4 Vectors - F_level - Ii_mr. Kishore Sharma_solutions_fina

IIT MATHS

47 I I T AKASH MULTIME

UPTO TRANSFORMATIONSUPTO TRANSFORMATIONSCLASSWORK SOLUTIONS

Exercise -1 Trignometric Ratios

1.2tan ( )bx a c

a c

2 2cos 2 sin cos siny a x b x x c x ... (i)and

2 2sin 2 sin cos cosz a x b x x c x ...(ii)Adding eqs. (i) and (ii)

y z a c [Alternate. (b) ]

andcos 2 2 sin 2 cos 2y z a x b x c x

cos 2 2 sin 2a c x b x

2 2

2

2 2

2 2

4 81

4 41 1

b ba c a c

a cb b

a c a c

2 2 2

2 22 2

4 8

4 4

a c a c b b a c

a c b a c b

2 2

2 2

4

4

a c a c b

a c b

= ( a - c)y z a c

[ Alternate. (c) ]

2.sin costansin cos

tan 1tan 1

tan tan4

/ 4, 1n n

or 2 2 22

n

sin 2 sin 2 cos 22

and cos 2 cos 2 sin 22

and sin cos 2 sin2

2 sin

2 sin

n

and sin cos 2 sin4

2 sin2

n

2 cos n

2 cos3. Given, cos sec 2x x

2cos 2 cos 1 0x x

2cos 1 0x

cos 1x

sec 1x

cos sec 1 1n nn nx x

2,2,if n is even

if n is odd

4. We have, 1 sin 1 cos,

cos sinx y

Multiplying, we get

1 sin 1 coscos sin

xy

1 sin cos sin cos sin cos1

cos sinxy

and

1 sin sin cos 1 coscos sin

x y

2 2sin sin cos coscos sin

sin cos 1 1cos sin

xy

Thus, 1 0xy x y

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IIT MATHS


UPTO TRANSFORMATIONS

1 11 1

y xx and yy x

5. Given , sec tan 1 sec tan 1

2 2[ sec tan 1] 2sec 2

sec 1

cos 1 Clearly, 1 is a root of given quadratic equation.

sec cosand are roots of given quadraticequation.

6. The given relation can be written as

2 tan 2 1 2 1 secm m m

2 2

2

2 tan 2 2 2 1

tan 2 1

m m m

m

2 22 1 1 tanm

2 2 22 2 1 tan 2 2m m m

2 22 1 tan 2 1 2 1 0m m m

2 2 23 1 tan 4 6 4

tan 8 0

m m m

m

23 tan 4 1 tan 2 0m m Which is true if

2tan 4 / 3 tan 2 1 .or m m

7. 0 0sin 1/ 2 30 150x y x y or

(1) and 0 0cos 1/ 2 60 300x y x y or

(2)Since x and y lie between 00 and 0180 , (1)and (2) are simultaneously true when

045x , 015y , or0165x , 0135y .But , for the values given

by (b) or (c), (1) and (2) do not hold simulta-neously.

8.

2 0 0 0 0

0 0 0

8 tan1 tan89 tan 2 tan88 ...

tan 44 tan 46 tan 45

x

0 0 0 0

0 0 0

tan1 cot1 tan 2 cot 2 ...

tan 44 cot 44 tan 45

21 9 3x x .9 Since 1 radian lies between 570 and 580 and

0 0sin 57 sin1 , so 0sin1 sin1 . Again 1radian is an acute angle and 2 radian is anobtuse angle, tan1 0 , tan 2 0 , so that tan1> tan2.

10. tan tanab

2

2 2 tan tantan tan1 tan tan

=

2

2

4 /

1 /

a a b

a b

2

4ab aba b

as 2tan 0, 4ab .

