TRIGONOMETRY (WORKSHEET-CW1)
1 INFOMATHS/MCA/MATHS/
1.Ans. (c)If is a rational no.
Then As,
Now, as is rational
is rational .
tan is rational,
Similarly
sin, cos are rational numbers as well.
2.Ans. (a)
As
is in IIIrd quadrant sin is negative.
[As, sin is -ve]
|sin + cos | = - (sin + cos )
as in 3rd quadrant both sin, cos are –ve
sin + cos is –ve
3.Ans. (c)
dividing Nr/Dr by cos 37
= tan(82) = cot 8
4.Ans. (d)
Dividing Mr/Dr by cos 11
= tan56
5.Ans. (a)
Max value of sin + cos is at
max value
is at
6.Ans. (a)
On squaring both sides:-
1 + sin2 = 2cos2 = 2 – 2sin2
1 + sin2 = 2 – 2sin2
2sin2 = 1 – sin2
cos - sin = sin
7.Ans. (d)
Expanding determinant along R3
sin = cos2 + sin2
sin = 1
8.Ans. (a)(1 + tanA) (1 – tanB)
= 1 + tanA – tanB – tanA tanB …* Eq.
= ?
Also,
tan(A – B) = 1
tanA – tanB = 1 + tanA.tanB
tanA – tanB – tanAtanB = 1.
* EQ. – gives (1 + tanA) (1 – tanB) = 2
9.Ans. (b)As 1radian = 57 (approximately)
And we know sin is strictly increasing curve in
sin 1 radian > sin1
10.Ans. (d)
in AMP in BMQ
h1 : h2 = 3 : 1
11.Ans. (a)tan2 = tan( + + - )
Also, and
as
So,
12.Ans. (b)sin2x = 1 – sinx
sinx = 1 – sin2x = cos2x
cos4x + cos2x = sin2x + sinx = 1
13.Ans. (b)cot-121 + cot-113 + cot-1 (-8)
=
14.Ans. (a)sin ( cos ) = cos ( sin )
Put n = 0
15.Ans. (d)
16.Ans. (c)
{As cos ( - ) = - cos}
Using cosA. Cos2A. cos22A ....
We have,
here n = no. of factors.
A = smallest angle.
17.Ans. (b)Remember. In ABC
{Since, cosA + cosB + cosC }
{as }
18.Ans. (b)As sinx + sin2x = 1
sinx = 1 – sin2x
= cos2x
cos12x + 3cos10x + 3cos8x + cos6x
= cos6x [cos6x + 3cos4x + 3cos2x + 1]
{as (a + b)3 = a3 + b3 + 3ab (a + b)}
= cos6x [(cos2x + 1]3 sin3x [sinx + 1]3
(sin2x + sinx)3 {as a3.b3 = (ab)3}
13 {as sin2x + sinx = 1}
19.Ans. (d)In ABE
y = x tan
In BEF
2h + y = xtan
Now, height of cloud above the surface
This can also be read as on replacing
Also and
20.Ans. (d)In BCD
h = a tan
in ADC
x + h = a tan
x + a tan = a tan
x = a (tan - tan )
21.Ans. (b)As
Given equation can be written as
22.Ans. (c)As
23.Ans. (d)
24.Ans. (a)As
But
take
10t = 4 + 4t2
2t2 – 5t + 2 = 0
(2t – 1) (t – 2) = 0
,
25.Ans. ()sin30cos45 + cos3045= sin(30 + 45)
= sin 75
Using sin(A + B) = sinAcosB + cosA sinB.
26.Ans. (a)As B = 45, C = 105, c
Then third angle of ABC will be A = 30 {Since, sum of all angles of ABC = 180}
Now, using sine Rule:-
as
27.Ans. (b)a cos2 + bsin2
= a
28.Ans. (c)(no choice satisfies)
Dividing both sides by
i.e.
, where n Z.
29.Ans. (c)As
30.Ans. (c)As sinx, cosx, tanx are in GP
sin.x tanx = cos2x
sin2x = cos3x
Also cot2x = secx cot6x - cot2x = sec3x – secx
=secx (sec2x – 1) = secx.tan2x
= 1 {as sin2x = cos3x}
31.Ans. (d)As, the greatest angle of a is opposite to the longest side always.
Here x2 + x + 1 is the longest
Side (By observation)
Angle opposite to this side will be largest.
{After solving the equation}
32.Ans. (d)2sin2 - 3sin - 2 = 0
2sin2 - 4sin + sin - 2 = 0
2sin(sin - 2) + (sin - 2) = 0
(2sin + 1) (sin - 2) = 0
, sin = 2
Whereas, sin = 2 is rejected.
