Trigonometric Identities and Equation Eng

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MEASUREMENT OF ANGLES

There are two systems used for the measurement of angles.

Sexagesimal system:

Here a right angle is divided into 90 equal parts known as degrees. Each degree is divided into 60 equal parts

called minutes and each minute is further divided into 60 equal parts called seconds.

60 seconds (or 60”) = 1 minute (or 1’)

60 minutes (or 60) = 1 degree (or 1°)

90 degrees (or 90°) = 1 right angle

Circular Measurement:

In this system a unit called ‘Radian’ is defined as follows:

One Radian (1c) =

)r (circleof Radius

'r 'magnitudeof lengtharc

i.e. one radian corresponds to the angle subtended by arc of length ‘r’

at the centre of the circle.

Since the ratio is independent of the size of a circle it follows that ‘radian’ is a

constant quantity. The circumference of a circle is always equal to times its

diameter or 2 times its radius.

1

A

BC

D

O r

r

Fig.1

For a general angle, e.g. AOD =r

' AD'arcradian(s).

Remember that angles at the centre of a circle are in proportion to the arc.

i.e.

2

2

r

r

ACarc

ABarc

AOC

AOBwhere AOB = 1

c=

180)angleright(2 AOC2

Note:

is a real number whereas cstands for 180°.

Remember the relation radians = 180° = 200g

1 Radian =2

a right angle =

0180

= 1800 0.3183098862….. = 57.2957795

0= 57

01744.8 nearly.

Illustration 1:

(i). Express 45 20 10 in radian system

(ii). The interior angles of a polygon are in A.P. and the smallest angle is 120 and common

difference 5. Find the number of sides of the polygon.

(iii). Reduce 94g2287 to sexagesimal measure.

Solution:

(i) 0.7C

(ii) n = 9 (iii) 8404853.388



Trigonometric Functions

In a right angled triangle ABC, CAB = A and BCA = 90° = /2. AC is the base, BC the altitude and AB is the

hypotenuse. We refer to the base as the adjacent side and to the altitude as the opposite side. There are sixtrigonometric ratios, also called trigonometric functions or circular functions. With reference to angle A, the

six ratios are:

BC opposite side= .

AB hypotenuseis called sine of A, and written as sin A

AC adjacent side= .

AB hypotenuseis called the cosine of A, and written as cos A

BC opposite side=

AC adjacent sideis called the tangent of A, and written as tan A.

B

C A Fig. 2

Obviously,sinA

tan A =cosA

. The reciprocals of sine, cosine and tangent are called the cosecant, secant and

cotangent of A respectively. We write these as cosec A, sec A, cot A respectively.

Important Notes:

Since the hypotenuse is the greatest side in a right angle triangle, sin A and cos A can never be

greater than unity and cosec A and sec A can never be less than unity. Hence sin A 1, cos A1,cosec A 1, sec A1, while tan A and cot A may have any numerical value lying between - to + .

All the six trigonometric functions have got a very important property in common that is periodicity.

Remember that the trigonometrical ratios are real numbers and remain same so long as the angle

remains same.

SOME BASIC RESULTS

cos2A + sin

2A = 1 cos

2A = 1 - sin

2A or sin

2A = 1 - cos

2A

1 + tan

2

A = sec

2

A sec

2

A - tan

2

A = 1 cot

2A + 1 = cosec

2A cosec

2A-cot

2A = 1

sinA cosA

tanA = and cotA =cosA sinA

Fundamental inequalities: For 0 < A <

2, 0 < cosA <

sinA

A<

1

cosA.

It is possible to express trigonometrical ratios in terms of any one of them

e.g.2 2

1 cotθ 1sinθ = , cosθ = , tanθ =

cotθ1+cos θ 1+ cot θ

22 1+cot θ

cosecθ = 1+cot θ, secθ = cotθ



TRIGONOMETRIC RATIOS OF ANY ANGLE

Consider the system of rectangular co-ordinate axes dividing the plane into four quadrants. A line OP makes

angle with the positive x-axis. The angle is said to be positive if measured in counter clockwise direction

from the positive x-axis and is negative if measured in clockwise direction. The positive values of the

trigonometric ratios in the various quadrants are shown, the signs of the other ratios may be derived. Note

that xoy = /2, xox' = , xoy' = 3/2

Oxx

y

y

P1

Q1

P4P3

P2

Q2

Q3 Q4

Oxx

y

y

quadrant I (A) All ratio + ve

quadrant IV (C)

cos, sec + ve

quadrant II (S)

sin, cosec + ve

quadrant III (T)

tan, cot + ve

PiQ i is positive if above the x-axis, negative if below the x-axis, OPi is always taken positive. OQ i is positive if

along positive x-axis, negative if in opposite direction.

i

iiii

OP

QPOPQsin

i

iii

OP

OQOPQcos

i

iiii

OQ

QPOPQtan (Where i = 1, 2, 3, 4 )

Thus depending on signs of OQ i and PiQ i the various trigonometrical ratios will have different signs.

TABLE

equals sin cos tan cot Sec cosec

– – sin cos –tan – cot sec –cosec 90° – cos sin cot tan cosec sec90° + cos – sin –cot – tan –cosec sec

180°– sin – cos – tan – cot – sec cosec180°+ – sin – cos tan cot – sec –cosec

360°–

– sin cos – tan – cot sec –cosec360°+ sin cos tan cot sec cosec

Note:

Angle and 90°– are complementary angles, and 180°– are supplementary angles

sin(n + (–1)n) = sin, n

cos(2n ± ) = cos, n tan(n + ) = tan, n

i.e. sine of general angle of the form n + (–1)n will have same sign as that of sine of angle and so on. The

same is true for the respective reciprocal functions also.

BASIC FORMULAE



TABLE 1

sin (A + B) = sin A cos B + cos A sin B

sin (A – B) = sin A cos B – cos A sin B

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

tanA + tanB

tan A+B = 1- tanAtanB

tanA - tanB

tan A -B =1+ tanAtanB

sin (A + B) sin (A – B) = sin2

A – sin2

B = cos2

B – c o s2

A.

cos (A + B) cos (A – B) = cos2

A – sin2

B = cos2

B – sin2

A.

sin2A = 2sinA cosA =2

2tanA

1+ tan A

cos2A = cos2A - sin

2A = 1-2 sin

2A = 2cos

2A-1 =

2

2

1- tan A

1+ tan A

tan2A =2

2tanA

1- tan A

sin3A = 3sinA - 4sin3A = 4sin(60° - A) sinAsin(60° + A)

cos3A = 4cos3A - 3cosA = 4cos(60° - A) cosAcos(60°+A)

3

o o

2

3tanA - tan Atan3A = = tan 60 - A tanAtan 60 + A

1-3tan A

TABLE 2

A +B A -BsinA + sinB = 2sin cos2 2

A -B A +B

sinA - sinB = 2sin cos2 2

A -B A +B

cosA +cosB = 2cos cos2 2

B - A A +B

cosA - cosB = 2sin sin2 2

(Here notice (B – A)!)

tanA + tanB = sin A +B

cosA.cosB 2sinAcosB = sin(A + B) + sin (A - B)

2cosAsinB = sin(A + B) - sin (A - B)

2cosAcosB = cos(A + B) + cos(A - B)

2sinAsinB = cos(A - B) - cos (A + B)

Illustration 2: If in a ABC, cos3 A + cos

3B + cos

3C = 3cosA cosB cosC, then prove that the triangle is

equilateral.