11. We are given that

4 4sin cos 1a b

a b

4 4 4 4sin cos sin cos 1b aa b

22 2sin cos

4 2 2 4sin 2sin cos cos 0b aa b

2

2 2sin cos 0b aa b

4 4sin cosb aa b

4 4

2 2

sin cos k saya b

Since 4 4sin cos 1 ,

a b a b

we get




IIT MATHS



21 1ak bk k

a b a b

2 22 28 8

3 3 3 3

sin cos a k b ka b a b

2 2 2ak bk a b k

4 3

1 1. .a ba b a b

12. 6 42 3 1P P

6 6 4 42 cos sin 3 cos sin 1

2 2 2 2

2 2

cos sin 3 sin cos2

cos sin

2 2 2 23 cos sin 2sin cos 1

2 22 1 3sin cos

2 23 1 2sin cos 1 0

Again for 4n , we have 2n nP P

2 2cos sin cos sinn n n n

2 2 2 2cos cos 1 sin sin 1n n

2 2 2 2sin cos cos sinn n


2 24sin cos nP

10 8 66 15 10 1P P P

10 8 8 6 6 4 4 26 9P P P P P P P P

2 26 4 2 0sin cos 6 9P P P P

2 26 4 2 03sin cos 6 9P P P P

2 0[ 1, 2]P P

2 2 2 23sin cos 1 3sin cos 0 .

6 4[ 2 3 1 0 ]P P as proved

13. We have 4 4 2 2sin cos sin cos ,x x x x as

sin 1x and cos 1x

1a (1)Next, 4 4sin cosx x a

2 2 2 2sin cos 2sin cosx x x x a

21 sin 2 12

x a

1 1/ 2a 2[ sin 2 1]x

1/ 2 a (2)From (1) and (2) we get 1/ 2 1a . Notethat 1/ 2a for / 4x and a = 1 for

/ 2x 14. We have 4 4 2 2sin cos sin cos ,x x x x as

sin 1x and cos 1x

1a (1)Next, 4 4sin cosx x a

2 2 2 2sin cos 2sin cosx x x x a

21 sin 2 12

x a

1 1/ 2a 2[ sin 2 1]x

1/ 2 a (2)From (1) and (2) we get 1/ 2 1a . Notethat 1/ 2a for / 4x and a = 1 for

/ 2x

15

2 422

22 2

2 11sec 11 1

xxx x

2

4

1cos2 1

x

x

2

4

1sin2 1

x

x




IIT MATHS



16 sin 1 so 2sin 5 / 4 ,

22 2

2cos 1 0ab a ba b

which is

true.

2 2

22 2cos 1 0m nec a b

m n

which

is not possible and sin 2.375 1 againnot possible.

Compound Angles

17. Let tan tan tan1 2 3

x y z say

tan , tan 2 , tan 3x y zx y z

3

2

tan tan tan tan tan tan6 6

6 1 0

1 0

tan 1, tan 2, tan 3tan tan tan 6

x y z x y z

x y zx y z

18. 3sin sin 2

2sin sin 2 sin

2cos sin

sin cos sin

....(i)

.( ) ,

sin sin 1

Alternate b is correct Also

sin cos cos sin

....(ii)From eqs. (i) and (ii)

sin sin cos sin

2sin sin cos

.( )Alternate c is correct

Alternate. (a) :

cot cos cot 3cot 2

sin 2 3cos 2cossin .sin sin sin 2

3cos 23sin cossin .sin sin 3sin

LHS

3sin sin 2

cos cos 23sinsin .sin sin

2sin sin3sinsin .sin sin

= 6Alternate. (d) :

19. tan tan tan 6A B C tan tan tan 6A B C 2 tan 6C

tan 3C 2

22

tan 9 9sin1 tan 1 9 10

CCC

From eq. (i),tan tan 3 tan tan 2A B and A B

2

tan tan

tan tan 4 tan tan )

1

A B

A B A B

we get, tan A = 2, 1 and tan B = 1,2

2 4 1sin ,1 4 1 1

A

and 2 1 4sin ,1 1 1 4

B

2 28 5 5 8sin , sin ,10 10 10 10

A and B

2 2 2sin : sin : sin 8 :5 :9 5 :8 :9A B C or




IIT MATHS



tan 2 tan

sin 2sincos cos

sin cos 2cos sin

sin cos cos sin

cos sin

sin cos sin [ Alternate.(b) ]