33.Ans. (c)cosec A . [sinBcosC + cosB sinC]
= 1 As sin ( - ) = sin
34.Ans. (d)tan9 - tan27 - tan63 + tan81
= (tan9 + tan81) – (tan27 + tan63)
4
35.Ans. (a)As
{ is in quad III
{as
36.Ans. (b)In ABC,
x = 60 cot 15
37.Ans. (c)sinx + cosx = 1
Dividing both sides by :-
x = 2n,
(check by taking a = even / odd in (c)
38.Ans. (c)sin = sin = n + (-1)n.
= n + (-1)n
39.Ans. (a)As we know
cos > sin for
cos < sin for
and cos = sin for =
here for = 10
cos > sin
cos10 - sin10 is positive.
40.Ans. (c)8R2 = a2 + b2 + c2…* Equation
We know = R(circum radius)
a = 2R sin A
b = 2R sin B
c = 2R sin C
* Eq gives: - 8R2 = (2R sinA)2 + (2R sin B)2 + (2 Rsin C)2
R2.8 = 4R2 (sin2A + sin2B + sin2C)
2 = sin2A + sin2B + sin2C
2 = 2(1 + cos A cos B cos C)
cos A. 2cosB. cos C = 0
either cosA = 0, or cos B = 0, cos C = 0
A = 90, B = 90, C = 90
41.Ans. (d)We know
2tan-1a – 2tan-1b = 2tan-1x
tan-1a – tan-1b = tan-1x
42.Ans. (a)
= 2 cos 60= 1
43.Ans. (d)...*Eq.
Given cot B – cot A = y … Eq. (1)
also, tanA – tan B = x
y = x. cot A . cot B
… Eq. (2)
From Eq. (1) and (2)
44.Ans. (a)cos ( - ) = a
cos ( - ) = b
Then sin2( - ) + 2ab cos ( - )
= sin2[( - ) - ( - )] + 2ab. cos [( - ) - ( - )]
= [sin ( - ) – cos ( - ) – cos ( - ) sin ( - )]2 + 2ab[cos (-) - cos ( - )+ sin ( - ) . sin ( - )]
= a2 (1 – b2) + b2(1 – a2) + 2a2b2= a2 + b2
45.Ans. (d)cos ( - ) = 1
- = 2n
cos ( + ) =
+ = 2m cos-1 ()
2 = 2(m+n) cos-1 ()
= (m+n)cos-1 ()
2 = 2(m-n) cos-1 ()
= (m-n)cos-1 ()
(1)If m = 0, n = 0
= -cos-1 ()= cos-1 ()
two solutions as =
(2)If m = 1, n = 0
= -cos-1 ()= - cos-1 ()
Also = , + ) = 2 cos-1
(3)If x = - 1, n = 0
, Also - = 0
+ = - 2
one solution
four solutions are these
46.Ans. (b)
47.Ans. (a)tan( cos ) = cot ( sin )
Dividing both sides by
48.Ans. (a)
cot-1x + sin-1 =
cot-1x + tan-1=
cot-1x + tan-1+ tan-1(1) =
cot-1x + tan-1(3) =
So, x= 3
49.Ans. (a)
50.Ans. (c)a = 2, b = 3,
Using Since Rule:-
sin B = 1.
B = 90 .
51.Ans. (a)= 0
52.Ans. (d)In ABC
x = 100 cot 60
Also, in ACD
x + y = 100 cot 45
x + y = 100
But
y = 100 – x
53.Ans. (c)a = 5, b , c = 5.
We know, the angle opposite to the longest side is largest. So B is the largest angle.
B = 120
54.Ans. (b)
55.Ans. (d)Let the height of the tower be ‘h’ mts.
Then in ABC
m.
56.Ans. (a)Smallest angle is opposite to the smallest side.
A = 20
57.Ans. (c)
58.Ans. (c)
Sol.x (x + 1) 0 and
x2 + x + 1 1
x2 + x 0
x (x + 1) 0
combing both equations.
x(x + 1) = 0
So the equation has two roots. i.e x = 0 and x = - 1.
