triginometry by abhishek jain for ssc exams

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TRIGINOMETRY by ABHISHEK JAIN If one angle of a triangle is 90 degrees and one of the other angles is known, the third is thereby fixed, because the three angles of any triangle add up to 180 degrees. The two acute angles therefore add up to 90 degrees: they are complementary angles. The shape of a triangle is completely determined, except for similarity, by the angles. Once the angles are known, the ratios of the sides are determined, regardless of the overall size of the triangle. If the length of one of the sides is known, the other two are determined. These ratios are given by the following functions of the known angle A, where a, b and c refer to the lengths of the sides in the accompanying figure: Sine function (sin), defined as the ratio of the side opposite the angle to the hypotenuse. Sin A = opposite/ hypotenuse = a/c Cosine function (cos), defined as the ratio of the adjacent leg to the hypotenuse. Cos A = adjacent/ hypotenuse = b/c

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Triginometry by Abhishek Jain for SSC exams by ABHISHEK JAIN (Study IQ)

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Page 1: Triginometry by Abhishek Jain for SSC exams

TRIGINOMETRY by ABHISHEK JAIN If one angle of a triangle is 90 degrees and one of the other angles is known, the third is thereby fixed, because the three angles of any triangle add up to 180 degrees. The two acute angles therefore add up to 90 degrees: they are complementary angles. The shape of a triangle is completely determined, except for similarity, by the angles. Once the angles are known, the ratios of the sides are determined, regardless of the overall size of the triangle. If the length of one of the sides is known, the other two are determined. These ratios are given by the following functions of the known angle A, where a, b and c refer to the lengths of the sides in the accompanying figure:

Sine function (sin), defined as the ratio of the side opposite the angle to the hypotenuse.

Sin A = opposite/ hypotenuse = a/c Cosine function (cos), defined as the ratio of the adjacent leg to the hypotenuse. Cos A = adjacent/ hypotenuse = b/c

Page 2: Triginometry by Abhishek Jain for SSC exams

Tangent function (tan), defined as the ratio of the opposite leg to the adjacent leg. Tan A = opposite/ adjacent = a/b = sinA/ cosA

The hypotenuse is the side opposite to the 90 degree angle in a right triangle; it is the longest side of the triangle, and one of the two sides adjacent to angle A. The adjacent leg is the other side that is adjacent to angle A. The opposite side is the side that is opposite to angle A. The terms perpendicular and base are sometimes used for the opposite and adjacent sides respectively. The reciprocals of these functions are named the cosecant (cosec or csc), secant (sec), and cotangent (cot), respectively: Cosec A = 1/sin A = c/ a Sec A = 1/ cos A = c / b Cot A = 1/ tan A = cos A / sin A = b/a Complementary angles in Trigonometry Complementary angles in trigonometry: Two angles are said to be complementary, if their sum is 900. It follows from the above definition that θ and (90 - θ ) are complementary angles in trigonometry for an acute angle θ In ΔABC, ∠B = 900 ∴ ∠A + ∠C = 900 ∠C = 900 - ∠A sin (900 - A) = cos A tan(900 - A) = cot A sec(900 - A) = cosec A cos (900 - A) = sin A cot(900 - A) = tan A cosec (900 - A) = secA Trigonometric Equations An equation involving trigonometric ratios of an angle θ (say) is said to be a trigonometric equations, if it is satisfied for all values of θ for which the given trigonometric ratios are defined. Some Trigonometric equations (Identities) are as follows: 1) Sin²θ + cos2θ= 1 2) sin2θ = 1 - cos2θ 3) cos2θ = 1 - sin 2θ 4) 1 + tan2θ = sec2θ 5) tan2θ = sec2θ - 1 6) sec2θ - tan2θ = 1 7) 1 + cot2θ = cosec2θ 8) cot2θ = cosec2θ - 1 9) cosec2θ - cot2θ = 1

Page 3: Triginometry by Abhishek Jain for SSC exams

These Trigonometric equations are true for any angle θ for which the trigonometric ratios are meaningful. Quadrants Below is a simple diagram to help you determine the sign (positive or negative) of the trig ratio in their respective quadrants. We call this diagram the ‘ASTC' diagram

For example we need to find value of sine 1300

θ Is in the 2nd quadrant and the basic angle of θ is 1300 .The basic angle is measured as the acute angle which OP makes with the x-axis. Since sine is positive in the 2nd quadrant as seen in the ‘ASTC' diagram, .Thus the trig ratio would be sine

Page 4: Triginometry by Abhishek Jain for SSC exams

Trigonometric values

Radians In trigonometry we consider π = 180°

The radian measure, θ, of the angle AOB is defined by:

S = length of arc r = radius of circle

Page 5: Triginometry by Abhishek Jain for SSC exams

To convert between degrees and radians:

1. Multiply a degree measure by0180

rad and simplify to convert to radians.

2. Multiply a radian measure byrad

1800

and simplify to convert to degrees.

