Wednesday 31 July 2019

TRIGONOMETRICAL IDENTITIES - PART – 88

122. If 4 cos2 A – 3 = 0 then prove that
(i) sin 3 A = 3 sin A – 4 sin3 A.
(ii) cos 3 A = 4 cos3 A – 3 cos A

TRIGONOMETRICAL IDENTITIES - PART – 87

121. (i) If 2 sin A – 1 = 0, 
show that sin 3A = 3 sin A – 4 sin3 A.
(ii) If 4 cos2 A – 3 = 0, 
show that cos 3A = 4 cos3 A – 3 cos A.


TRIGONOMETRICAL IDENTITIES - PART – 86

120. If tan A = n tan B and 
sin A = m sin B, 
prove that: cos2 A = (m2 – 1) / (m2 + 1).

Tuesday 30 July 2019

Monday 29 July 2019

TRIGONOMETRICAL IDENTITIES - PART – 82

116. If x = r cos A  cos B, 
y = cos A sin B and then 
z = r sin A, then prove that 
x2 + y2 + z2 = r2

TRIGONOMETRICAL IDENTITIES - PART – 81

115. If sin A + cos A  = m and
sec A + cosec A = n show that 
n (m2 – 1) = 2 m

TRIGONOMETRICAL IDENTITIES - PART – 80

114. If x = r sin A  cos B, 
y = sin A sin B and then 
z = r cos A, then prove that 
x2 + y2 + z2 = r2

TRIGONOMETRICAL IDENTITIES - PART – 79

113. If m = a sec A + b tan A and 
n = a tan A + b sec A, then prove that 
m2 – n2 = a2 – b2

TRIGONOMETRICAL IDENTITIES - PART – 78

112. If p = sin A + cos A and 
q = sec A + cosec A, then prove that 
q (p2 – 1) = 2 p

Saturday 27 July 2019

TRIGONOMETRICAL IDENTITIES - PART – 77

111. If x cos A + y sin A = m and 
            x sin A – y cos A = n, then prove that: 
            x2 + y2 = m2 + n2


TRIGONOMETRICAL IDENTITIES - PART – 76

110. If x = a sec A cos B, 
y = b sec A sin B and z = c tan A 
show that
x2/a2 + y2/b2 + z2/c2 = 1


TRIGONOMETRICAL IDENTITIES - PART – 75

109. If tan A + sin A = m and tan A – sin A = n; 
prove that
 m2 – n2 = 4 (mn)1/2

Friday 26 July 2019

TRIGONOMETRICAL IDENTITIES - PART – 74

108. [(cos A) / (cosec A + 1)]
+
[(cos A )/ (cosec A –  1)]
= 2 tan A

TRIGONOMETRICAL IDENTITIES - PART – 73

107. [cot A – tan A]  [(2 cos^2 A – 1) / (sin A cos A)]


TRIGONOMETRICAL IDENTITIES - PART – 72

106. (cosec A + 1) / (cosec A – 1)
=
(sec A + tan A)2

TRIGONOMETRICAL IDENTITIES - PART – 71

105. (sin Ꝋ + 2sec Ꝋ)2 + (cos Ꝋ + cosec Ꝋ)2
=
(1 + sec Ꝋ cosec Ꝋ)2


TRIGONOMETRICAL IDENTITIES - PART – 70

104. (sin Ꝋ – 2sin3 Ꝋ) / (2cos3 Ꝋ – cos Ꝋ)
=
      tanꝊ   
                         

TRIGONOMETRICAL IDENTITIES - PART – 69

103. (cosec A – sin A) (sec A – cos A)
=
1 / tan A + cot A


TRIGONOMETRICAL IDENTITIES - PART – 68


TRIGONOMETRICAL IDENTITIES - PART – 67


TRIGONOMETRICAL IDENTITIES - PART – 66


TRIGONOMETRICAL IDENTITIES - PART – 65


TRIGONOMETRICAL IDENTITIES - PART – 64


TRIGONOMETRICAL IDENTITIES - PART – 63


TRIGONOMETRICAL IDENTITIES - PART – 62




TRIGONOMETRICAL IDENTITIES - PART – 61

95. (sinꝊ – cosꝊ + 1) / (sinꝊ + cosꝊ – 1)
=
(1/ (secꝊ – tanꝊ)  