Solution: Given that cos3A + cos

3B + cos

3C – 3cosA cosB cosC = 0

(cosA + cosB + cosC) (cos2A + cos

2B + cos

2C – cosAcosB – cosB cosC – cosCcosA) = 0

cos2

A + cos2

B + cos2

C – cosAcosB – cosBcosC – cosCcosA = 0(as cosA + cosB + cosC = 1 + 4 sinA/2 sinB/2 sinC/2 0)

(cosA – cosB)2

+ (cosB – cosC)2

+ (cosC – cosA)2

= 0



cosA = cosB = cosC

A = B = C, 0 < A, B, C < .

ABC is equilatral.

Illustration 3: If

3 3sin θ cos θ =

sin 2 θ + cos 2θ + , prove that tan2 = 2tan(3 + ).

Solution:

3 3si n θ cos θ=

si n 2θ + α cos 2θ + α= k

4 4si n θ cos θ= = k

si nθsi n 2θ + α cosθcos 2θ + α

=

4 4cos θ- sin θ

cosθcos 2θ + α - sinθsin 2θ+ α=

cos2θ

cos 3θ+α

Again

3 3si n θcosθ cos θsi nθ= = k

si n 2 θ + α cosθ c os 2θ + α si nθ

3 3si n θcosθ + cos θsi nθ

si n 2θ+α cosθ+ cos 2θ+ α si nθ

=

3sin2

2sin

3sin

cossin

3sin2

2sin

3cos

2cos

tan2 = 2 tan(3 + ).

Illustration 4: For any real , find the maximum value of cos2( cos ) + sin

2(sin ) .

Solution: The maximum value of cos2( cos) is 1 and that of sin

2( sin) is sin

21, both exists

for = /2. Hence maximum value is 1+ sin21.

Illustration 5: If t an3θ si n3θ

t anθ si nθ= 4,then find the value of .

Solution:an3θ

= 4t anθ

tan2 = 1/11

now,sin3θ

sinθ= 3 – 4 sin

2 = 3 - 4

2

1 8=

1+ cot θ 3

Illustration 6: If A, B,C and D are angles of a quadrilateral and sinA

2 sin

B

2 . sin

C

2 . sin

D

2 =

1

4 , prove

that A = B = C = D = /2.

Solution: GivenA B C D

2sin .sin . 2sin .sin 12 2 2 2

A B A B C D C Dcos cos cos cos 1

2 2 2 2

Since, A + B = 2 - (C + D), the above equation becomes,

A B A B C D A B

cos cos cos cos2 2 2 2

= 1

2 A B A B A B C Dcos cos cos cos 1

2 2 2 2

A B C Dcos cos

2 2

= 0.



This is quadratic equation in cos which has real roots.

2

A B C D A B C Dcos cos 4 1 cos .cos

2 2 2 2

0

2

A B C Dcos cos

2 2

4

A B C Dcos cos 2,2 2 Now both cos A B

2 and cos C D

2 1

A B C D

cos 1 cos2 2

A B C D0

2 2

A = B, C = D.

Similarly, A = C, B = D A = B = C = D = /2.

Illustration 7: If A, B and C are angles of a triangle, prove that

B C C A A Bcos cos cos

2 2 2

E 6B C C A A Bcos cos cos2 2 2

Solution: Since A + B + C = B C C A A B

cos cos cos2 2 2

EB C C A A B

cos cos cos2 2 2

B C C A A Bcos cos cos

2 2 2 A B C

sin sin sin2 2 2

=

B C A C A B A B C 2cos cos 2sin cos2cos cos2 2 2 22 2

sinA sinB sinC

=sinB sin C sinC sin A sin A sinB

sinA sinB sinC

=sinB sin A sinC sinB sin A sinC

sin A sinB sinB sinC sinC sin A

as A, B, C are angles of 0 < A, B, C <

sin A, sin B, sin C > 0

E 2 + 2 + 21

as x 2, if x 0x

E 6

Illustration 8: If cos(A + B) sin (C + D) = cos(A – B) sin(C – D),

prove that cotA cotB cotC = cotD.

Solution: We have cos (A + B) sin(C + D) = cos(A – B) sin(C – D)

i.e.cos(A B) sin(C D)

cos(A B) sin(C D)



cos(A B) cos(A B) sin(C D) sin(C D)

cos(A B) cos(A B) sin(C D) sin(C D)

or2cos A cosB 2sinCcosD

2sinA sinB 2cosCsinD

cotA cotB = tanC cotD

or cotA cotB cotC = cotD.

Illustration 9: Show that 2 3 4 5 1

cos cos cos cos cos =11 11 11 11 11 32

Solution:2 3 4 5 1

cos cos cos cos cos =11 11 11 11 11 32

LHS =2 3 4 5

cos cos cos cos cos11 11 11 11 11

Let11

=

= cos cos 2 cos 3 cos 4 cos 5= – cos cos 2 cos 4 cos 8 cos 5= – cos 2

0 cos 21 cos 2

2 cos 23 cos 5

= –

4

4

sin2

2 sincos 5 = –

16 5sin cos

11 11

16sin11

=

5 52sin cos

111 11 =32

32sin11

.

Illustration 10: Prove that cot 7

0 1

2 = 2 + 3 + 4 + 6

Solution: Let = 7

01

2 2 = 15

0

Now cot =

0

0

1+cos2θ 1+ cos15=

sin2θ sin15

=

3 +11+

2 2 + 3 +12 2 =3 -1 3 -1

2 2

= 2 + 3 + 4 + 6 .

Illustration 11: If 2tan2 tan

2 tan

2 + tan

2 tan

2 + tan

2 tan

2 + tan

2 tan

2 = 1, prove that sin

2

+ sin2 + sin

2 = 1.

Solution: We have, 2tan2 tan

2tan2 + tan

2tan2 + tan

2tan2 + tan

2tan2 = 1

2 + cot2 + cot

2 + cot2 = cot

2 cot2 cot

2 cosec

2 + cosec2 + cosec

2 – 1

= (cosec2 – 1) (cosec

2 – 1) (cosec2 – 1)

cosec2 + cosec2 + cosec2 – 1

= – 1 + cosec2 + cosec

2 + cosec2 – (cosec

2 cosec2 + cosec

2 cosec2 +

cosec2cosec

2 + cosec2 cosec

2 cosec2



cosec2 cosec

2 + cosec2 cosec

2 + cosec2 cosec

2= cosec

2 cosec2 cosec

2 sin2 + sin

2 + sin2 = 1

IDENTITIES

A trigonometric equation is an identity if it is true for all values of the angle or angles involved. A givenidentity may be established by reducing either side to the other one, or reducing each side to the same

expression, or any convenient modification of these.