Chart of some popular radians

Page 6: Triginometry by Abhishek Jain for SSC exams

Some important points:-

1. The lengths of the sides of a 450- 450- 900 triangle are in the ratio of 1: 1:√2.

2. The lengths of the sides of a 30°- 60°- 90° triangle are in the ratio of 1:√3:2

3. radian =

4. If

=2 then = 1 & If

= 2 then = 1

5. Pythagoras triplet : (3,4,5) , (5,12,13) , (8,15,17) , (7,24,25) , (9,40,41) , (20,21,29)

6. Two angles are said to be complementary, if their sum is 900.

7. Two angles are said to be supplementary, if their sum is 1800.

8. If tanA .tanB = 1 then A & B are complementary angles it means A + B = 900

Page 7: Triginometry by Abhishek Jain for SSC exams

Example 1:- cosec (90o − A) = a. cosec A b. sin A c. cos A d. tan A e. sec A Solution : cosec (90o − A) = 1/sin (90o − A) = 1/cos A = sec A. Example 2:- sec (90o − A) = a. sec A b. sin A c. cos A d. cosec A e. tan A Solution : sec (90o − A) = 1/cos (90o − A) = 1/sin A = cosec A. Example 3:- cot (90o − A) = a. tan A b. cos A c. sin A d.sec A e. cosec A Solution : cot (90o − A) = 1/tan (90o − A) = 1/cot A = tan A. Example 4:- sin (90o − A) / cos (90o − A) = a. tan A b. cot (90o − A) c. cot A d. sin A e. none Solution : We know sin (90o − A) = cos A. Similarly cos (90o − A) = sin A. Substituting these values in the expression, we get sin (90o − A)/cos (90o − A) = cos A/sin A = cot A. Example 5:- cos (90o − A) / sin (90o − A) = a. tan A b. cot A c. tan (90o − A) d. sec (90o − A) e. none Solution : Substituting the values cos (90o − A) = sin A and sin (90o − A) = cos A in the expression, we get cos (90o − A)/sin (90o − A) = sin A/cos A = tan A. Example 6:- [cosec2 A − 1] s cos (90o − A) / sin (90o − A) = a. tan A b. sin A c. cot A d. cos A Solution : The value of cosec2 A − 1 = cot2 A. Also the value of cos (90o − A) / sin (90o − A) = sin A / cos A = tan A. Substituting these values, the answer is found to be cot2A tan A = cot A. Example 7:- cot A [cos (90o − A) / sin (90o − A)] = a. tan A b. cot2 A c. tan2 A d. cot A e. 1 Solution: The value of cos (90o − A) / sin (90o − A) = sin A / cos A = tan A. Therefore the given expression reduces to cot A tan A which equals 1. Example 8:- The value of tan 45o − cos 45o sin 45o is a. 1 / 2 b. 1 c. 0 d. 3 / 4 e. 1 / 4 Solution: From the table of values of trigonometric functions, tan 45o − cos 45o sin 45o = 1 − (1/2) = 1/2. Example 9:- The value of sin2 30o + cos2 30o is a. 0 b. 1 c. 1 / 2 d. 3 / 2 e. 1 / 4 Solution: From the table of values of trigonometric functions, sin2 30o + cos2 30o = ¼ + ¾ = 1. But it should be remembered that sin2 A + cos2 A = 1, for all values of A.

Page 8: Triginometry by Abhishek Jain for SSC exams

Example 10:- The value of tan 45o + cos 0 + sin 90o is a. 2 b. 1 c. 1 / 2 d. 0 e. 3 Solution: From the table of values of trigonometric functions, tan 45o + cos 0 + sin 90o = 1 + 1 + 1 = 3. Example 11:- The value of tan 60o cos 30o − sin 60o tan 30o is a. 0 b. 1 / 2 c. 3 / 2 d. 1 e. 2 Solution: From the table of values of trigonometric functions, tan 60o cos 30o − sin 60o tan 30o = (3/2) − (1/2) = 1. Example 12:- The value of [1 + sin 60o + sin2 30o + sin2 60o + sin4 45o] [cos 30o − sin 60o] is a. 1 b. 13 / 12 c. 12 / 13 d. 1 / 2 e. 0 Solution : The value of the second term [cos 30o − sin 60o] is 0. Hence the value of the entire expression is 0, irrespective of the value of the first term. Example 13:- (sin A + cos A)2 − 2 sin A cos A = a. 2 b. 0 c. 1 d. tan A e. sin2 A − cos2 A Solution : (sin A + cos A)2 − 2 sin A cos A = sin2 A + cos2 A = 1. Example 14:- sin2 A − sec2 A + cos2 A + tan2 A = a. 1 b. 0 c. cot A d. cosec A e. cosec 2A Solution : Here the terms need to be grouped properly. The given expression can be written as (sin2 A + cos2 A) − (sec2 A − tan2 A) = 1 − 1 = 0. Example 15:- 1/(1 + cot2 A) + 1/(1 + tan2 A) = a. 0 b. sin2 A c. 1 d. cos2 A e. sin2 A/cos2 A Solution: The expression 1/(1 + cot2 A) + 1/(1 + tan2 A) = 1/cosec2 A + 1/sec2 A = sin2 A + cos2 A = 1. Example 16:- cot A tan A = a. sin A b. cos A c. sin A cos A d. 1 e. 1/(sin A cos A) Solution : cot A = 1 / tan A. Hence cot A tan A = 1. Alternatively cot A = cos A/sin A and tan A= sin A/cos A. So cot A tan A = (cos A/sin A) (sin A/cos A) = 1. Example 17:-

From the figure, the value of cosec A + cot A is a. (a + b)/c b. (b + c)/a c. a/(b + c) d. b/(a + c) e. (a + c)/b Solution:- We know cosec A = b/a and cot A = c/a. Hence cosec A + cot A = (b + c)/a.

Page 9: Triginometry by Abhishek Jain for SSC exams

Example 18:- Which of the following relationships is true: a. sin A cot A = 1 b. sin A + cosec A = 1 c. cos A sec A = 1 d. sec A - cos A = 1 e. sec A cot A = 1 Solution: cos A sec A = 1 By definition, sec A = 1 / cos A. So cos A sec A = 1 is true. Example 19:-

From the figure, the value of sin2 A + cos2 A is a. a/b + c/b b. b/a + c/b c. 1 d. (a/b + c/b)2 e. (b/a + c/b)2 Solution: This question is a bit tricky. We know sin A = a/b and cos A = c/b. So sin2 A + cos2 A = (a2 + c2) / b2. By Pythagoras Theorem, a2 + c2 = b2 for a right-angled triangle. Hence sin2 A + cos2 A = 1, which is a famous identity. Example 20:- From the figure, the value of cot C + cosec C is

a. (a + c)/b b. (a + b)/c c. (c + b)/a d. a/c + c/b e. c/a + b/c Solution: cot C is Base/Opposite Side and cosec C is Hypotenuse/Opposite Side. From these definitions, the values of cot C and cosec C are given by a/c and b/c respectively. Hence the answer is (a + b)/c. Example 21:- cosec A / sec A = a. tan A b. sin A c. cos A d. cot A e. sin A + cos A Solution: By definition, cosec A = 1 / sin A and sec A = 1 / cos A. So cosec A / sec A = cos A / sin A = cot A. Example 22:- For the figure given on the right, the value of cot A is

a. sin A / cos A b. tan C c. cos C / sin C d. a / c e. c / b Solution: The value of cot A is c/a. Similarly the value of tan C is c/a. Hence cot A = tan C.

