Trigonometric Identities and Equation Eng
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Transcript of 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
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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
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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
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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
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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.
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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)
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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
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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 )
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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
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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 ,
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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
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/ 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
.
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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.
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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
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Hence (A) is the correct answer.
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
Hence (A) is the correct answer.
8: The minimum value of 3tan2 + 12 cot2 is:
(A) 6 (B) 8
(C) 10 (D) None of these
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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 .
Hence (A) is the correct answer.
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
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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
.
Hence (A) is the correct answer.
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
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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
.
Hence (B) is the correct answer.
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
.
Hence (A) is the correct answer.
15: The value of ( )5
1
cos 2 111
r
r
p
=
-å is :
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(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
Hence (A) is the correct answer.
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.
Hence (D) is the correct answer.
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] .
Hence (A) is the correct answer.
18: The value of -1 -1cos[ tan (sin(tan x)) ]
x →∞li m is equal to
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(A) -1 (B) 2
(C)1
- 2
(D)1
2
Solution:2
1
x21
x1lim
2
2
x
.
Hence (D) is the correct answer.
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
Hence (C) is the correct answer.
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
Hence (D) is the correct answer.
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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
Hence (B) is the correct answer.
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
Hence (A) is the correct answer.
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
Hence (D) is the correct answer.
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.
Hence (A) is the correct answer.
25: The minimum value of cos(cosx) is
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(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.
Hence (C) is the correct answer.
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 .
Hence (B) is the correct answer.
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
Hence (A) is the correct answer.
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(
Hence (D) is the correct answer.
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
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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.
Hence (B) is the correct answer.
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
Hence (C) is the correct answer.
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
Hence (A) is the correct answer.
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
Hence (B) is the correct answer.
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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
Hence (A) is the correct answer.
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
Hence (D) is the correct answer.
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
.
Hence (A) is the correct answer.
36. If tan 2. tan = 1, then is equal to
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(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 ± .
Hence (B) is the correct answer.
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 = .
Hence (B) is the correct answer.
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 £ £
Hence (B) is the correct answer.
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.
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Hence (B) is the correct answer.
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 - £ £ .
Hence (B) is the correct answer.
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.
Hence (A) is the correct answer.
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 = .
Hence (C) is the correct answer.
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
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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
° ° - ° °= = = =
° ° °.
Hence (B) is the correct answer.
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
.
Hence (A) is the correct answer.
45. Number of solutions of 2 25cos 3sin 6sin cos 7q - q + q q = in the interval [0, 2] is
(A) 2 (B) 4
(C) 0 (D) None of these.
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.
Hence (C) is the correct answer.
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 + .
Hence (B) is the correct answer.
47. Number of solutions of 11 sin x = x is
(A) 4 (B) 6
(C) 8 (D) None of these.
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.
Hence (D) is the correct answer.
48. If 2 2 19
3 sin x cos x x x3 9
p + p = - + , then x is equal to
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(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.
Hence (B) is the correct answer.
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 + .
Hence (A) is the correct answer.
50. If tan and tan are the roots of 2x 3x 1 0- - = , then value of tan ( + ) is
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(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.
Hence (C) is the correct answer.
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.
Hence (C) is the correct answer.
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
-.
Hence (A) is the correct answer.
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=
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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 .
Hence (D) is the correct answer.
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.
Hence (A) is the correct answer.
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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
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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
Hence (D) is the correct answer.
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.
Hence (C) is the correct answer.
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
Hence (C) is the correct answer.
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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.
Hence (A) is the correct answer.
61. If tan ( cos x) = cot ( sin x), then cos4
x
pæ ö-ç ÷è ø
is
(A)1
2(B)
1
2 2
(C) 0 (D) None of these.
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
.
Hence (B) is the correct answer.