For any angles A, B, C

sin (A + B +C) = sinA cosB cosC + cosA sinB cosC + cosA cosB sinC- sinA sinB sinC

cos (A + B +C) = cosA cosB cosC- cosA sinB sinC - sinA cosB sinC - sinA sinB cosC

tanA + tanB+ tanC- tanA tanB tanC

tan (A +B + C) =1- tanA tanB - tan BtanC - tanA tanC

;

cotAcotBcotC - cotA - cotB - cotCcot (A +B +C) =cotA cotB + cot BcotC+ cotA cotC -1

If A, B, C are angles of a triangle (or A + B + C = ):

sinA cosB cosC + cosA sinB cosC + cosA cosB sinC = sinA sinB sinC

cosA sinB sinC + sinA cosB sinC + sinA sinB cosC = 1 + cosA cosB cosC

tanA + tanB + tanC = tanA tanB tanC

cotB cotC + cotC cotA + cotA cotB = 1

1=2

Btan

2

Atan+

2

Atan

2

Ctan+

2

Ctan

2

Btan

2

Ccot

2

Bcot

2

Acot

2

Ccot

2

Bcot

2

Acot

sin2A + sin2B + sin2C = 4sinA sinB sinC

cos2A + cos2B + cos2C = -1-4cosA cosB cosC

cos2A + cos

2B + cos

2C = 1 - 2cosA cosB cosC

2

Ccos

2

Bcos

2

Acos4CsinBsin Asin

2

Csin

2

Bsin

2

Asin41CcosBcos Acos

Illustration 12: If x + y + y = xyz, Prove that

2 2 2 2 2 2

x y z 4xyz + + = .

1 - x 1 - y 1 - z (1 - x )(1 - y )(1 - z )

Solution: Let x = tanA, y = tanB, z = tanC

tanA + tanB + tanC = tanA. tanB. tanC.

A + B + C = tan(2A + 2B) = tan(2 – 2C)

or tan(2A + 2B) = -tan2C

or tan2A + tan2B + tan2C = tan2A.tan2B.tan2C

or2 2 2 2 2 2

2x 2y 2z 8xyz

1 x 1 y 1 z (1 x )(1 y )(1 z )

2

2tanAtan2A

1 tan A

or,2 2 2 2 2 2x y z 4xyz .

1 x 1 y 1 z (1 x )(1 y )(1 z )



Illustration 13: If A + B + C = 180 0 , prove that

sin (B + C – A) + sin (C + A – B) + sin (A + B – C) = 4 sin A sin B sin C.

Solution: sin (B + C – A) + sin (C + A – B) + sin (A + B – C) = 4 sin A sin B sin C

LHS = sin (B + C – A) + sin (C + A – B) + sin (A + B – C)

= sin ( – A – A) + sin ( – B – B) + sin ( – C – C) ( A + B + C = )

= sin 2A + sin 2B + sin 2C

= 4 sin A sin B sin C

TRIGONOMETRIC SERIES

If we have a cosine series in its product form where the angles are in G.P. with common ratio 2 then multiply

both numerator and denominator by 2 sin (least angle).

Illustration 14: Simplify the product cosA cos2Acos22

A …. Cos2n–1

A.

Solution: cosA cos2A….cos2n–1

A = Asin2

1 (sin2Acos2A)cos2A….cos2

n–1A

= ) A2cos A2(sin Asin2

1 ……cos2

n–1A = ) A2cos A2(sin

Asin2

12

……cos2n–1

A

Continue like this, finally we have = Asin2

A2sinn

n

Note:

Asin2

A2sin A2cos

n

n1n

0r

r

where denotes products .

If we have a cosine series or a sine series in its sum form where the angles are in A.P. then multiply

both numerator and denominator with 2sin

2

differencecommon.

Illustration 15: Prove that cos7

2 + cos

7

4+ cos

7

6 = –1/2.

Solution: cos7

2+ cos

7

4+ cos

7

6=

7sin2

7sin2

7

6cos+

7

4cos+

7

2cos

=2

1

7sin2

75sinsin

73sin

75sin

7sin

73sin

.

Note:

2

Bsin

2

nBsinB

2

1n Asin

B1r Asinn

1r



2

Bsin

2

nBsinB

2

1n Acos

B1r Acosn

1r

. Where denotes summation.

Illustration 16: Sum to n–terms of the series

sin – sin( α + β ) + sin( α + 2β ) – sin( α + 3β )+ …Solution: sin( ) = –sin and sin(2 ) = sin

–sin( ) = sin( + )

sin( 2 ) = sin(2 + 2 ) – sin( 3 ) = sin(3 + 3 ) and so on. Using

these results, required sum is

S=sin +sin( + )+sin(2 + 2 ) +sin(3 + 3 )+… to n terms

= sin + sin ( + ) + sin( + 2 ) + sin( + 3 )+... to n terms

=

sinn.2 .sin n 1.

2sin2

nsin

n 12sin sin( ) sin( 2 ) .....ton terms sin2

sin2

TRIGONOMETRIC EQUATIONS

An equation involving one or more trigonometrical ratios of unknown angle is called a trigonometric

equation e.g. cos2

x – 4 sin x = 1

It is to be noted that a trigonometrical identity is satisfied for every value of the unknown angle where as

trigonometric equation is satisfied only for some values (finite or infinite) of unknown angle.

e.g. sec2

x – tan2

x = 1 is a trigonometrical identity as it is satisfied for every value of x R.

SOLUTION OF A TRIGONOMETRIC EQUATION

A value of the unknown angle which satisfies the given equation is called a solution of the equation e.g. sin

= ½ = /6 .

General SolutionSince trigonometrical functions are periodic functions, solutions of trigonometric equations can be generalized

with the help of the periodicity of the trigonometrical functions. The solution consisting of all possible solutions of

a trigonometric equation is called its general solution.

We use the following formulae for solving the trigonometric equations:

sin = 0 = n,

cos = 0 = (2n + 1)2

,

tan = 0 = n,

sin = sin = n + (–1)n, where [–/2, /2]

cos = cos = 2n , where [ 0, ]

tan = tan = n + , where ( –/2, /2)

sin2 = sin

2 , cos2 = cos

2 , tan2 = tan

2 = n ,



sin = 1 = (4n + 1)2

,

cos = 1 = 2n ,

cos = –1 = (2n + 1),

sin = sin and cos = cos = 2n + .

Note: Everywhere in this chapter n is taken as an integer, If not stated otherwise.

The general solution should be given unless the solution is required in a specified interval.

is taken as the principal value of the angle. Numerically least angle is called the principal value.

Method for finding principal value

Suppose we have to find the principal value of satisfying the equation sin = –1

2.

Since sin is negative, will be in 3rd or 4th quadrant. We can approach 3rd or 4th

quadrant from two directions. If we take anticlockwise direction the numerical value

of the angle will be greater than . If we approach it in clockwise direction the angle

will be numerically less than . For principal value, we have to take numericallysmallest angle.

So for principal value :

1. If the angle is in 1 st or 2nd quadrant we must select anticlockwise direction and if the angle if the

angle is in 3rd or 4th quadrant, we must select clockwise direction.

2. Principal value is never numerically greater than .

3. Principal value always lies in the first circle (i.e. in first rotation)

On the above criteria will be6

or

5

6

. Among these two

6

has the least numerical value.

Hence6

is the principal value of satisfying the equation sin = –

1

2.

Algorithm to find the principle argument:

Step 1: First draw a trigonometric circle and mark the quadrant, in which the angle may lie.

Step 2: Select anticlockwise direction for 1st and 2nd quadrants and select clockwise direction for

3rd and 4th quadrants.

Step 3: Find the angle in the first rotation.

Step 4: Select the numerically least angle among these two values. The angle thus found will be the

principal value.

Step 5: In case, two angles one with positive sign and the other with negative sign qualify for the

numerically least angle, then it is the convention to select the angle with positive sign as

principal value.