Page 10: Triginometry by Abhishek Jain for SSC exams

Example 23:- The angle of elevation of the top of a tower 30 m high, from two points on the level ground on its opposite sides are 45 degrees and 60 degrees. What is the distance between the two points? a. 30 b. 51.96 c. 47.32 d. 81.96 Solution: Let OT be te tower. Therefore, Height of tower = OT = 30 m Let A and B be the two points on the level ground on the opposite side of tower OT. Then, angle of elevation from A = TAO = 45o and angle of elevation from B = TBO = 60o Distance between AB = AO + OB = x + y (say) Now, in right triangle ATO, AO = OT = 30 & in right triangle BTO OB = 30/ √3 = 30√3/3 = 10√3 = 10×1.732 = 17.32 Hence, the required distance = x + y = 30 + 17.32 = 47.32 m

Example 24:- If Sin(A+B) = √ and Sin(A - B) =1/2 then what are the values of A and

B? (given that both A and B are acute angles and A>B)

a. 30, 60 b. 45, 45 c. 45, 15 d. none of above

Solution:

Sin(A+B)=root3/2 (this is given in the question itself).

If you look at the table, sin 60 = √ 2. That means A+B=60………eq1

Similarly we’ll get A-B=30……….eq2

So we’ve two equations:

A+B=60 & A-B=30

Now add these two equations eq1+eq2

(A+B)+(A-B)=60+30

2A=90

A=45

Put this value back in eq1 (or eq2). And you get B=15

Final answer C: 45,15

Page 11: Triginometry by Abhishek Jain for SSC exams

Example 25:-Find the value of cos2 × cos4 × cos6 × cos8 ×….× cos92

Solution:

As you can see the angles are increasing as per the multiplication table of 2. So in the chain, you’ll also get cos90 (because 2 x 45=90). So we can write the chain as

cos2 × cos4 × cos6 × cos8×….cos88 × cos90 × cos92

but we know that cos90=0, hence the whole multiplication will become zero.

Example 26:- The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Solution:

tan30 = p/30 or p = 30/√ = 10√ = 17.32

Example 27:- A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Solution:

Sin60 = p/h =60/

Or √ /2 = 60/ h or h = 120/√ = 40√

Example 28:- A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Solution : AB = 30m (height of building) DC= EG = 1.5m (height of boy) Angle D = 30 Angle E = 60 AF = 30-1.5 = 28.5m

In AFD , tan 30 = AF/FD = 28.5/FD or 1/√ = 28.5/ FD

Page 12: Triginometry by Abhishek Jain for SSC exams

FD= 28.5√

In AFE , tan 60 = AF/FE = 28.5/FE or 1/√ = 28.5/ FE

FD= 28.5/√

Required distance = ED = 28.5√ - 28.5/√ = 19√ Example 29:- From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Solution: Height of building = DB = 20m Angle DCB = 45 In DBC , tan 45 = 1 = DB/BC or DB = BC = 20m

In ABC, tan60 = √ = AB/ BC or AB = 20√

Now AD = AB – DB = 20√ = 14.64m Example 30:- A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Solution: Height of statute = AD = 1.6m Angle ACB = 60 Angle DCB = 45 In DBC , tan45 = 1= DB/BC Or BC = DB

In ABC , tan60 = √ = AB/BC

or BC = AB / √ = (BC +16)/ √

or BC √ - BC = 16 or BC = 16/0.732 = 2.18m

Page 13: Triginometry by Abhishek Jain for SSC exams

Example 31:- The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Solution : height of tower = AB = 50m Angle ACB = 60 Angle DBC = 60

In ABC, tan60 = √ = 50/ BC or BC = 50/√

In DCB, tan30 = 1/√ = DC/ BC or DC = 50/ m Example 32:- Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Solution : BD = width of road = 80m Angle ACB = 60 Angle ECD = 30 AB = ED = height of poles

In ABC, tan 60 = √ = AB/BC

or AB = BC √ ……..(i)

In EDC, tan30 = 1/√ = ED/CD = AB/ (80-BC)

Or AB = (80 - BC)/ √ ……(ii)

Or BC √ = (80 - BC)/ √ Or 3 BC = 80 – BC

Page 14: Triginometry by Abhishek Jain for SSC exams

Or 4 BC = 80

Or BC = 20 so AB = 20 √ Example 33:- A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.

Solution: width of canal = BC CD = 20m Angle ACB = 60 Angle ADB = 30

In ABC, AB = BC √

In ABD, AB = (BC+20)/√ or 3 BC = BC + 20 or 2BC = 20 or BC =10 (width of canal) Example 34:- From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the

tower. Solution : AB = height of building = 7m Angle ACB = 45 Angle EAD = 60 In ABC , AB = BC or AB = BC = AD = 7m,

In EAD, ED = √ AD = 7√

So height of tower = DC + ED = 7 + 7√ = 7(1+√ ) = 19.124m

Page 15: Triginometry by Abhishek Jain for SSC exams

Example 35:- As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Solution : AB = 75m Angle EAC = Angle ACB = 45 Angle EAD = Angle ADB = 30 In ABC , AB = AC = 75m,

In ABD , BD = AB √ = 75√

So distance between ships C & D = CD = 75√ - 75 = 75(√ ) = 54.9m Example 36:- A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

Solution : AC = GH = height of balloon = 88.2m BC = EF = height of Girl = 1.2m Angle GFB = 60 Angle AFB = 30

In ABF, BF = 87√

In GFI, FI = 87/√

Distance traveled by balloon = BI = 58√

Page 16: Triginometry by Abhishek Jain for SSC exams

Example 37:- The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Solution: BD = 4m angle Angle ADB = Angle ACB = 90 – In ABD , AB = tan 4 In ABC, tan (90- ) = AB/BC OR AB = Cot 9 or 4tan =9 Cot ot tan2 = 9/4 or tan = 3/2 = AB/4 or AB = 6m

Example 38:- A man is walking along a straight road. He notices the top of a tower subtending an angle A = 60o with the ground at the point where he is standing. If the height of the tower is h = 30 m, then what is the distance (in meters) of the man from the tower? Solution:

Let BC represent the tower with height h = 30 m, and A represent the point where the man is standing. AB = d denotes the distance of the man from tower. The angle subtended by the tower is A = 60o. From trigonometry,

tan A = tan 60o = h / d = √3

So d = 30 / √3 s m. Hence the distance of the man from the tower is 17.32 m.

