Question Paper Maths IIT

8/10/2019 Question Paper Maths IIT

1/17

1 The two legs of a right triangle are sin+ sin 32

and cos cos3

2

. The length of its hypotenuse

is

(A)1 (B) 2 (C*) 2 (D) some function of

2 If

4cos3cos2cos

4sin3sin2sin= tan k is an identity then the value k is equal to :

(A) 2 (B*) 3 (C) 4 (D) 6

[Sol.

3cos4cos2cos

3sin4sin2sin=

3coscos3cos2

3sincos3sin2=

3cos

3sin

= tan3= tanK K = 3 Ans ]

3 The expression

97cos157cos7cos23sin

98cos158cos8cos22sinwhen simplified reduces to :

(A*) sec(100) (B*) cosec

2

3(C) sin

2

7(D*) cot

4

5

[Sol.)790cos()23180cos(7cos23sin

)890cos()22180cos(8cos22sin

=

7sin23sin7cos23sin

8sin22sin8cos22sin=

30sin

30sin= 1 ]

4 If1

1

1

sin

sin

sin

cos cos

A

A

A

A A, for all permissible values of A, then A belongs to

(A*) First Quadrant (B) Second Quadrant (C) Third Quadrant (D*) Fourth Quadrant

[Hint : L.H.S =|Acos|

Asin1+

Acos

Asin=

Acos

1which is true only if | cos A| = cosA result ]

5 If secA =8

17and cosecB =

4

5then sec(A + B) can have the value equal to

(A*)36

85(B*)

36

85(C*)

84

85 (D*)

84

85

[Hint :A = I & IV quadrant and B = I & II quadrant and accordingly four cases.]

6 The exact value of

69sin51cos39cos21sin

66sin6sin6cos24sinis ______ . [ Ans. 1]

[Sol.

21sin39cos39cos21sin

24cos6sin6cos24sin=

18sin

18sin= 1 ]

7 If cos(+ ) = m cos( ), then tan is equal to :

(A)1

1

m

m

tan (B) 11

m

m

tan (C*) 11

m

m

cot (D) 11

m

m

cot

[Sol. m)cos(

)cos(

Applying componendo & dividendo

sinsin2coscos2

1m1m


2/17

3 1

9 Which of the following when simplified reduces to unity?

(A*)1 2

24 4

2

2

sin

cot cos

(B*)

sin

sin cos tan

2

+ cos ( )

(C)1

4

1

42 2

2 2

2sin cos

( tan )

tan

(D*)1 2

2

sin

(sin cos )

10 If [1 sin( +) + cos(+ )]2+ 1 32

32

2

sin cos = a + b sin

2 then find the value o

a and b. [ Ans. a = 4 & b = 2 ][Sol. [ (1 + sin cos)2+ (1 + cos sin)2]

= 1 + sin2+ cos2+ 2sin sin2 2cos+ 1 + cos2+ sin2+ 2cos sin2 2sin4 2sin2= a + bsin2a = 4 , b = 2 ]

11 Exact value of tan200 (cot10 tan10) is ______ . [ Ans. : 2 ]

12 The greatest value of the expression

sin2 158

4

x sin2 178

4

x for 0 x 8

is _________ __. [ Ans. 12

]

[Sol. sin

x48

17x4

8

15 sin

x48

17x4

8

15

= sin

x88

32sin

4

= )x8(sin2

1=

2

1Ans]

13 The expression

1 2

2 2 34

sin

cos . tan

1

4sin 2 cot cot

2

3

2 2

when simplified reduces to

:(A) 1 (B) 0 (C) sin2 (/2) (D*) sin2

[Sol.

2

tan2

cot4

2sin

4tan2cos

2sin1=

sin

cos

4

cossin

tan1

tan12cos

2sin1

=

22

cos

sincoscossin2cos

)cos(sin

2cos2cos

2cos = sin2 ]

14 Exact value of cos20 + 2 sin2 55 2 sin 65 is :

(A*) 1 (B)1

2(C) 2 (D) zero

[Hint : 1 + cos 20 cos 110 2 sin 65

1 + 2 sin 65 sin 45 2 sin 65 = 1 ]

15 Prove the identity, cos3

24

+ sin(3 8) sin(4 12) = 4 cos2cos4sin6.