59.Ans. (a)
5x = 1 – 6x2
6x2 + 5x – 1 = 0
6x2 + 6x – x – 1 = 0
6x(x + 1) – (x + 1) = 0
(6x – 1) (x + 1) = 0
60.Ans. (d)A = cos2 + sin4
= 1 – sin2 + sin4 = 1 – sin2 (1 – sin2)
= 1 – sin2.cos2
Also, - 1 sin2 1
0 (sin2)2 1
Adding 1 throughout :-
61.Ans. (d)By choices (a) choice fails because imaging roots
(b) gives x = 0, as is not differ so fails
(c) gives x = -1, 1/2 again cos-1 (1 – (-1) = cos-12 not possible so (d) is as (1 – x) = 0
62.Ans. (b)sin4x+cos4x+sin2x+ = 0
sin4x+(1- sin2x)2+ sin2x+ = 0
sin4x + sin4x + 1- 2sin2x+ sin2x+ = 0
2sin4x- sin2x+1+ = 0
sin4x- + 1 + = 0
sin4x- + +1+ = 0
(sin2x - )2 = -
0 (sin2x - )2 1
0 -1
- -
(1+ ) -
-
only (b) choices satisfies
63.Ans. (d)acos2x + bsin2x + c = 0
acos2x + bsin2x = -C
= -C
Sin(2x + ) = -, where tan =
-1 <sin(2x + ) < 1
-1 < < 1
can be satisfies by infinite triples (a, b, c)
64.Ans. (d)
Expanding the determinant along R.
= cosC . (-sinB tanA) – tanA . (sin B. cos C)
= - 2 tan A sin B cos C
65.Ans. (c)
There are four solutions.
66.Ans. (b)The formula 2sin'x holds for.
i) – 1 x 1
ii)
2 sin-1x
sin-1x
So, x
67.Ans. (d) ONE as
68.Ans. (c)cos-1x > sin-1x sin-1x – cos-1x < 0
Also cos-1x > sin-1x for -1 x < 0
as
(2)
69.Ans. (a)
n = even 2x1 – 2x2 = n
x1 – x2as n = even
x1 – x2 = n, n Z
70.Ans. (a)If A and B are position of person at lower and higher floor respectively and S is the position of statue, then by the above figure.
AO = OS = 100…(i)
and
…(ii)
The distance between the two persons, AB = OB – OA
71.Ans. (a)As cot, cot2…….cotn = 1
cos cos2…….cosn =sin sin2……sinn
cos2 cos22…..cos2n=[sin21]
cos cos2…….cosn
72.Ans. (d)cos + cos = a
sin + sin = b....(*)
Also,
But + = 2( is A.M. of , )
Hence
73.Ans. (c)(1 + tan1)(1 + tan2)(1 + tan3) …..(1 + tan44) (1 + tan45) = 2n….. * Eq.
Considering
(1 + tan1) . (1 + tan44)
=1 + tan1 + tan 44 + tan1 tan 44 =1+(tan45)=2
{Since,
1 – tan 44 tan 1 = tan44+ tan1
tan44 + tan1 + tan44 tan 1 = 1. }
Now Eq. * gives 222 . (tan45 + 1) = 2n
223 = 2n
n = 23.
74.Ans. (c)sin12 sin 4854
75.Ans. (a)
as sin (2 - ) = - sin
So
So they will cancel out.
76.Ans. (b)a = 5, b = 4,
9 – 9 cosC = 7 + 7cos C
2 = 16cos C
5 = 41 – c2
c2 = 36
c = 6
77.Ans. (a)
Also, the triangle is right angled at A
b2 + c2 = a2
= tan-1(1) 45
78.Ans. (d)3a = b+c
tantan=
= = as
= as b + c = 3a
79.Ans. (c)cos-1(cosx) = x
is true for all x [0, ]
Since, cos x [-1, 1]
80.Ans. (c)
Since
81.Ans. (b)Let the height of tower be h.
h = 100 mt.
82.Ans. (d)in ABE
in ACD,
h = 120 mt.
height of tower = 360 – 120 = 360 – 120 = 240 mt.
83.Ans. (c)Consider, a = 7, b = 5, c = 3.
A = 120
84.Ans. (d)
X = 30
85.Ans. (b)We know, l = r
radians.
86.Ans. (b)radius of metallic ring = 1 foot [circumference = 2r = 2 foot] since the ring has been moulded into a circular arc ℓ = 2, radius = 80 ft.
radius
Area formed by the sector
= 80 . sq. feet.