Example 39:- A little boy is flying a kite. The string of the kite makes an angle of 30o with the ground. If the height of the kite is h = 18 m, find the length (in meters) of the string that the boy has used. Solution:

Page 17: Triginometry by Abhishek Jain for SSC exams

If the kite is at C and the boy is at A, then AC = l represents the length of the string and BC = h represents the height of the kite. From the figure, sin A = sin 30o = h / l = 1 / 2. Hence the length of the string used by the little boy is l = 2 h = 2 (18) = 36 m.

Example 40:- Two towers face each other separated by a distance d = 45 m. As seen from the top of the first tower, the angle of depression of the second tower's base is 60oand that of the top is 30o. What is the height (in meters) of the second tower? Solution:

The first tower AB and the second tower CD are depicted in the figure on the left. First consider the triangle BAC. Angle C = 60o. tan BCA = tan 60o = AB / AC. This gives AB = d tan 60o. Similarly for the triangle BED, BE = d tan 30o. Now height of the second tower CD = AB − BE = d (tan 60o − tan 30o) = 45 (√3 − 1/ √3) = 45 × 2 / √3 = 51.96 m.

Example 41:- A ship of height h = 12 m is sighted from a lighthouse. From the top of the lighthouse, the angle of depression to the top of the mast and the base of the ship equal 30o and 45o respectively. How far is the ship from the lighthouse (in meters)? Solution: Let AB represent the lighthouse and CD represent the ship. From the figure, tan BCA = tan 45o = AB / AC. Similarly for the triangle BED, tan BDE = tan 30o = BE / ED. Now, AC = ED = d. Height of the ship = CD = AB − BE = d (tan 45o − tan 30o) = 12 m. Thus distance of the ship from the lighthouse d = 12 / (1 − 1 / √3 ) = 28.39 m

Example 42:- Two men on opposite sides of a TV tower of height 28 m notice the angle of elevation of the top of this tower to be 45o and 60o respectively. Find the distance (in meters) between the two men. Solution:

Page 18: Triginometry by Abhishek Jain for SSC exams

The situation is depicted in the figure with CD representing the tower and AB being the distance between the two men. For triangle ACD, tan A = tan 60o = CD / AD. Similarly for triangle BCD, tan B = tan 45o = CD / DB. The distance between the two men is AB = AD + DB = (CD / tan 60o) + (CD / tan 45o) = (28 / √3) + (28 / 1) = 44.17 m.

Example 43:- Two men on the same side of a tall building notice the angle of elevation to the top of the building to be 30o and 60o respectively. If the height of the building is known to be h =50 m, find the distance (in meters) between the two men. Solution:

In the figure, A and B represent the two men and CD the tall building. tan A = tan 30o = DC / AC = h / AC; and tan B = tan 60o = DC / BC = h / BC. Now the distance between the men is AB = x = AC − BC = (h / tan 30o) − (h / tan 60o) = (50 √3 ) − (50 / √3 ) = 57.73 m.

Example 44:- A pole of height h = 40 ft has a shadow of length l = 40.00 ft at a particular instant of time. Find the angle of elevation (in degrees) of the sun at this point of time. Solution:

In the figure, BC represents the pole and AB its shadow. tan A = BC / AB = h / l = 40 / 40.00 = 1.000 From trigonometric tables, we note that tan A = 1.000 for A =45o. Hence the angle of elevation of the sun at this point of time is 45o.

Example 45:- You are stationed at a radar base and you observe an unidentified plane at an altitude h = 6000 m flying towards your radar base at an angle of elevation = 30o. After exactly one minute, your radar sweep reveals that the plane is now at an angle of elevation = 60o maintaining the same altitude. What is the speed (in m/s) of the plane? Solution:

Page 19: Triginometry by Abhishek Jain for SSC exams

In the figure, the radar base is at point A. The plane is at point D in the first sweep and at point E in the second sweep. The distance it covers in the one minute interval is DE. From the figure, tan DAC = tan 30o = DC / AC = h / AC. Similarly, tan EAB = tan 60o = EB / AB = h / AB. Distance covered by the plane in one minute = DE = AC − AB = (h / tan 30o) − (h / tan 60o) = (6000 √3) − (6000 / √3 ) = 6928.20 m. The velocity of the plane is given by V = distance covered / time taken = DE / 60 = 115.47 m/s.

Example 46:- cos2 (

) - sin2 (

) is equal to

Solution:- We know that cos2A – sin2B = cos (A + B) cos (A – B), therefore

cos2 (

)– sin2 (

) = cos (

) cos (

)

= cos

cos x =

√ cos x.

Example 47:- If sec θ - tan θ = ½, then θ lies in which quadrant? Solution:- sec θ – tan θ = ½ sec θ + tan θ = 2 ( sec2 θ – tan2 θ = 1) sec θ = 5/4 and tan θ = ¾. cos θ = 4/5 and sin θ = 4/5 ¾ = 3/5 As both sin θ and cos θ are +ve the angle θ lies in the 1st quadrant.

Example 48:- If sin (A + B + C) = 1, tan (A – B) = 1/√ , sec(A + C) = 2, then Solution:- As sin (A + B + C) = 1 A + B + C = 90o

As, tan (A – B) = 1/√ & sec (A + C) = 2 A – B = 30 & A + C = 60o From the above statements we can conclude, A = 60o, B = 30o, C = 0o OR As sin (A + B + C) = 1

A + B + C = 900 {Note: we can easily rule out options (1) & (3) as A + B + C > 900} Now checking option (2)

sin 900 = 1, tan 30 = 1/√ , sec 60 = 2, thus satisfy.

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Example 49:- Find the value of tan0× tan1 × tan2 × tan3×..….tan89?

Solution:-

From the table we know that tan 0 = 0. So no matter what you multiply with zero, final answer will always be zero.

Example 50:-Find value of tan1 x tan 2 x tan3 x ….x…tan88 x tan89

Solution:-

you make pairs of complimentary numbers: 1+89=90, 2+88=90…. There is only one angle

left who doesn’t get a pair (45)

So it’ll look like this = (tan1 x ta89) (tan2 x tan88)x…xtan45

In each of those pairs, you convert one tan into its complimentary cot.