Example 1: If tan = – 1, then will lie in 2nd or 4th quadrant.

/ 6/ 6

Y

Y'

X' X

B

A



/ 4

3 / 4

Y

Y'

X' X

For 2nd quadrant we will select anticlockwise and for 4th quadrant. we will select

clockwise direction.

In the first circle two values4

and

3

4

are obtained.

Among these two,4

is numerically least angle. Hence principal value is

4

.

Example 2: If cos =1

2

, then will lie in 1st

or 4th

quadrant.

/ 3

Y

Y'

X' X

/ 3

O

For 1st quadrant, we will select anticlockwise direction and for 4th quadrant, we will

select clockwise direction.

In the first circle two values3

and

3

are thus found.

Both3

and –

3

have the same numerical value. In such case

3

will be selected as

principal value.

Illustration 17: Solve cot (sinx + 3) = 1.

Solution: sinx + 3 =4

n

44

n2

n = 1 sinx = 34

x =

3

45sin1n 1n or

3

43sin1n 1n

Illustration 18: If sin 5x + sin 3x + sin x = 0, then find the value of x other than zero, lying between 0 x

2

.

Solution: sin 5x + sin 3x + sin x = 0 (sin 5x + sin x) + sin 3x = 0

2 sin 3x cos 2x + sin 3x = 0 sin 3x(2 cos 2x + 1) = 0

sin 3x = 0; cos 2x = –2

1 3x = n, 2x = 2n

3

2

The required value of x is3

.



Illustration 19: Find all acute angle such that cos cos 2 cos 4 =8

1.

Solution: It is given that cos cos2 cos4 =8

1

2sin cos cos2 cos4 =4

sin 2sin2 cos2 cos4 =

2

sin

2sin4 cos4 = sin sin8 – sin = 0

2sin7

2

cos

2

9= 0

Either sin2

7= 0

n

2

7 =

7

n2

For n = 0 = 0 which is not a solution.

=7

n2 n = 1, i.e. =

7

2

or cos 02

9

2

9= (2n + 1)

2

= (2n + 1)

9

=

3,

9

Hence =3

,9

,72 .

Illustration 20: Solve for x: 0 )2 ( log )2 ( log 2 )2 ( log )2 ( log x cos x sin x cossin 2 2 2

x 2 .

Solution: 0xcoslogxsinlog

2

xcoslog

1

xsinlog

12

22

22

22

2

0xcoslogxsinlog

2xsinlogxcoslog2

22

2

22

22

2xcos.xsinlog 222

22

x2sinlog

2

2

4

1

2

x2sin2

2

1

2

x2sin

sin2x = 1 2x = (2n + 1)2

x = (2n+1)

4

, n I

OBJECTIVE ASSIGNMENT

1: The general value of satisfying both 2

1

sin

and 3

1

tanθ is :

(A) 2n (B) 2n + 7/6

(C) n + /4 (D) 2n + /4

Solution: Let us first find out lying between 0 and 360°.

Since2

1sin

= 210° or 330°

and3

1tan = 30° or 210°

Hence = 210° or6

7 is the value satisfying both.



The general value of

n,

6

7n2

Hence (B) is the correct answer.

2: 3 cosec20° - sec20° =

(A) 1 (B) 2

(C) 3 (D) 4

Solution: Given =

20cos20sin

20sin20cos3

20cos

1

20sin

3

40sin

20sin60cos20cos60sin.4

20cos20sin2

20sin2

120cos

2

3.4

=sin 40

4 4sin 40

°=

°Hence (D) is the correct answer.

3: tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A =

(A) Cot A (B) tan 6A

(C) cot 4A (D) None of these

Solution: tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A

= tanA + 2tan2A + 4tan4A + 8 A4tan2

A4tan1 2

4 A2tan2 Atan A4cot4 A2tan2 Atan A2tan2

A2tan1 2

Atan

Atan1 Atan A2cot2 Atan

2 = cot A

Hence (A) is the correct answer.

4: The value of sin 12°. sin48°.sin54° =

(A) 1/8 (B) 1/6

(C) 1/4 (D) 1/2

Solution: sin 12°. sin48°.sin54° =

54sin2

154sin36cos

2

154sin60cos36cos

2

1

= 54sin18sin90sin4

154sin54sin36cos2

4

1

= 36cos18sin214

118sin54sin14

1

=

18cos

36cos36sin1

4

136cos18cos

18cos

18sin21

4

1

=1 2sin 36 cos 36 1 sin 72 1 1 1

1 1 14 2 cos18 4 2 sin 72 4 2 8

é ù é ù é ù- = - = - =ê ú ê ú ê úë ûë û ë û

Alternative Method

Let = 12°

sin 12°. sin48°.sin54° =

54sin72sin48sin12sin72sin

1

36cos8

36cos

36cos36sin8

54sin36sin

72sin

54sin123sin

4

1

=

1

8




5: The smallest positive value of x (in degrees) for which

tan(x + 100°) = tan(x + 50°) tan x tan(x - 50°) is :

(A) 30° (B) 45°

(C) 60° (D) 90°

Solution: The relation may be written as xtan50xtan50xtan

100xtan

xcos50xcos

xsin50xsin

100xcos50xsin

50xcos100xsin

50x2cos50cos

50x2cos50cos

150sin50x2sin

150sin50x2sin

50x2cos

50cos

150sin

50x2sin cos50°+ 2sin(2x + 50°) cos(2x + 50°) = 0

cos50°+ sin (4x + 100°) = 0 cos50° + cos(4x + 10°) = 0

cos(2x + 30°) cos(2x – 20°) = 0 x = 30°, 55°

The smallest value of x = 30°Hence (A) is the correct answer.

6. The most general value of satisfying 3 – 2cos –4sin –cos2 + sin2 =0:

(A) 2n (B) 2n + /2

(C) 4n (D) 2n + /4

Solution: 3 – 2cos – 4 sin – cos 2 + sin 2 = 0

3 – 2cos – 4 sin – 1 + 2sin2 + 2sin cos = 0

2sin2 – 2cos – 4sin + 2sin cos + 2 = 0

(sin2 – 2sin + 1) + cos (sin – 1) = 0

(sin – 1)[sin – 1 + cos ] = 0

either sin = 1

= 2n + /2 where n I

or, sin + cos =1

cos( – /4) = cos(/4) – /4 = 2n /4

= 2n, 2n + /2 where n I

Hence = 2n, 2n + /2.

Hence (A, B) is the correct answer.

7: If sin = 3sin( + 2), then the value of tan ( + ) + 2tan is:

(A) 0 (B) 2

(C) 4 (D) 1Solution: Given sin = 3sin ( + 2)

sin ( + ) = 3sin ( + + )

sin ( + ) cos – cos( + ) sin= 3sin ( + ) cos + 3cos ( + ) sin –2sin ( + ) cos = 4cos ( + ) sin

si n ( θ + α ) 2si nα

=cos ( θ + α ) cosα

tan(+) + 2tan = 0


8: The minimum value of 3tan2 + 12 cot2 is:

(A) 6 (B) 8

(C) 10 (D) None of these



Solution: A.M. G.M 1

2(3tan

2 +12 cot2 ) 6

3 tan2 +12cot

2 has minimum value 12.

Hence (D) is the correct answer.