TRIGONOMETRICAL IDENTITIES - PART – 60

94. (cos2 A + tan2 A – 1) / sin2 A = tan2

TRIGONOMETRICAL IDENTITIES - PART – 59

93. [(1 + cot A) / cos A] + [(1 + tan A) / sin A]
=
2 (sec A + cosec A)

TRIGONOMETRICAL IDENTITIES - PART – 58

92. sin2 A tan A + cos2 A cot A + 2 sin A cos A
=
tan A + cot A

Tuesday 23 July 2019

TRIGONOMETRICAL IDENTITIES - PART – 57

91. sin A (1 + tan A) + cos A (1 + cot A)
=
Sec A + cosec A

TRIGONOMETRICAL IDENTITIES - PART – 56

90. (1 + sec A) / sec A = [sin2 A / (1 – cos A)]



TRIGONOMETRICAL IDENTITIES - PART – 55

89. [(1 + cos A)/sin A]2
=
[(1 + cos A) / (1 – cos A)]


TRIGONOMETRICAL IDENTITIES - PART – 54

88. [1 / cos A + sin A – 1] 
[1 / cos A + sin A + 1] 
cosec A + sec A

TRIGONOMETRICAL IDENTITIES - PART – 53

87. [tan A + 1 / cos A]2 + [tan A – 1 / cos A)]2
=
[1 + sin A / 1 – sin A]

TRIGONOMETRICAL IDENTITIES - PART – 52

86. [(1 – 2 sin2A)2 / (cos4 A + sin4 A)]
=
2 cos2A – 1

TRIGONOMETRICAL IDENTITIES - PART – 51

85. [(sin A – 2 sin3A) / (2cos2 A + cos A)] 
tan A

TRIGONOMETRICAL IDENTITIES - PART – 50

84. [(cos3A + sin3A) / (cos A + sin A)] 
+
 [(cos3A – sin3A) / (cos A – sin A)] 
2

TRIGONOMETRICAL IDENTITIES - PART – 49

83. [sin A – sin B / cos A + cos B] 
+
[cos A + cos B / sin A + sin B]
0


TRIGONOMETRICAL IDENTITIES - PART – 48

82. [cot A / 1 – tan A] + [tan A / 1 – cot A]
=
sec A cosec A + 1


TRIGONOMETRICAL IDENTITIES - PART – 47

81. [Cot A / 1 – tan A] + [tan A / 1 – cot A]
=
1 + tan A + cot A

TRIGONOMETRICAL IDENTITIES - PART – 46

80. (sin A + cos A) (tan A + cot A) 
sec A + cosec A 

TRIGONOMETRICAL IDENTITIES - PART – 45

79. (sin A + cos A) (sec A + cosec A)
=
2 + sec A cosec A


TRIGONOMETRICAL IDENTITIES - PART – 44

 78. (1 + tan A tan B)2 
(1 – tan A tan B)2
                                             =
Sec2A – sec2

TRIGONOMETRICAL IDENTITIES - PART – 43

77. (sin A + cos A / sin A –  cos A) 
+
(sin A –  cos A / sin A +  cos A) 
=
(2 /2sin2 A – 1)

TRIGONOMETRICAL IDENTITIES - PART – 42

76. (cos A cot A / 1 + sin A) 
1 – cosec A 

TRIGONOMETRICAL IDENTITIES - PART – 41

75. (sin A tan A / 1 – cos A) 
1 + sec A

TRIGONOMETRICAL IDENTITIES - PART – 40

74. (cosec A – 1 / cosec A + 1)
   =
       (cos A / 1 + sin A)2

TRIGONOMETRICAL IDENTITIES - PART – 39

73. (cosec A – sin A) (sec A – cos A) sec2 A
tan A

TRIGONOMETRICAL IDENTITIES - PART – 38

72.  sec2 A + cosec2
tan A + cot A


TRIGONOMETRICAL IDENTITIES - PART – 37

71. cos A (1 + cot A) + sin A (1 + tan A) 
sec A + cosec A

TRIGONOMETRICAL IDENTITIES - PART – 36

70. (sin A – cos A) (1 + tan A + cot A) 
[(sec A / cosec2 A) – (cosec2 A / sec2 A)]


APPLICATION OF TRIGONOMETRY – PART – 33

PROBLEM – 38
For the top of a hill, the angles of depression are found to be 30 degrees and 45 degrees respectively. Find the distances of the two stones form the foot of the hill.