= (tan1 x tan(90-89)) x (tan2 x cot(90-88))…..x1 ; because tan45=1

= (tan1xcot1) x (tan2xcot2)x…..x1 = 1 because tan and cot are inverse of each other.

Example 51:-. Find value of tan48 x tan23 x tan42 x tan67

Solution:-

When you get this type multiplication chain question, you’ve to find out the pair of

complimentary angles. Here 48 + 42=90 and 23 +67=90 so I’ll club them together

In both parts, we’ll convert any one tan into cot, then tan x cot = 1 (because they’re inverse

of each other).

Final answer =1.

Example 52:- On the level ground, the angle of elevation of the top of the tower is 30o. On moving 2 meters nearer, the angle of elevation becomes 60o. What is the height of the tower? Solution:-

= tan 600

Page 21: Triginometry by Abhishek Jain for SSC exams

= tan 300

= 3 x = 1 h = √

Example 53:- The angle of elevation of the top of the tower observed from each of the three points A, B and C on the ground, forming a triangle is the same angle α, If R is the circum-radius of the triangle ABC, then the height of the tower is Solution:- Since the tower makes equal angles at the vertices of the triangle, therefore foot of the tower is at the circum centre. From Δ OAP, we have tan α = OP/A

or OP = OA tan A or OP = R tan OP = OA tan A or OP = R tan α Example 54:- If the length of a chord of a circle is equal to that of the radius of that circle, then the angle subtended (in radians) at the centre of the circle by the chord is Solution:- In this case, the chord and the two radii joining the centre to the ends of the chord make an equilateral triangle; so the angle subtended is 60o = π/3 radians. Example 55:- In a triangle ABC, A = 45°, then the value of (1 + cot B) (1 + cot C) is equal to Solution:- A = 45o or B + C = 180o – 45 = 135o tan (B + C) = – 1 or tan B + tan C = – 1 + tan B tan C

or

+

= -1 +

or cot B + cot C + cot B cot C = 1 or 1 + cot B + cot C + cot B cot C = 2. or (1 + cot B)(1 + cot C) = 2.

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or A = 450 A + B + C = 1800 B + C = 180 – 45o = 135o Lets take, B = 900 & C = 450 or (1 + cot B) (1 + cot C) = (1 + cot 900) (1 + cot 450) = (1 + 0) (1 + 1) = 1 × 2 = 2 Example 56:- If cos( A - B) = 3/5 and tanA tanB = 2, then a. cos A cos B = 1/5 b. sin A sin B = – 2/5 c. cos (A + B) = 1/5 d. sin A sin B = 4/5 Solution:- cos (A – B) = 3/5 and tan A tan B = 2 or cos A cos B + sin A sinB = 3/5 and sin A sin B = 2 cos A cos B or 3 cos A cos B = 3/5 cos A cos B = 1/5. Example 57:- cos 1° cos 2° cos 3° …. cos 179° is equal to Solution:- The given product contains the factor cos90o = 0. Thus the value of the product becomes 0. Example 58:- If sin2 θ + 3 cos θ – 2 = 0, then cos3 θ + sec3 θ is equal to Solution:- Given sin2 θ + 3 cos θ – 2 = 0 cos2 θ – 3 cos θ + 1 = 0 or cos2 θ + 1 = 3cos θ

or cos θ +

= 3

Cubing both sides, we get

cos3 θ +

+ 3 (cos θ +

) = 27

or cos3 θ + sec3 θ = 27 – 3 × 3 = 18

Example 59:-. Find the value of cos18°/sin72°

Solution:-

We know that cos and sin are complimentary. cosA=sin(90-A)

So for Cos18, You can write cos18=sin(90-18)=sin72. Let’s use it

Cos18/sin72 =sin72/sin72 (because cos18=sin72) =1 (because numerator and denominator

are same so they’ll cancel each other.)

Example 60:- Find value of sec70° x sin20° – cos20° x cosec70°

Solution:-

Page 23: Triginometry by Abhishek Jain for SSC exams

If you convert all four (sin, cos, cosec, and sec) into their complimentary (cos, sin, sec and

cosec) then you’ll run into infinite loop.

In the part A, if I convert sin into its complimentary cos, then sec x cos =1 because sec and

cos are inverse of each other. Same way in part B if I convert cosec into its complimentary

sec, that too will lead to 1. Let’s see

=1-1 = 0 is the final answer.

So, tan and cot are complimentary

1. tan A = cot (90-A) but not valid for 90 degrees because tan90 is not defined. 2. cotA=tan(90-A) but this is not valid for 0 degree because cot0 is not defined.

The tan and cot’s complimentary relationship is also import for multiplication chain

questions because tan and cot are also inverse of each other. That is tan A x cot A= tan A x

1/tan A=1.

Page 24: Triginometry by Abhishek Jain for SSC exams

Exercise

Q1. If Cos A = 1 2sin230°, then find value of A ?

a. 30 b. 45 c. 60 d. 90

Q2. If cos 60° = cos2A sin230°, then find value of A?

a. 30 b. 45 c. 60 d. 90

Q3. If sinA=2sin30° cos30°, then find value of A?

a. 30 b. 45 c. 60 d. 90

Q4. If sinA=2tan30°/ (1 tan230°), then find value of A?

a. 30 b. 45 c. 60 d. 90

Q5. If cosA=(1 tan230°)/( 1 tan230°), then find value of A?

a. 30 b. 45 c. 60 d. 90

Q6. If cosA=4cos330° 3cos30°, then find value of A?

a. 30 b. 45 c. 60 d. 90

Q7. Find value of (5cos260° 4sec230° tan245°)/(sin230° cos230°)?

a. 67 b. 12 c. 67/12 d.12/67

Q8. Find value of 3cos230° sec230° 2cos0° 3sin90 tan260°?

a. 67 b. 12 c. 67/12 d.12/67

Q9. If sin A= cos A, what is the value of 2tan2A sin2A 1

a. 2 b.7 c.7/2 d.2/7

Q10. What is the value of sin Acos B+ cosA sinB, if A=30° and B=60°

a. 0 b. ½ c.2 d. 1

Q11. What is the value of cosAcosB sinAsinB, if A=30° and B=60°

a. 0 b. ½ c.2 d. 1

Q12. secA = cosec60°, what is the value of 2cos2A 1

Page 25: Triginometry by Abhishek Jain for SSC exams

a. 0 b. ½ c.2 d. 1

Q13. (tan10° x tan15° x tan75° x tan80°)

a. 0 b. 1 c. 3 d. 2

Q14. What is the value of 5cos90° cot30°+(sin60°/cos245°)

a. 1 b. 3 c.0 d. 5

Q15. Find the value of (cos30°+sin60°)/(sin30°+cos60°+1)

a. ½ b. √ c. √ /2 d. 2/√

Q16. Find the value of tan260°/(sin245°+cos245°)

a. 3 b. 1/2 c.2/3 d. 1/3

Q17. If cot(A+B)=1/√ and Cot(A B)= √ . Find the values of A and B

a. 30,60 b. 60,30 c. 15, 45 d. 45,15

Q18. Find the value of cos10° × cos20° × cos30° ×………× cos90°

a. 1 b. ½ c. 0 d. Not defined.