87.Ans. (b)A, B, C, D are the angles of the cyclic quadrilateral
A + C = B + D = 180
cosA + cosB + cosC + cosD
= (cosA + cosC) + (cosB + cosD)
= 0
88.Ans. (c)x.cos - ysin =
x.sin + ycos =
Squaring & adding:-
x2cos2 + y2sin2 - 2xysincos = 2
x2sin2 + y2cos2+ 2xysincos = 2
x2(cos2 + sin2) + y2(sin2 + cos2) = 2+ 2
x2 + y2 = 2+ 2
89.Ans. (b)Matrix A is non-invertible
If |A| = 0
i.e.
sin2 - cos2= 0
-cos2 = 0
cos2 = 0
90.Ans. (c)
91.Ans. (a)A + C = B
tan (A + C) = tan B
tanA + tanC = tanB . (1 – tanA.tanC)
tanA + tanC = tanB – tanA. tanB. tanC
tanA . tanB. tanC = tanB – tanA – tanC
92.Ans. (c)Consider
x
But
93.Ans. (a)
As, lies in 1st quadrant.
94.Ans. (b)sin2 - sin2 - 15cos2 = 0
1 – cos2 - sin2 - 15cos2 = 0
1 – sin2 - 16cos2 = 0
(cos - sin)2 – 16cos2 = 0
(cos - sin)2 = 16cos2
cos - sin = 4 cos.
cos - sin = 4cos, cos - sin = - 4cos
- 3cos = sin5 cos = sin
tan = - 3 tan = 5
= m - tan-1 (3) = n + tan-1(5)
95.Ans. (b)In a , if length of base is equal to the length of perpendicular.
Then the angle of evaluation is 45 always.
96.Ans. (b)tanA + cotA= 4
square both sides
tan2A+cot2A+2tanAcotA = 16
tan2A+cot2A= 16-2=14
Again squaring both sides
tan4A+cot4A+2tan2Acot2A =196
tan4A+cot4A = 196 – 2 = 194
97.Ans. (a)Let the sides be
a = x, b = 2x
A = , B = + 60
According to sin Rule:-
(say)
From the 1st and 2nd members:-
sin ( + 60) = 2sin
sin . cos60 + cos.sin60 = 2sin
= 30, so other angle + 60 = 90
so triangle is right angle
98.Ans. (c)sinA = sinB, cosA = cosB
sin A = sinB
A = n + (-1)nB…………(i)
CosA = CosB
A= 2 n B……..(ii)
From (i) and (ii)
A = 2 n + B
99.Ans. (a)tan-1x + tan-1y
x + y = 1 – xy
x + y + xy = 1
100.Ans. (d)The equation 3cosx + 4sin x = 6 has no solution as the maximum value of 3cosx + 4sinx is 5.
i.e
Which is never equal to RHS.
No solution does exist.
101.Ans. (a)sinx(1 + cosx) = z
= - sin2x + cosx + cos2x
= - (1 – cos2x) + cosx + cos2x
= 2 cos2x + cosx – 1
For max./min Put
2cos2x + cosx – 1 = 0
2cos2x + 2cosx – cosx – 1 = 0
2cosx (cosx + 1) – (cosx + 1) = 0
(2cosx – 1) (cosx + 1) = 0
2 cosx – 1 = 0, cosx = - 1
, cosx = - 1
x = 60,x = ,
Double derivative test:-
= - 4sinx.cosx – sinx.
gives maxima.
function f(x) is maximum at
102.Ans. (c)
103.Ans. (b)cos40 + cos80 + cos160
= (cos40 + cos80) + cos160
= 2cos60. cos20 + cos160
= cos20 + cos160
= cos20- cos20 = 0
as cos 160 = cos (180 - 20)
= - cos20
104.Ans. (d)2B = A + C, A+B+C = 180
3B = 180 B = 60
Also A + B + C =
A = - (B + C)
sin A = sin (B + C)
105.Ans. (d) Let the altitude of the aeroplane be _____.
in AMCin MCB,
h = x.
mt.
106.Ans. (c)
107.Ans. (c)cosec x + cotx = 2sinx,
1 + cosx = 2sin2x
2(1 – cos2x) = 1 + cosx
2 – 2cos2x = 1 + cosx
2cos2x + cosx – 1 = 0
(2cosx – 1) . (cosx + 1) = 0
- 1
,
108.Ans. (d)ABCD is a cyclic quad.
A + C = 180 =
sin(A + C) = sin = 0.
109.Ans. (a)max. value of
3cosx + 4sinx + 5 is
= 5 + 5 10.
110.Ans. (c)The greatest angle is always opposite to the longest side of a triangle.Among a, b and . The largest side is
Using the cosine rule.
gives C = 120.
111.Ans. (a)
Take tan-1y =
y = tan
Take
112.Ans. (b)= (let)
ac + a2 + b2 + bc = 2ab
= bc + ab + c2 + ac
a2 + b2 = c2 + ab
1 INFOMATHS/MCA/MATHS/
1 INFOMATHS/MCA/MATHS/