APPLICATION OF TRIGONOMETRY – PART – 32

PROBLEM – 37
An aeroplane flying horizontally 1 km above the ground and going away from the observer is observed at an elevation of 60 degrees. After 10 seconds, its elevation is observed to be 30 degrees; find the uniform speed of the aeroplane in km per hour.

APPLICATION OF TRIGONOMETRY – PART – 31

PROBLEM – 36
The length of the shadow of a tower standing on level plane is found to be 2y metres longer when the sun’s altitude is 30 degrees then when it was 45 degrees. Prove that the height of the tower is y (√3 + 1) meters.

APPLICATION OF TRIGONOMETRY – PART – 30

PROBLEM – 35
The horizontal distance between two towers is 75 m and the angular depression of the top of the first tower as seen from the top of the second, which is 160 m high, is 45 degrees. Find the height of the first tower.


APPLICATION OF TRIGONOMETRY – PART – 29

PROBLEM – 34
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60 degrees. When he moves 40 m away  from the bank, he find the angle of elevation to be 30 degrees. Find
(i) the height of the tree, correct to 2 decimal places.
(ii) the width of the river.



Wednesday 17 July 2019

APPLICATION OF TRIGONOMETRY – PART – 28

PROBLEM – 33
A man in a boat rowing away from a lighthouse 150 m high, takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60 degrees to 45 degrees. Find the speed of the boat.

APPLICATION OF TRIGONOMETRY – PART – 27

PROBLEM – 32
A man on a cliff observes a boat, at an angle of depression 30 degrees, which is sailing towards the shore to the point immediately beneath him. Three minutes later, the angle of depression of the boat is found to be 60 degrees. Assuming that the boat sails at a uniform speed, determine:
(a) how much more time it will take to reach the shore?
(b) the speed of the boat in metre per second, if the height of the cliff is 500 m.

Tuesday 16 July 2019

APPLICATION OF TRIGONOMETRY – PART – 26

PROBLEM – 31
From the top of the cliff, 60 m high, the angles of depression of the top and bottom of a  tower are observed to be 30 degrees and 60 degrees. Find the height of the tower.



APPLICATION OF TRIGONOMETRY – PART – 25

PROBLEM – 30
The angle of elevation of the top of a tower is observed to be 60 degrees. At a point, 30 m vertically above the first point of observation, the elevation is found to be 45 degrees. Find
(i) the height of the tower
(ii) its horizontal distance from the points of observation.

APPLICATION OF TRIGONOMETRY – PART – 24

PROBLEM – 29
From the figure, given below, calculate the length of CD.


APPLICATION OF TRIGONOMETRY – PART – 23

PROBLEM – 28
Two pillars of equal heights stand on either side of a roadway, which is 150 m wide. At a point in the roadway between the pillars the elevation of the tops of the pillars are 60 degrees and 30 degrees; find the height of the pillars and position of the point.


Sunday 14 July 2019

APPLICATION OF TRIGONOMETRY – PART – 22

PROBLEM – 27
From the top of a light house 100 m high, the angle of depression of two ships are observed as 48 degrees and 36 degrees respectively. Find the distance between the two ships if:
(a) the ships are on the same side of the light house.
(b) the ships are on the opposite sides of the light house.



APPLICATION OF TRIGONOMETRY – PART – 21

PROBLEM – 26
Find the height of a building, when it is found that on walking towards it 40 m in a horizontal line through its base the angular elevation of its top changes from 30 degrees to 45 degrees.

Saturday 13 July 2019

APPLICATION OF TRIGONOMETRY – PART – 20

PROBLEM – 25
Find the height of a tree when it is found that on walking away from it 20 m, in a horizontal line through its base, the elevation of its top changes from 60 degrees to 30 degrees.


APPLICATION OF TRIGONOMETRY – PART – 19

PROBLEM – 24
In the fig, given below, it is given that AB is perpendicular to BD and is of length X metres. DC = 30 m. Triangle ADB = 30 degrees and Triangle ACB = 45 degrees. Without using tables, find X.


APPLICATION OF TRIGONOMETRY – PART – 18

PROBLEM – 23
A vertical pole and a vertical tower are on the same level ground. From the top of the pole the angle of elevation of the top of the tower is 60 degrees and angle of depression of the foot of the tower is 30 degrees. Find the height of the tower if the height of the pole is 20 m.