Q19. If secA cosecA=0 then find value of secA + cosecA, given that A is an acute angle.

a. b. 2√ c. 0 d. Not defined.

Q20. Given that 4A is an acute angle and sec4A = cosec (A 20), what is the value of A?

a. 22 b. 23 c. 24 d. 25

Q21. Find value of cosec32 - sec58 ?

a. 1 b. 2 c. 3 d. 0

Q22. Find value of cot12 x cot48 x cot52 x cot60 x cot78

a. 1 b. 0 c. √ d. 1/√

Q23. (tan10° x tan45° x tan40° x tan50° x tan80°)

a. 0 b. 1 c. 3 d. 2

Q24. Cos48 °x cosec42° + sin48° x sec42°

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a. 0 b. 1 c. 2 d. 3

Q25. (cos70°/sin20°)+(cos59° x cosec31°)

a. 1 b. 2 c. 3 d. 0

Q26. (sec20°/cosec70°)+[(cos55° x cosec35°)/(tan5° x tan25° x tan45° x tan65° x tan85°)]

a. 1 b. 2 c. 3 d. 0

Q27. (sin70°/cos20°)+(cosec20°/sec70°)-(2cos70° x cosec20°)

a. 1 b. 2 c. 3 d. 0

Q28. If sin3A = cos(A 26), then what is the value of A? given that 3A is an acute angle.

a. 21 b. 23 c.24 d. 29

Q29. If sec2A = cosec(A 42), then what is the value of A? given that 2A is an acute angle.

a.22 b.33 c.44 d. 55

Q30. If sinA = cos30°, what is the value of 2tan2A tan45, given that A is an acute angle. a.3 b.4 c.5 d.7

Q31. If sin7x = cos11x then the value of tan9x + cot9x is a.1 b.2 c. 3 d.4

Q32. If cot2A = tan(A+6), find the value of A. given that 2A and A+6 are acute angles? a.18 b.28 c.30 d. none of above

Q33. tan 10 tan 20 tan 880 tan 890

a. 0 b. 1 c. 3 d. 1/√ Q34. tan 40 tan 430 tan 470 tan 860

a. 0 b. 1 c. 3 d. 1/√

Q35. If θ θ

θ θ = 2 then the value of sinθ is

a. 2/√ b. √ /2 c. 1/2 d. 1

Q36. If cosec 390 = x then the value of

+ sin2 39 + tan2 51

is

a. √ b. √ c. d. Q37. If tan A = 2/3 find the value of (3sin A + 4 cos A)/ (3sin A – 4 cosA) a. -3 b. - 4 c. - 5 d. None of these

Page 27: Triginometry by Abhishek Jain for SSC exams

Q38. sin2 1 + sin2 3 + sin2 5 + sin2 7 + …. + sin2 89 = a.22.5 b.23 c. 24 d. 25 Q39. sin4 A cos4 A = a. 1 b. 0 c. sin2 A − cos2 A d. tan2 A Q40. [(sec A − tan A)(sec A + tan A)] + [(cosec A − cot A)(cosec A + cot A)] a. 1 b. 0 c. 2 d. 1 / 2

Q41. The value of

a.2 b. 1 c. 1/4 d. 1/2 Q42. If sin A = 1/3 then cos A cosec A + tan A sec A =

a. √

b.

c. d.

Q43. If sin A = 3/5, then find the value of 4 tanA + 3 sin A is equal to a. 6 cos A b. 6 sec A c. 6 sec A d. tan A

Q44. If cos A = 0.96 then

is equal to

a. 0.98 b. 2 c. 4 d. 7 Q45. If sec - cos = 3/2 ( is a positive acute angle), then sec is equal to a. 2 b. – 1/ 2 c. 0 d. 1/2 Q46. The value of [1 + sin 60o + sin2 30o + sin2 60o + sin4 45o] [cos 30o − sin 60o] is a. 1 b. 13 / 12 c. 12 / 13 d. 0 Q47. If tan2 tan4 = 1 then the value of tan3 =

a. 1/√ b. 0 c. 1 d. √

Q48. If tan 22

= x then cos67

a.

√ b.

√ c. d. √

Q49. If cos 43 =

√ , then the value of tan 47

a.

√ b.

√ c.

d.

Q50.

=

√ then the value of in the circular measure will be

a. /6 b. /4 c. /12 d. /3 Q51. If tan + cot = 2 where 0 < < 900 ; find the value of tan17 + cot19 . a. 1 b. 2 c. 3 d. 4

Page 28: Triginometry by Abhishek Jain for SSC exams

Q52. sin2 2 + sin2 4 + sin2 6 + sin2 8 + ……… + sin2 88= a. 22.5 b. 23 c.24 d. 22 Q53. tan A / sin A = a. cosec A b. sin A c. sec A d. 1 / sin A Q54. (sin A / tan A) + cos A = a. 2 sec A b. sec A c. 2 cos A d. 2 cosec A Q55. tan 10 tan 20 …………………….tan 880 tan 890 a. 0 b. 1 c. 2 d.3 Q56. If tan(x+y) tan(x-y) = 1 then the value of tanx =

a. 1/√ b. 0 c. 1 d. √ Q57. 1/(tan 10 tan 20 …………………….tan 1880 tan 1890 ) = a. 0 b.1 c. 2 d. 3 Q58. 2 cosec2230 cot2 670 – sin2230- sin2670 – cot2670 = a. 0 b. 1 c. sec2230 d. tan2230 Q59. The length of the shadow of a vertical tower on level ground increases by 10 m when the altitude of the sun changes from 450 to 300. Then the height of the tower is

a. 10√ b. 5√ c. 10(√ d. 5(√ +1)