9: If A + B + C = o180 then the value of tanA + tanB + tanC is :

(A) 3 3 (B) 2 3

(C) > 3 3 (D) > 2 3

Solution: tan(A + B) = tan(180

– C)

or,tanA tanB

1 tanA tanB

= tanC

or, tanA + tanB + tanC = tana tanB tanC

3tanA tanB tanCtanA tanBtanC

3

[since A.M. G.M.]

or, tanA tanB tanC 3

tanA tanB tanCor,

2tan A2tan B

2tan C 27 [cubing both sides]

or tanA tanB tanC 3 3

tanA + tanB + tanC 3 3 .


10: Let 0 < A, B <2

satisfying the equalities 3

2 sin A + 22 sin B = 1 and 3sin2A – 2sin2B = 0.

Then A + 2B = :

(A)

p

4 (B)

p

3

(C)2

(D) None of these.

Solution: From the second equation, we have

sin2B =3

2sin2A …(1)

and from the first equality

32sin A = 1 –2

2sin B = cos2B …(2)

Now cos (A + 2B) = cosA. cos2B – sinA . sin2B

= 3 cosA .2

sin A –

3

2 . sinA . sin2A

= 3cosA.2sin A – 3

2sin A . cosA = 0

A + 2B =2

or

3

2

Given that 0 < A <2

and 0 < B <

2

0 < A + 2B < +

2

Hence A + 2B =2

.

Hence (C) is the correct answer.

11: If a cos3 + 3a cos sin

2 = x and a sin3 + 3a cos

2 sin = y, then (x + y)2/3

+ (x – y)2/3

=

(A) 2a2/3

(B) a2/3

(C) 3a2/3

(D) 2a1/3



Solution: a cos3 + 3a cos sin

2 = x

a sin3 + 3a cos

2 sin = y

x + y = a[sin3 + cos

3 + 3 sin cos (sin + cos )] = a(sin + cos)3

3/1

a

yx

= sin + cos ……(1)

x – y = a[cos3 – sin

3 + 3 cos sin2 – 3 cos

2 sin ] = a[cos – sin]3

3/1

a

yx

= cos – sin ……(2)

(sin + cos )2

+ (cos – sin )2

=3/2

3/23/2

a

)yx()yx(

2 (sin2 + cos

2 ) =3/2

3/23/2

a

)yx()yx(

(x + y)2/3

+ (x – y)2/3

= 2a2/3

.


12: If + 1 + a t anα = 1 + 1 - a , then sin4 =

(A) a/2 (B) a

(C) a2/3

(D) 2a

Solution: Let a = sin 4 1+ a = cos 2 + sin 2 and 1- a = cos 2 – sin 2

(1 + 1+ a ) tan = (1 + 1- a )

(1 + cos 2 + sin 2) tan = 1 + cos 2 – sin 2

2cosθ cosθ + sinθ

2cosθ cosθ - sinθ= cot

cosθ+sinθ

cosθ-sinθ = cot

1+ tanθ

= -cotα1- tanθ

tan +θ4

= tan +2

= -4

a = sin 4 = sin ( – 4) = sin 4 Hence (B) is the correct answer.

13: If cos2 = 2 1

a - 13

and tan2 θ

2 = tan

2/3, then cos2/3 + sin

2/3 =

(A) 2a2/3

(B)

2/32

a

(C)æ öç ÷è ø

1/32

a(D) 2a

1/3

Solution: cos2 =

3

1a2 , tan

2

2

= tan

2/3

tan3

2

= tan

cos

sin

2cos

2sin

3

3



cos2

cos

sin2

sin 33

= k

sin3

2

= k sin ……(1)

cos3

2

= k cos ……(2)

k2/3

sin2/3 + k

2/3cos = 1

sin2/3 + cos

2/3 =3/2k

1

Squaring and adding (1) and (2)

k2(sin

2 + cos2 ) = sin

6

2

+ cos

6

2

=

2

cos2

sin2

cos2

sin32

cos2

sin 2222

3

22

k

2

= 1 – 4

3

sin

2

= 1 – 4

3

+ 4

3

cos

2

k2

=4

a2

k =2

a

sin2/3 + cos

2/3 =

3/2

a

2

.


14: If 3 sin2 + 2 sin

2 = 1 and 3 sin 2 –2 sin 2 = 0, where , are positive acute angles, then

+ 2 =

(A)2

(B)3

p

(C)4

p(D)

6

p

Solution: 3 sin2 + 2 sin

2 = 1 ……(1)

3 sin 2 = 2 sin 2 ……(2)

3 sin2 = 1 – 2 sin

2 = cos 23 sin sin = cos 2 ……(3)

from equation (2)

3 . 2 sin cos = 2 sin 2

3 sin =

cos2sin

from equation (3)

sin

cos

2sin= cos 2

cos cos 2 – sin sin 2 = 0

cos ( + 2) = 0

+ 2 =2

.


15: The value of ( )5

1

cos 2 111

r

r

p

=

-å is :



(A)1

2(B)

1

3

(C)1

4(D)

1

6

Solution:

5

1r 111r 2cos

11

9cos

11

7cos

11

5cos

11

3cos

11cos

=

11

9cos

11

7cos

11

5cos

11

3cos

11cos

11sin2

11sin2

=

11sin2

11

8sin

11

10sin

11

6sin

11

8sin

11

4sin

11

6sin

11

2sin

11

4sin

11

2sin

=2

1

11sin2

11sin

11sin2

11

10sin


16: The number of solutions of sin3

x cos x + sin2

x cos2

x + sin x cos3

x = 1 in [0, 2] is

(A) 4 (B) 2

(C) 1 (D) 0

Solution: sin x cos x [sin2x + sin x cos x + cos2 x] = 1 sin x cos x + (sin x cos x)

2= 1

sin2

2x + 2 sin 2x –4 = 0 sin 2x = 512

1642

, which is not possible.


17: The number of solutions of the equation x3

+2x2

+5x + 2cosx = 0 in

[0, 2] is:

(A) 0 (B) 1

(C) 2 (D) 3

Solution: Let f(x) = x

3

+ 2x

2

+ 5x +2 cosx f (x) = 3x2

+4x + 5 – 2 sinx

= 3 xsin23

11

3

2x

2

Now 0xsin23

11 x ( as -1 sinx 1)

f (x) > 0 x

f(x) is an increasing function.

Now f(0) = 2

f(x) = 0 has no solution in [ 0, 2] .


18: The value of -1 -1cos[ tan (sin(tan x)) ]

x →∞li m is equal to



(A) -1 (B) 2

(C)1

- 2

(D)1

2

Solution:2

1

x21

x1lim

2

2

x

.


19: sinnx=n

r 0 r

r a .sin , where n is an odd natural number, then:

(A)0 a = 1,

1a = 2n (B)0 a = 1,

1a = n

(C)0

a = 0,1

a = n (D)0

a = 0,1

a = -n

Solution: sin nx = Im(ein x

) = Im ((cosx + i sinx)n)

n n 1 n n 3 3 n n 5 5

1 3 5C cos x .sinx C cos sin x C cos x.sin x …..

Since n is odd, let n = 2 + 1

sin nx =

n 2

1C (cos x) sinx

–

n 2 1 3

3C (cos x) sin

+ ….=

n 2

1C (1 sin x) sinx –n 2 1 3

3C (1 sin x) .sin +n 2 2 5

5C (1 sin x) .sin x + ….

=n n n 3

1 1 1 3C sinx ( C . C C )sin x ...

