planes at that instant is [NSEJS-2011]
(A) 1050 m (B) 2100 m (C) 4200 m (D) 5250 m

1
6. If sin x + sin y = a and cos x – cos y = b. Then find the value of (2 – a2 – b2) [NSEJS-2012]
2
(A) cos (x + y) (B) cos (x – y) (C) sin (x + y) (D) sin (x – y)

7. If cot2 (1 – 3sec + 2sec2) = 1 [NSEJS-2012]

(A) 120° (B) 210° (C) 300° (D) 330°

8. We all know that the sky appears to touch the ground at a distance. The distance at which we
perceive the sky to touch the ground is called horizon. The reason for the perception is due to
the fact that the Earth is a sphere (almost) and not a flat surface. Which of these pictures
below accurately depict the horizon for a person standing on a high rise building like Burj Khalifa
in Dubai? (Here, ‘h’ represents the height of the building while line ‘H’ represents the horizon):
[NSEJS-2012]

(A) (B) (C) (D)

14 Trigonometry
9. If sin  and cos  are roots of the equation px2 + qx + r = 0, then: [NSEJS-2018]
(*A) p2 – q2 + 2pr = 0 (B) (p + r)2 = q2 – r2
(C) p + q – 2pr = 0
2 2
(D) (p – r)2 = q2+ r2

10. An observer standing at the top of a tower, finds that the angle of elevation of a red bulb on
the top of a light house of height H is a. Further, he finds that the angle of depression of
reflection of the bulb in the ocean is b. Therefore, the height of the tower is
[NSEJS - 2019]
H (tan  − tan  ) H sin(  −  ) H (cos  − cos  )
(A*) (B) (C) (D) H
(tan  + tan  ) cos ( +  ) (cot  + cot  )

11. In the convex quadrilateral ABCD, the diagonals AC and BD meet at 0 and the measure of angle
AOB is 30°. If the areas of triangle AOB, BOC, COD and AOD are 1, 2, 8 and 4 square units
respectively, what is the product of the lengths of the diagonals AC and BD in sq. units ?
[NSEJS - 2019]
(A*) 60 (B) 56 (C) 54 (D) 64

12. If tan + sec  = 1.5, then value of sin( ) is [NSEJS - 2019]

5 12 3 2
(A*) (B) (C) (D)
13 13 5 3

13. If sin2 x + sin2 y + sin2 z = 0 , then which of the following is NOT a possible value of
cos x + cos y + cos z ? [NSEJS - 2019]
(A) 3 (B) –3 (C) –1 (D*) –2

PART – II PREVIOUS YEAR KVPY PROBLEMS

14. In given figure, AB = 12 cm, CD = 8 cm, BD = 20 cm, ABD = AEC = EDC =90°. If BE = x, then

[KVPY-2013]
(A) x has two possible values whose difference is 4
(B) x has two possible values whose sum is 28
(C) x has only one value and x ≥ 12
(D) x cannot be determined with the given information

Trigonometry 15
15. In the figure given below, ABCDEF is a regular hexagon of side length 1, AFPS and ABQR are
squares. Then, the ratio ar(APQ)/ar(SRP) equals [KVPY-2013]

2+1 3 3
(A) (B) 2 (C) (D) 2
2 4

16. The sides a, b, c of a triangle satisfy the relations c2 = 2ab and a2 + c2 = 3b2 . Then the measure
of BAC, in degrees, is [KVPY – 2019]
(A) 30 (B*) 45 (C)60 (D) 90

17. Let ABC be a triangle in which AB = BC. Let X by a point on AB such that AX : XB = AB : AX. If AC
= AX, then the measure of ABC equals [KVPY – 2019]
(A) 18° (B*) 36° (C) 54° (D) 72°

PART – III PREVIOUS YEAR PRMO PROBLEMS

18. In a triangle ABC, right-angled at A, the altitude through A and the internal bisector of  A have
lengths 3 and 4, respectively. Find the length of the median through A. [PRE-RMO-2018]
Ans. 24.

19. If x = cos1°cos2°cos3°…..cos89° and y = cos2°cos6°cos10°…..cos86°, then what is the integer

2
nearest to log2(y/x) ? [PRE-RMO-2019]
7
Ans. 19

20. How many distinct triangles ABC are there, up to similarity, such that the magnitudes of angles
A, B and C in degrees are positive integers and satisfy
cos A cos B + sin A sin B sin kC = 1
for some positive integer k, where kC does not exceed 360°? [PRE-RMO-2019]
Ans. 6

21. Let ABC be an isosceles triangle with AB = BC. A trisector of  B meets AC at D. If AB, AC and
BD are integers and AB – BD = 3, find AC. [PRE-RMO-2019]
Ans. 26

22. In a triangle ABC, it is known that  A = 100° and AB = AC. The internal angle bisector BD has
length 20 units. Find the length of BC to the nearest integer, given that sin 10°  0.174.
[PRE-RMO-2019]
Ans. 27

16 Trigonometry