Q60. If 5 tan =4 then

=

a. 1/3 b. 2/3 c.1/4 d. 1/6 Q61. If 2(cos2 sin2 ) = 1( is a positive acute angle), then cot is equal to

a. √ b. √ c. 1/√ d. 1 Q62. If tan C = 11, then find value of sin2 C + cos2 C is equal to a. 2 b. 0 c. 1 d. 1/2

Q63. In a right angle triangle ABC, right angled at B, the ratio of AB to AC is 1: √ then 2 tan A / (1- tan2 A) is equals to: a. 2 b. 1 c. 3 d. undefined Q64. If Sin A =( a2 - b2 )/(a2 + b2) ; then tan A = a. a2 + b2/2ab b. 2ab/(a2 - b2) c. a2 - b2/2ab d. tan A Q65. cos (90o − A) / sin (90o − A) = a. tan A b. cot A c. tan (90o − A) d. sec (90o − A) Q66. [cosec2 A − 1] cos (90o − A) / sin (90o − A) =

Page 29: Triginometry by Abhishek Jain for SSC exams

a. tan A b. sin A c. cot A d. cos A Q67. cot A [cos (90o − A) / sin (90o − A)] = a. tan A b. cot2 A c. tan2 A d. 1 Q68. The value of tan 45o − cos 45o sin 45o is a. 1/2 b. 1 c. 0 d. 3/4 Q69. sin2 A − sec2 A + cos2 A + tan2 A = a. 1 b. 0 c. cot A d. cosec A Q70. cot A tan A = a. sin A b. cos A c. sin A cos A d. 1 Q71. Find the value of sin2 1 + sin2 2 + sin2 88 + sin2 89 a. 3 b. 2 c. 3 d. none of these Q72. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

a. 8 b. 16 c. 8 √ d. None of these Q73. Upper part of a tree broken over by the wind makes an angle of 45° with the ground, and the horizontal distance from the foot of the tree to the point where the top of the tree touches the ground is 12m. Find the height of the tree before it was broken. a. 12m b. 12+12/√3 c. 12(1+√3) d. None of these Q74. From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60oand the angle of depression of the foot of the tower is 30o. Find the height of the tower. a. 15 b. 28 c. 32 d. None of these Q75. A man standing on the deck of a ship, which is 16m above the water level, observe the angle of elevation of the top of cliff as 60° and the angle of depression of the base of the cliff as 30° .Calculate the distance of the cliff from the ship and the height of the cliff. a. 16 √3 m, h=64m b. 20 √3 m, h=32 m c. 15 √3 m, h=64m d. none of these Q76. An aeroplane when flying at a height of 5000m from the ground passes vertically above another aero plane at an instant when the angles of the elevation of the two planes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the aero planes at the instant. a. 2116.5 b. 2115 c. 2113.5 d. none of these Q77. If distance between two pillars of length 16 & 9 m is x meters. If two angle of elevation of their respective top from a point on ground of other are complementary to each other, then value of x is

Page 30: Triginometry by Abhishek Jain for SSC exams

a. 7 b. 16 c. 12 d. 9 Q78. Evaluate: sin2 25o + sin2 65o. a. 2 b. 0 c. 1 d. -1 Q79. The angle of elevation of the top of a tower 30 m high, from two points on the level ground on its opposite sides are 45 degrees and 60 degrees. What is the distance between the two points? a. 30 b. 51.96 c. 47.32 d. 81.96 Q80. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

a. 10 b. 10 √ c. 11.32 d. 41.96 Q81. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

a. 20 (√ ) b. 20 √ c. 20/√ d. 41.96 Q82. A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60°with the wall, find the height of the wall. a. 7.5√3 b. 7.5 c. 5√3 d. None of these Q83. A pole 12 m high casts a shadow 4 √3 m long on the ground. Find the angle of elevation a. 30° b. 60° c. 90° d. None of these Q84. The angle of elevation of the top of a tower from a point on the ground is 30° if on walking 30m towards the tower, the angle of elevation becomes 60°.Find the height of the tower. a. 15√3 b. 7√3 c. 5√3 d. None of these Q85. The angle of elevation of the aero plane from a point on the ground is 60°.After 15 seconds flight, the angle of elevation changes to 30° .If the aero plane is flying at a height of 1500 √3m .Find the speed of the plane a. 200m/s b. 450m/s c. 250 m/s d. None of these Q86. An observer 1.5m tall is 20.5m away from a tower 22m high. Determine the angle of elevation of the top of the tower from the eye of the observer. a. 75° b. 60° c. 45° d. None of these

Q87. If 3 sin2 + 7 cos2 = 4, then the value of tan -is (where 0 < < 90°)

a. √ b. √ c. √ d. √

Q88. If tan – cot = 0, 0° < < 90°, the value of (sin – cos) is

a. 1 b. 2 c. –2 d. 0

Page 31: Triginometry by Abhishek Jain for SSC exams

Q89. If 0 < <

, then sin + cos is always

a. greater than 1 b. less than 1 c. equal to 1 d. greater than 2

Q90 If sec2 + tan2 = 7, then the value of , when 0° < < 90°, is

a. 60° b.30° c. 0° d. 90°

Q91. If

= 3, then the vlaue of sin4 – cos4 is

a.

b.

c.

d.

Q92 If 2 cos – sin =

√ , (0° < < 90°) the value of 2 sin + cos is

a.

√ b. √ c.

√ d.

Q93. If 0° < < 90° and sin + cos =

then the vlaue of sin – cos is

a.

b.

c.

d.

Q94. If

= 7, then the value of tan is equal to:

a.

b.

c.

d.

Q95. If

= 2, (0 < < 90°), then the value of sin is

a.