0 1a 0, a n


20: If tanx = n. tany, n R , then maximum value of 2 sec (x – y) is equal to:

(A)

2 (n+1)

2n(B)

2 (n+1)

n

(C)

2

(n+1)2

(D)

2

(n+1)4n

Solution: tanx = n tany, cos(x – y)

= cosx. cosy + sinx.siny.

cos(x – y) = cosx.cosy(1 + tanx.tany)

= cosx. cosy (1 + n tan2y)

2 22

2 2

sec xsec ysec (x y)

(1 n tan y )

2 2

2 2

(1 tan x)(1 tan y)

(1 ntan y)

2 2 2

2 2

(1 n tan y)(1 tan y)

(1 ntan y)

2 2

2 2

(n 1) tan y1

(1 ntan y)

Now,

22

21 ntan yntan y.

2

2

2 2

t an y 1

(1 n tan y) 4n

2 2

2 (n 1) (n 1)sec (x y) 1

4n 4n




21: If 3sin + 5cos = 5, then the value of 5sin – 3cos is equal to

(A) 5 (B) 3

(C) 4 (D) none of these

Solution: 3sin = 5(1 – cos) = 5 2sin2

/2 tan/2 = 3/5

5sin – 3cos =

2tan1

2tan1

3

2tan1

2tan2

52

2

2

= 3

25

91

25

913

25

91

5

32

5


22: In a ABC, if cotA cotB cotC > 0, then the is

(A) acute angled (B) right angled(C) obtuse angled (D) does not exist

Solution: Since cotA cotB cotC > 0

cotA, cotB, cotC are positive is acute angled


23: If < 2 <2

3, then 4cos2 2 2 equals to

(A) –2cos (B) –2sin(C) 2cos (D) 2sin

Solution: |2cos|22)4cos1(22 = )2cos1(2

= 2 | sin | = 2sin as4

3

2


24: If tan = n for some non-square natural number n, then sec2 is

(A) a rational number (B) an irrational number

(C) a positive number (D) none of these

Solution:n1

n1

tan1

tan12sec

2

2

where n is a non-square natural number so 1 – n 0.

sec2 is a rational number.


25: The minimum value of cos(cosx) is



(A) 0 (B) –cos1

(C) cos1 (D) –1

Solution: cos x varies from –1 to 1 for all real x.

Thus cos(cosx) varies from cos1 to cos0 minimum value of cos(cosx) is cos1.


26: If sin x cos y = 1/4 and 3 tan x = 4 tan y, then find the value of sin (x + y).

(A) 1/16 (B) 7/16

(C) 5/16(D) none of these

Solution: 3 tan x = 4 tan y 3 sin x cos y = 4 cos x sin y

3/4 = 4 cos x sin y cos x sin y = 3/16

sin (x + y) = sin x cos y + cos x sin y =16

7

16

3

4

1 .


27: The maximum value of 4sin2

x + 3cos2x +

2

x cos

2

x sin is

(A) 2 4 (B) 2 3

(C) 9 (D) 4

Solution: Maximum value of 4sin2x + 3cos

2x i.e. sin

2x + 3 is 4 and that of sin

2

x+ cos

2

xis

2

1

2

1 =

2 , both attained at x = /2. Hence the given function has maximum value 24


28: If and are solutions of sin2

x + a sin x + b = 0 as well as that of cos2x + c cos x + d = 0, then

sin( + ) is equal to

(A)2 2 d b

bd 2

(B)ac 2

c a 2 2

(C)bd 2

d b 2 2

(D)2 2 c a

ac 2

Solution: According to the given condition, sin+sin = –a and cos +cos= -c.

c2

cos2

cos2&a2

cos2

sin2

c

a

2tan

222 ca

ac2

2tan1

2tan2

)sin(


29: If sin, sin and cos are in G.P, then roots of the equation x2

+ 2x cot + 1 = 0 are always.

(A) equal (B) real

(C) imaginary (D) greater than 1



Solution: sin, sin, cos are in G.P.

sin2 = sin cos cos2 = 1 – sin2 0

Now, the discriminant of the given equation is

4cot2 – 4 = 4 cos2 cosec

2 0 Roots are always real.


30: If ,n

)1n( cos

n

2 cos

ncosS 2 2 2

then S equals

(A) )1n( 2

n (B) )1n(

2

1

(C) )2 n( 2

1 (D)

2

n

Solution:

n

)1n(cos

n

2cos

n

cosS 222

=

n)1n(2cos1

n

6cos1

n

4cos1

n

2cos1

2

1

=

n

k2cos1n

2

1 1n

1k

= 2n2

111n

2

1


31: If in a ABC, C =90°, then the maximum value of sin A sin B is

(A)2

1(B) 1

(C) 2 (D) None

Solution: sinA sinB = Bsin Asin22

1

= )B Acos()B Acos(2

1 = 90cos)B Acos(

2

1= )B Acos(

2

1

2

1

Maximum value of sinA sinB =

2

1


32: If in a ABC, sin2A + sin

2B + sin

2C = 2, then the triangle is always

(A) isosceles triangle (B) right angled

(C) acute angled (D) obtuse angled

Solution: sin2

A + sin2

B + sin2C = 2 2 cos A cos B cos C = 0

either A = 90o

or B = 90o

or C = 90o




33. Maximum value of the expression 2sinx + 4cosx + 3 is

(A) 2 5 + 3 (B) 2 5 - 3

(C) 5 + 3 (D) none of these

Solution: Maximum value of 2sinx + 4cosx = 2 5 .

Hence the maximum value of 2sinx + 4cosx +3 is 35 2


34: If sin = 3sin( + 2), then the value of tan ( + ) + 2tan is

(A) 3 (B) 2

(C) 1 (D) 0

Solution: Given sin = 3sin ( + 2)

sin ( + ) = 3sin ( + + )

sin ( + ) cos – cos( + ) sin

=3sin ( + ) cos + 3cos ( + ) sin

–2sin ( + ) cos = 4cos ( + ) sin

cos

sin2

)cos(

)sin(

tan(+) + 2tan = 0


35: If cos =

coscos1

coscos

, then one of the values of tan

2

is

(A) tan2

cot

2

(B) tan

2

cot

2

(C) sin2

sin

2

(D) none of these

Solution: tan2

2

=

cos1

cos1=

coscos1

coscos

1

coscos1

coscos1

=

coscoscoscos1

coscoscoscos1

=)cos1(cos)cos1(

)cos1(cos)cos1(

=)cos1)(cos1(

)cos1)(cos1(

= tan2

2

cot

2

2

.

tan2

= tan

2

cot

2

.


36. If tan 2. tan = 1, then is equal to



(A) n6

p p + (B) n

6

p p ±

(C) 2n6

p p ± (D) None of these.

Solution: tan 2 . tan = 12

22

2tan 11 tan1 tan 3

qÞ = Þ q =- q

2 2tan tan n6 6

p pÞ q = Þ q = p ± .


37. If is the root of 252cos 5cos 12 0, / 2q + q - = p < a < p , then sin 2 is equal to

(A)24

25(B)

24

25

-

(C)13

18(D)

13

18

-

Solution: Since, is the root of 225cos 5cos 12 0q + q - =

225cos 5cos 12 0\ a + a - =

5 25 1200 5 35cos cos

50 35

- ± + - -\ a = Þ a =

4 24 5 35cos sin2 2sin cos cos

5 25 35

- - - -Þ a = \ a = a a = a = .