[Hint: LHS = sin4+ sin8+ sin12; = 2 sin8cos 4+ sin8


3/17

[Hint : Nr= 2cos2

x9cos

2

x= 2cos

2

x

2

x3cos3

2

x3cos4 3 =

32

x3cos4

2

x3cos

2

xcos2 2

=2

x3cos

2

xcos2 [2(1 + cos 3x) 3] =

2

x3cos

2

xcos2 [2 cos3x 1]

r

r

D

N=

2

x3cos

2

xcos2 = cos 2x + cos x ]

17 Prove the identity, sin 2(1 + tan 2. tan) +1

1

sin

sin

= tan 2+ tan2

4 2

.

[Sol. LHS = sin2

2coscossin2sincos2cos +

sin1sin1

or sin2

2coscos

cos+

22

)2cos()2sin(

)2cos()2sin(

or tan2+2

)2tan(1

)2tan(1

or tan2+2

)2tan()4tan(1

)2tan()4tan(

or tan2 + tan2

24= RHS ]

18 Prove that

tan

8tan= (1 + sec2) (1 + sec4) (1 + sec8)

[Sol. RHS =

8cos

8cos1

4cos

4cos1

2cos

2cos1=

8cos4cos2cos

4cos22cos2cos2 222

8cos

cos]4cos2coscos8[=

8cos

cossin

8sin

]

19 If sin2= 4 sin2, show that 5 tan( ) = 3 tan(+ ).

[Sol. 42sin

2sin

35

2sin2sin2sin2sin (using C and D)

3

5

)sin()cos(2

)cos()sin(2

3

5

)tan(

)tan(

proved ]

20 If 0 < x sin 3 ]

53 The value of x satisfying the equation, x = x222 is

(A) 2 cos 10 (B) 2 cos 20 (C*) 2 cos 40 (D) 2 cos 80[Sol. Note that x [2, 2]

Let x = 2 cos where [0, ]

x = cos2222

2 cos =2

cos222 =

2cos122 =

4sin22 =

42cos22

=

42

cos12

2 cos = 2 cos

84; =

84

8

9=

4

=

9

2

Hence x = 2 cos

9

2= 2 cos 40 (C) Ans. ]

54 a, b, c are the sides of a triangle ABC which is right angled at C, then the minimum value of

2

b

c

a

c

is

(A) 0 (B) 4 (C) 6 (D*) 8[Hint: a = c sin

b = c cos

E =

2

b

c

a

c

=2

cos

1

sin

1

=

2sin

)2sin1(42

=

2sin

12sin

142

where 0 < < 2


11/17

n = n = n = n =

[Hint:

6cos96cos

6cos96cos=

45sin51sin2

45cos51cos2

= cot51= tan(90 + 51) = tan141 = tan(180 + 141) = tan(321) = tan(3107) n = 3 ]

56 If A + B + C =2

then prove that

Asin 2 + 2 Asin = 1.[Sol. LHS sin2A + sin2B + sin2C + 2 Asin

1 (cos2A sin2B) + sin2C + 2 Asin

1 cos(A + B) cos(A B) + sin2C + 2 Asin (A + B = C2

; cos(A + B) = sinC)

1 sinC [cos(A B) sinC] + 2 Asin1 sinC [cos(A B) cos(A + B)] + 2 Asin1 sinC[2 sinA sinB] + 2 Asin1 2

Asin + 2

Asin = 1 Hence proved. ]

57 If x is eliminated from the equation, sin(a + x) = 2b and sin(a x) = 2c, then find the eliminant.

[Ans.asin

)cb(2

2+

acos

)cb(2

2= 1]

[Sol. adding sin(a + x) + sin(a x) = 2(b + c)

2 sin a cos x = 2(b + c) cos x =asin

cb ....(1)

sub sin(a + x) sin(a x) = 2(b c)

2 cos a sin x = 2(b c) sin x = acos

cb

....(2)squaring and adding both equation (1) and (2), we get

asin

)cb(2

2+

acos

)cb(2

2= 1 Ans. ]

58 If is eliminated from the equations x = a cos( ) and y = b cos ( ) then

)cos(ab

xy2

b

y

a

x2

2

2

2

is equal to

(A) sec2( ) (B) cosec2( ) (C) cos2( ) (D*) sin2( )[Sol. ( ) = ( ) ( )

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

cos( ) = 2

2

2

2

b

y1.

a

x1

a

x.

b

y

2

2

2

22

b

y1

a

x1)cos(

ab

xy

)cos(ab

xy2)(cos

ba

yx 222

22

=22

22

2

2

2

2

ba

yx

a

x

b

y1

)cos(ab

xy2yx2

2

2

2

= sin2( ) ]


12/17


13/17

apply componendo and dividendo,

2cos1

2cos1=

2cos1

2cos1

2

1or tan2=

2

1tan2.