APPLICATION OF TRIGONOMETRY – PART – 17

PROBLEM – 22
Form a point on the ground, the angle of elevation of the top of a vertical tower is found to be such that its tangent is 3 / 5. On walking 50 m towards the tower, the tangent of the new angle of elevation of the top of the tower is found to be 4 / 5. Find the height of the tower.

APPLICATION OF TRIGONOMETRY – PART – 16

PROBLEM – 21
The angle of elevation of a stationary cloud from a point 25 m above a lake is 30 degrees and the angle of depression of its reflection in the lake is 60 degrees. What is the height of the cloud above the lake-level.


Wednesday 10 July 2019

APPLICATION OF TRIGONOMETRY – PART – 15

PROBLEM – 20
A man on the top of the vertical observation tower observes a car moving at a uniform speed directly towards it. If it takes 12 minutes for the angle of depression to change from 30 degrees to 45 degrees, how soon after this will the car reach the observation tower.

APPLICATION OF TRIGONOMETRY – PART – 14

PROBLEM – 19
An observer on the top of the cliff; 200 m above the sea-level, observes the angles of depression of the two ships to be 45 degrees and 30 degrees respectively. Find the distance between the ships, if the ships are
(a) on the same side of the cliff
(b) on the opposite sides of the cliff.

Tuesday 9 July 2019

APPLICATION OF TRIGONOMETRY – PART – 13

PROBLEM – 18
From the top of a cliff 92 m high, the angle of depression of a buoy is 20 degrees. Calculate to the nearest metre, the distance of the buoy from the foot of the cliff.

APPLICATION OF TRIGONOMETRY – PART – 12

PROBLEM – 17
A man stands 9 m away from a flag pole. He observes that angle of elevation of the top of the pole is 28 degrees and the angle of depression of the bottom of the pole is 13 degrees. Calculate the height of the pole.

APPLICATION OF TRIGONOMETRY – PART – 11

PROBLEM – 16
Two climbers are at points A and B on a vertical cliff face. To an observer C, 40 m from the foot of the cliff, on the level ground. A is at an elevation of 48 degrees and 57 degrees. What is the distance between the climbers?



APPLICATION OF TRIGONOMETRY – PART – 10

PROBLEM – 15
Two vertical poles are on either side of a road. A 30 m long ladder is placed between the two poles. When the ladder rests against one pole. When the ladder rests against one pole, it makes angle 30 degree with pole and when it is tuned to rest against another pole, it makes angle 30 degrees with the road. Calculate the width of the road.



APPLICATION OF TRIGONOMETRY – PART – 09

PROBLEM – 14
At a particular time, when the sun’s altitude is 30 degrees, the length of the shadow of a vertical tower is 45 m. calculate (a) the height of the tower (b) the length of the shadow of the same tower, when the sun’s altitude is (i) 45 degrees (ii) 60 degrees.



APPLICATION OF TRIGONOMETRY – PART – 08

PROBLEM – 13
The angle of elevation of the top of an unfinished tower at a point distance 80 m from its base is 30 degrees. How much higher must the tower be raised so that its angle of elevation at the same point may be 60 degrees.



Thursday 4 July 2019

APPLICATION OF TRIGONOMETRY – PART – 07

APPLICATION OF TRIGONOMETRY – PART – 07
PROBLEM – 12
The upper part of a palm tree, broken over by the wind makes an angle of 45 degrees with the ground; and the  distance from the root to the point where the top of the tree touches the ground is 15 m. What was the height of the tree before it was broken.

APPLICATION OF TRIGONOMETRY – PART – 06

APPLICATION OF TRIGONOMETRY – PART – 06
PROBLEM – 11
The length of the shadow of a vertical tower on level ground increases by 10 m, when the altitude of the sun changes from 45 degrees to 30 degrees. Calculate the height of the tower.


APPLICATION OF TRIGONOMETRY – PART – 05

APPLICATION OF TRIGONOMETRY – PART – 05
PROBLEM – 09
Two people standing on the same side of a tower in a straight line with it, measure the angle of elevation of the top of the tower as 25 degrees and 50 degrees respectively. If the height of tower is 70 m, find the distance between the two people.
PROBLEM – 10
A boy 1.5 m tall , is 25 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45 degrees and (ii) 60 degrees. Find the height of the tower in each case.