38. The equation k sinx cos2x 2k 7+ = -

possesses a solution if

(A) k > 6 (B) 2 k 6£ £(C) k > 2 (D) None of these.

Solution: We have k2sinx (1 2sin x) 2k 7+ - = -

22 sin x k sin x 2 (k 4) 0Þ - + - =2k k 16k 64

sinx4

± - +Þ =

k (k 8) 1(k 4), 2

4 2

± -= = -

But sin 2¹ , therefore,

1sinx (k 4)2= -

Now,k 4

1 sinx 1 1 1 2 k 62

-- £ £ Þ - £ £ Þ £ £


39. The general solution of the equation tan 3x = tan 5x is

(A) x = n/2, n Z (B) x = n, n Z

(C) x = (2n + 1) , n Z (D) None of these.

Solution: We have tan 3x = tan 5x

5x n 3x, n Z x n / 2, n ZÞ = p + Î Þ = p Î

if n is odd, then x = n/2, gives the extraneous solutions. Thus, the solution of the given

equation will be given by x = n/2, where n is even say n = 2 m, m Z. Hence, the required

solution is x = m , m Z.




40. The equation4 2 2sin x 2cos x a 0- + = is solvable if

(A) 3 a 3- £ £ (B) 2 a 2- £ £(C) 1 a 1- £ £ (D) None of these.

Solution: We have4 2 2sin x 2cos x a 0- + =

2 2y 2(1 y) a 0Þ - - + = where 2sin x y=2 2y 2y a 2 0Þ + + - =

2y 1 3 aÞ = - ± -for y to be real.

Discriminant2 20 4 4(a 2) 0 a 3³ Þ - - ³ Þ £ . . . (1)

But2sin x y= , therefore 0 y 1£ £

2 20 1 3 a 1 1 3 a 2Þ £ - + - £ Þ £ - £2 2 21 3 a 4 2 a 0 a 2Þ £ - £ Þ - ³ Þ £ . . . (2)

From (1) and (2), 2a 2 2 a 2£ Þ - £ £ .


41. The set of values of x for whichtan3x tan2x

11 tan3x tan2x

-=

+is

(A) (B) /4

(C) n ; n 1, 2, 3 .....4

pì ü p + =í ý

î þ(D) 2n ;n 1, 2,3 .....

4

p p + =

Solution: tan(3x 2x) tanx 1- = =

x n ( / 4)\ = p + pbut this value does not satisfy the given equation as

tan2x tan( / 2)= p = ¥ and it reduces to indeterminate form.


42. If tan sec 3; 0q + q = < q < p , then is equal to

(A) /3 (B) 2/3

(C) /6 (D) 5/8

Solution: 3 cos sin 1q - q =

or3 1 1

cos sin2 2 2

q - q =

cos cos6 3

p pæ ö\ q + =ç ÷è ø

6 3 6

p p p\ q + = \ q = .


43. The value of the expression1 4sin10 sin 70

2sin10

- ° °

°is

(A) 1/2 (B) 1

(C) 2 (D) None of these.

Solution: Given expression is



11 2 cos 80

1 2[cos60 cos80 ] 2

2 sin10 2 sin10

é ù- - °ê ú- ° - ° ë û= =

° °2 cos 80 cos(90 10 ) sin 10

12sin10 sin10 sin10

° ° - ° °= = = =

° ° °.


44. If cos 3 sin 2q + q = , then (only principal value) is

(A) /3 (B) 2/3

(C) 4/3 (D) 5/3

Solution: cos 1 cos03 3

p pæ öq - = = ° \ q =ç ÷è ø

.


45. Number of solutions of 2 25cos 3sin 6sin cos 7q - q + q q = in the interval [0, 2] is

(A) 2 (B) 4


Solution:2 25cos 3sin 6sin cos 7q - q + q q =

1 cos2 1 cos25 3 3sin2 7

2 2

+ q - qæ ö æ öÞ - + q =ç ÷ ç ÷è ø è ø

4cos 2 3 sin2 6Þ q + q = ,

but2 24cos2 3sin2 4 3 5q + q £ + =

Solution does not exist.


46. If sin cos 2 cosq + q = q , then general solution for is

(A) 2n8

p p ± (B) n

8

p p +

(C)nn ( 1)

8

p p + - (D) None of these.

Solution: sin cos 2 cos tan 2 1q + q = q Þ q = -

n8 8

p pÞ q = Þ p + .


47. Number of solutions of 11 sin x = x is

(A) 4 (B) 6


Solution: 11 sin x = x

. . . (1)

On replacing n by –, we have 11 sin (–x) = –x

11sinx xÞ =So for every positive solution, we have negative solution also and x = 0 is satisfying (1), so

number of solution will always be odd. Therefore, (d0 is appropriate choice.


48. If 2 2 19

3 sin x cos x x x3 9

p + p = - + , then x is equal to



(A)1

3- (B)

1

3

(C)2

3(D) None of these.

Solution: L.H.S. 3 sin x cos x 2sin x 2

6

pæ ö= p + p = p + £ç ÷

è øand equality holds for

1x

3=

and R.H.S.

2

2 2 19 1x x x 2 2

3 9 3

æ ö= - + = - + ³ç ÷è ø

equality olds if 1

x3

= .

Thus L.H.S. = R.H.S. for1

x3

= only.


49. General solution for if 5

sin 2 cos 26 6

p pæ ö æ öq + + q + =ç ÷ ç ÷è ø è ø

, is

(A)7

2n6

p p + (B) 2n

6

p p +

(C)7

2n6

p p - (D) None of these.

Solution:5

sin 2 cos 26 6

p pæ ö æ öq + + q + =ç ÷ ç ÷è ø è ø

. . . (1)

sin 2 16

pæ öq + £ç ÷è ø

and5

cos 16

pæ öq + £ç ÷è ø

(1) may holds true iff sin 26

pæ öq +ç ÷è ø

and5

cos6

pæ öq +ç ÷è ø

both equal to 1 simultaneously.

First common value of is7

6

pfor which

5sin 2 sin sin 1

6 2 2

p p pæ öq + = = =ç ÷è ø

and 5 7 5cos cos cos2 16 6 6 p p pæ ö æ öq + = + = p =ç ÷ ç ÷è ø è ø

and since periodicity of sin 26

pæ öq +ç ÷è ø

is

and periodicity of 5

cos6

pæ öq +ç ÷è ø

is 2, therefore, periodicity of

5sin 2 cos

6 6

p pæ ö æ öq + + q +ç ÷ ç ÷è ø è ø

is 2. Therefore, general solution is7

2n6

pq = p + .


50. If tan and tan are the roots of 2x 3x 1 0- - = , then value of tan ( + ) is



(A)1

2(B) 1

(C)3

2(D) None of these.

Solution: tan , tana b are the roots of 2x 3x 1 0- - =

tan tan 3\ a + b =and

tan tan 1\ a + b = -.

tan tan 3tan ( )

1 tan tan 2

a + b\ a + b = =

- a b.


51. Number of solutions of the equation tan x = sec x = 2 cos x lying in the interval [0, 2] is

(A) 0 (B) 1

(C) 2 (D) 3

Solution: The given equation can be written as2sinx 1 2cos x+ =

2

sinx 1 2(1 sin x)Þ + = -22sin x sinx 1 0Þ + - =

1 1 8 1 3 1sinx

4 4 2

- ± + - ±= = = or –1

5x ,

6 6

p pÞ =

Hence, the required number of solutions is 2.