√ b.

c.

d. 1

Q96. If sec (4 – 50°) = cosec(50° – ), then the value of , when 0° < < 90°, is

a. 33

b.18° c. 3

d.30°

Q97. If cos + sec = √ , then the value of cos3 + sec3 is

a. –1 b. √ c. 0 d. 1

Q98. The maximum value of 24 sin + 7 cos is

a. 24 b. 25 c. 7 d. 17

Q99. If tan 2. tan 4 = 1, then the value of tan 3 is

a.

b. 2 c. 0 d. 1

Q100. In "ABC, A is a right angle and AD is perpendicular to BC. If AD = 4cm, Bc = 12 cm,

then the value of (cost B cot C) is

a. 4 b.

c. 6 d. 3

Q101. If + = 90° and : = 2 : 1, then the value of sin : sin is:

Page 32: Triginometry by Abhishek Jain for SSC exams

a. 1 : 1 b. √ : 1 c. √ : 1 d. 2 : 1

Q102. If sec + tan = 2, then sec is equal to

a.

b.

c.

d.

Q103. The value of [0° < < 90°] for which

= 4 is

a. 45° b. 60° c. 30° d. none of these

Q104. If 2y cos = x sin and 2x sec – y cosec = 3, then the relation between x and y is

a. 2x2 + y2 = 2 b. x2 + 4y2 = 4 c. x2 + 4y2 = 1 d. 4x2 + y2 = 4

Q105. In a right-angled triangle ABC, AB = 2.5cm, cos B = 0.5, ACB = 90°.

The length of side AC, in cm, is

a.

√ b.

√ c. 5√ d.

Q106. If tan – cot = a and cos – sin = b, then the value of (a2 + 4)(b2 – 1)2 is:

a. 1 b. 2 c. 3 d. 4

Q107. If x = cosec = sin and y = sec – cos, then the value of x2y2 (x2 + y2 + 3) is:

a. 0 b. 1 c. 2 d. 3

Q108. In a right-angled triangle ABC, B is the right angled and AC = 2√ cm.

If AB – BC = 2cm, then the value of (cos2A – cos2C) is:

a.

b.

c.

d.

Q109. If 0 <

, 2y cos = x Sin and 2x sec – y cosec = 3, then the value of x2 + 4y2 is:

a. 1 b. 2 c. 3 d. 4

Q110. If cos. cosec 23° = 1, the value of is

a. 23° b. 37° c. 63° d. 67°

Q111. If sin(3x – 20°) = cos(3y + 20°), then the value of (x + y) is

a. 20° b. 30° c. 40° d. 45°

Q112. If cos =

, then the value of

is

a.

b.

c.

d.

Q113. If sin 2 =

, then the value of cos (90 – ) is

a. 1 b.

c.

d.

Q114. If

= 2 with 0 < < 90°, then what is equal to?

Page 33: Triginometry by Abhishek Jain for SSC exams

a. 30° b. 45° c. 60° d. 75°

Q115. If 5 tan = 4, then the value of

is:

a.

b.

c.

d.

Q116. If x, y are positive acute angles, x + y < 90° and sin (2x – 20) = cos (2y + 20°), then the value of sec (x + y) is

a. √

√ c.

d.

Q117. If (a2 +b2) sin + 2ab cos = a2 + b2, then the value of tan is

a.

(a2 – b2) b.

(a2 – b2) c.

(a2 + b2) d.

(a2 + b2)

Q118. If A + B = 90°, then the value of sec2A + sec2B – sec2A.sec2B is:

a. 0 b. 1 c. 2 d. 3

Q119. If cos + sec = √ then the value of cos3 + sec3 is:

a. 0 b. 1 c. –1 d. √

Q120. If sin(x + y) = cos[3(x + y)] then the value of tan [2(x + y)] is

a. √ b. 1 c. 0 d.

Q121. If X + 1/X= then the value of cos2 + sec2 is:

a. 0 b. 1 c. 2 d. √

Q122. The least value of 4 cosec2a + 9 sin2a is:

a. 14 b. 10 c. 11 d. 12

Q123. If sec – cosec = 0, then the value of (sec + cosec) is:

a. √

b.

√ c. 0 d. 2√

Q124. If cosx = siny and cot (x – 40) = tan (50 – y) then the value of x and y are:

a. 1 b. 1 c. 1 d. 2

Q125. If xsin3 + ycos3 = sin cos, and xSin – ycos = 0 then value of (x2 + y2)

a. 1 b. sin – cos c. sin + cos d. 0

Q126. If sin + cos = √ cos (90 – ), then cot is

a. √ + 1 b. 0 c. √ d. √ – 1

Q127. If sec + cosec (90 –) = 4, (0 < < 90°) then the value of tan is:

a.

√ b. 1 c. √ d.

Page 34: Triginometry by Abhishek Jain for SSC exams

Q128. If √ .cos(5x + 5°) = cot 45°, then the value of x in degree is:

a. 10 b. 8 c. 11 d. 0

Q129. If sin – cos = 0, find the vlaue of sin(

) + cos(

)

a. 0 b. 1 c. √ d. 2√

Q130. If sin =

, then the value of sec + cos will be:

a. √ a b. a c.

√ d.

Q131. If sin (x – 2y) = cos (4y – x), then the value of cot 2y is:

a. 0 b. 1 c.

√ d. undefined

Q132. If tan (

) √ , value of cos is:

a. 0 b.

√ c.

d. 1

Answers

1. C 2. A 3. C 4. C 5.C 6. D 7. C 8. C 9. C 10. D

11. A 12. B 13. B 14. C 15. A 16. A 17. D 18. C 19.B 20. A

21. D 22.D 23. B 24. C 25. A 26. B 27. D 28. D 29. C 30. C

31. B 32. B 33. B 34. B 35. B 36. C 37. A 38. A 39. C 40. C

41. B 42. A 43. A 44. D 45. A 46. D 47. C 48. B 49. C 50. D

51. B 52. D 53. C 54. C 55. B 56. C 57. A 58. C 59. D 60. D

61. C 62. C 63. D 64. B 65. A 66. C 67. D 68. A 69. B 70. D

71. B 72. C 73. D 74. B 75. A 76. C 77. C 78. C 79. C 80. B

81. A 82. B 83. B 84. A 85. A 86. C 87. A 88. D 89. A 90. A

91. C 92. C 93. D 94. B 95. B 96. D 97. C 98. B 99. D 100.D

101. C 102. B 103.B 104. B 105.A 106.D 107.B 108.A 109.D 110.D

111.B 112.C 113.B 114.B 115.A 116.A 117.B 118.A 119. A 120. B

121.C 122.D 123.D 124.B 125.A 126.D 127.C 128.B 129.A 130.B

131.D 132.C

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