But given tan= k tan whereby k2= 2, and since and are positive acute angles k = 2 ]

65 Prove that : sin7

2+ sin

7

4 sin

7

6= 4 sin

7

sin

7

3sin

7

5

[Sol. RHS = 4 sin7

sin

7

2sin

7

4

7

4sin

7

3sinand

7

2sin

7

5singsinu

2 sin7

7

4sin

7

2sin2 or 2 sin

7

7

6cos

7

2cos =

7cos

7

2cos

7sin2

sin7

2+ 2 sin

7

cos

7

2 sin

7

2+ sin

7

3 sin

7

sin7

2+ sin

7

4 sin

7

6 = L.H. S. proved ]

66 Let)sin(

)sin(

=b

a &

)cos(

)cos(

=d

c then prove that cos( ) =

bcad

bdac

.

[Hint :a

sin ( ) =

b

sin ( ) = k1 ;

c

cos ( ) =

d

cos ( ) = k2

=a c bd

a d b c

=

k k

k k

1 2

1 2

)(cos)(sin)(cos)(sin

)(cos)(sin)(cos)(sin

= )(sin)(2sin)(sin)(2sin)(2sin)(2sin

=

)(2sin2

)(cos)(2sin2

= cos ( ) ]

67 Find the value of the continued product

17

1k 18

ksin [Ans.

162

9]

[Sol. sin10 sin20 sin30........sin90 sin100.......sin160 sin170(sin10 sin20 sin30........sin80)2

= (sin10 cos10)2(sin20 cos20)2(sin30 cos30)2(sin40 cos40)2

=2

80sin60sin40sin20sin16

1

=2

256

3

= 162

9 Ans. ]

68 If x=48 , find the value of

xxxx

xxxx

3cos9cos5cos7cos3sin9sin5sin7sin

. [Ans. 12 ]

[Sol. If x =48

E =x3cosx9cosx5cosx7cos

x3sinx9sin)x5sinx7(sin

= x3cosx6cos2xcosx6cos2

x3cosx6sin2xcosx6sin2

E =)x3cosx(cosx6cos2

)x3cosx(cosx6sin2

= tan 6x = tan

48

6 = tan8

= 12 Ans. ]

69 If X = sin

12

7+ sin

12 + sin

12

3, Y = cos

12

7+ cos

12 + cos

12

3then


14/17

=

12

3sin

3cos

4sin2 =

4

sin4

sin =

4

sin2

Y = cos

12

7+ cos

12

+ cos

12

3

=

4

cos3

cos4

cos2 =

4

cos2

XY

YX =

4sin

4cos

4cos

4sin=

sincossincos

sincossincos

=

tan1

tan1

tan1

tan1=

2

22

tan1

)tan1()tan1(= 2tan2 Hence proved ]

70 The exact value of96 80 65 35

20 50 110

sin sin sin

sin sin sin

is equal to

(A) 12 (B*) 24 (C) 12 (D) 48

[Hint : Asin = 4 2Acos in Dr. as A + B + C = ]71 The value of cotx + cot(60 + x) + cot(120 + x) is equal to :

(A) cot3x (B) tan3x (C) 3 tan3x (D*)3 9

3

2

3

tan

tan tan

x

x x

[Sol. cotx +)60xsin(

)60xcos(

)x60sin(

)x60cos(

= )60xsin()60xsin(

)x2sin(

xsin

xcos

= 3xsin4

xcosxsin8

xsin

xcos2

=xsin3xsin4

xcosxsin8xcos3xcosxsin43

22

=xsin

]xcos4xcos3[33

3 = 3 cot3x

xtanxtan3

]xtan31[33

2

Ans ]

72 , , & are the smallest positive angles in ascending order of magnitude which have their sines equal tothe positive quantity k. The value of

4 sin

2

+ 3 sin

2

+ 2 sin

2

+ sin

2

is equal to :

(A) 2 1 k (B*) 2 1 k (C) 2 k (D) 2k

[Hint : = ; = 2 + ; = 3 where 0 < 0 now 3400lies in IV quadrant. Hence sinA

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