52. If tan m + cot n = 0, then the general value of is

(A) (2r 1)2(m n)

+ p-

(B) (2r 1)2(m n)

+ p+

(C)r

m n

p

+(D)

r

m n

p

-Solution: The given equation can be written as

tan m cot n tan( / 2 n )q = - q = p + q

m r n , r 2

pæ ö\ q = p + + q Î Iç ÷è ø

or1

(m n) (2r 1) , r

2

- q = + p Î I

(2r 1), r

2(m n)

+ p\ q = Î I

-.


53. The general solution of the equation ( 3 1)sin ( 3 1)cos 2- q + + q = is

(A)nn ( 1)

4 12

p p p + - - (B) 2n

4 12

p p p ± -

(C)nn ( 1)

4 12

p p p + - + (D) 2n

4 12

p p p ± +

Solution: Let 3 1 r cos , 3 1 r sin+ = a - = a2 2 2r ( 3 1) ( 3 1) 8\ = + + - = or r 2 2=



and

11

3 1 3tan

13 1 13

--

a = =+ +

or tan tan(45 30 ) tan15a = ° - ° = °

1512 p\ a = ° =

Using these in the given equation, we get

r cos( ) 2q - a =

or2 2 1

cos cos12 r 42 2 2

p pæ ö æ öq - = = =ç ÷ ç ÷è ø è ø

2n12 4

p p\ q - = p ± or 2n , n

12

pq = p ± Î I .


54. One solution of the equation 2 24cos sin 2sin 3 sinq q - q = q is

(A)n 3

x n ( 1)10

- pæ ö= p + - ç ÷è ø

(B)n 3

x n ( 1)10

pæ ö= p + - ç ÷è ø

(C) x 2n6

p= p ± (D) None of these.

Solution: The given equation can be written as2sin [4(1 sin ) 2sin 3] 0q - q - q - =

or2sin [1 2 sin 4 sin ] 0q - q - q =

or2sin [4 sin 2 sin 1] 0q q + q - =

Either sin = 0 which gives = n or

24sin 2sin 1 0q + q - = which gives

2 4 16 2 2 5 1 5sin

2 4 8 4

- ± + - ± - ±q = = =

´

1 5 1 5,

4 4

- + - -=

Now,1

sin ( 5 1) sin18 sin4 10

pæ öq = - = ° = ç ÷è ø

n

n ( 1) 10

pæ ö

\ q = p ± - ç ÷è ø

Again1

sin ( 5 1) cos364

q = - - = - °

cos(90 54 ) sin54= - ° - ° = - °

3sin( 54 ) sin

10

- pæ ö= - ° = ç ÷è ø

n 3n ( 1)

10

pæ ö\ q = p + - -ç ÷è øThus, one solution of given equation is

n 3n ( 1)

10

pæ öq = p + - -ç ÷è ø.




55. Solve for x and y, the equations:

x3cos y + 3x cosy.

2 sin y = 14

x3sin y + 3x.

2 cos y siny = 13

(A) y =1 1

tan ,x 5 52

where 2n < y < 2n +

2

(B) y =1 1

tan ,x 5 52

where 2n + < y < 2n +3

2

(C) both

(D) None of these

Solution: Clearly, x 0 dividing the equations, we get3 2

3 2

cos y 3cosy sin y 14

sin y 3cos ysiny 13

by componendo and dividenodo, we get

3

3

(cosy siny) 14 13

(cosy siny) 14 13

or,

3

cosy siny

cosy siny

= 27 =3

(3)

or, =cosy siny 3

cosy siny 1

dividing numerator and denominator by cosy, we get

1 tan y 3

1 tan y 1

or,

2 tan y 2

2 4 .

siny =1

5, cosy =

2

5(when y is in 1st quadrant)

and siny = -

1

5 and cosy = -

2

5 (when y is in 3rd quadrant)

When y is in first quadrant.

8 2 1x 3 , 14, x 5 5

55 5 5

When y is in 3rd quadrant.

8 2 1x 3 . 14 x 5 5

55 5 5

Hence y =1 1

tan ,x 5 52

where 2n < y < 2n +2

and y =1 1

tan ,x 5 52

where 2n + < y < 2n +3

2

56. The solution of sinx + 3 cosx = 2 is :

(A) 2n +12

5(B) 2n -

12

(C) 24

n

p p ± (D) None of these

Solution: Given, 3 cosx + sinx = 2

2

3 cos x +2

1 sinx =2

2



cos4

cos6

x

4

n26

x

x = 2n 64

.

x = 2n +12

5 , 2n -12 where n I.

Hence (A, B) is the correct answer.

57. The solution of the equation tan . tan 2 = 1 is :

(A) n +12

5(B) n -

12

(C) 24

n

p p ± (D) n

6

Solution: Given tan . tan 2 = 1

2

2

tan1

tan2

= 1

2 tan2 = 1 –tan

2 3 tan2 = 1

tan =3

1 = n

6


58. Find the general solution of the equation

sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x:

(A)5

2 12

n p p+ (B) n -

12

(C)82

n (D) n 8

p

Solution: Given sin x – 3 sin 2x + sin 3x = cos x –3 cos 2x + cos 3x

2 sin 2x cos x – 3 sin 2x = 2 cos x cos 2x – 3 cos 2x

sin 2x (2 cos x –3) = cos 2x (2 cos x –3) sin 2x = cos 2x

(

cos x 3/2)

tan 2x = 1 2x = n +4

x =

82

n

, n I.


59. Solve for x, the equation sin3x + sin x cos x + cos

3x = 1:

(A) 2m (B) (4n + 1)2

(C) Both (D) None of these

Solution: The given equation is sin3

x + cos3

x + sin x cos x = 1

(sin x + cos x) (sin2

x – sin x cos x + cos2

x ) + sin x cos x – 1 = 0

(1 – sin x cos x)[sin x + cos x – 1] = 0

Either 1 – sin x cos x = 0 sin 2 x = 2 which is not possible

Or, sin x + cos x – 1 = 0 cos (x – /4) =2

1

m2

4x

4

x = 2m and x = (4n + 1)2




60. The equation esinx

– e –sinx

– 4 = 0 has:

(A) no real solution (B) one real solution

(C) two real solutions (D) can't be determined

Solution: The given equation can be written as

e2 sin x

– 4esin x

– 1 = 0 esin x

=

2

4164 = 2 + 5

sin x = ln (2 + 5 ) (ln (2 – 5 ) not defined as (2 – 5 ) is negative)

Now, 2 + 5 > e ln (2 + 5 ) > 1 sin x > 1

Which is not possible. Hence no real solution.


61. If tan ( cos x) = cot ( sin x), then cos4

x

pæ ö-ç ÷è ø

is

(A)1

2(B)

1

2 2


Solution: Given that tan ( cos x) = cos ( sin x)

or tan ( cos ) tan sin2

x x

p p p

æ ö= -ç ÷è ø

cos sin2

x x

p p pÞ = -

1cos sin

2 x xÞ + =

1 1 1cos sin

2 2 2 2

x xÞ + =

1cos

4 2 2 x

pæ öÞ - =ç ÷è ø

.


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Trigonometric Identities and Equation Eng