20. We have,4 4 2 2cos sin cos sin cos 2

2 22 4 4 4 2b b

2 2 0b b

1, 2b

21.cos cosA B

x y

thentan tan sin sin tan

cos cos 2x A y B A B A B

x y A B

tan tan sin sin tancos cos 2

x A y B A B A Bx y A B

sin sin sin cos cos sinsin sin sin cos cos sinsinsin

y A x B A B A By A x B A B A B

A BA B

22 2 2 2 2 2tan tan 1 2 tan tan tan . 2 2 2

2 2 2

2 2 2

sin sin sintan tan tan

1 tan 1 tan 1 tan

2 2 2 2 2

2 2 2 2 2 2

tan 1 tan tan tan tan

1 tan tan tan tan tan tan

2 2 2 2 2 2

2 2 2 2

tan 2 tan tan 3tan tan tan1 tan 1 tan tan tan

2 2 2 2

2 2 2 2

2 tan tan tan tan1

2 tan tan tan tan

2 2 2cos cos cos 2

cos 2 cos 2 cos 2 1

and 2

2 2 2

cos cos cos

cos sin cos

2 2 2cos cos cos 1 which is notcorrect.

23. 2 2 2 2sec sec 1/2 sin sec sec 0

2 2cos cos 1/ 2 sin 0

sin sin 1/ 2 sin 0

sin 0 sin 1/ 2Either or .Multiple and sub multiple angles

24. nP u be a polynomial in u of degree n.

sin 2 2sin cosnx nx nx

2 1 2 1sin cos cos sinn nxP x or xP x

25. 0

tan / 2 1 sec 2n

rn

r

f

0

1 cos 2tan / 2

cos 2

rn

rr

2 1

0

2cos 2tan / 2

cos 2

rn

rr

2 11

0

cos 22 .tan / 2

cos 2

rnn

rr

1 2

0

cos 22 .tan / 2 .cos / 2 .

cos 2

rnn

nr

sin 22 .sin .

2 .sin .cos 2

nn

n n

tan 2n

2.( ): tan 116 4

Alternate a f




IIT MATHS



3.( ): tan 132 4

Alternate b f

4.( ): tan 164 4

Alternate c f

5.( ): tan 1128 4

Alternate d f

26. Given, tan 3tan

A kA

...(i)tan 3 tan 1

tanA A k

A

sin 2 1cos3 sin

A kA A

2 cos 1cos3

A kA

cos 1cos3 2

A kA

(a) is incorrect. Again , tan 3tan

A kA

sin 3 cos.cos3 sin

A A kA A

sin 3 2 2.sin 1 1

A kkA k k

(b) is correct.33sin sin 2

sin 1A A k

A k

2 22 33 4sin 4sin1 1

k kA or Ak k

30 4[sin 0 1]1

k A ork

Now,3 01

kk

,1 3,k

... (ii)

and 3 41

kk

3 1 01

kk

,1/ 3 1,k

From eqs. (ii) and (iii) , we get...(iii)

1 33

k and k

27. 3 cos3 3coscos .sin 2 sin 24

x xx x x

1 3sin 5 sin sin 3 sin8 8

x x x x

1 3 1sin sin 3 sin 5 .4 8 8

x x x

1 2 3

4 5

1 35, , 0, ,4 8

10, .8

n a a a

a a

28 Applying 1 1 3 2 2 3R R R and R R R onthe LHS, the given equation can be written as

2 2

1 0 10 1 1 0sin cos 1 4sin 4

Expanding the LHS along 1R , we get2 21 4sin 4 cos sin 0

4sin 4 2 sin 4 1/ 2

4 7 / 6 11 / 6or

0 0 4 22

7 / 24 11 / 24.or

29

2

2

cos 2 sin 2 1

cos 2 sin 2 1

A Ay

A A

cos 2 sin 2 1cos 2 sin 2 1

A Ay

A A

Which gives us four values of y, asy

1y , 2y , 3y and 4y . We have

1

1 cos 2 sin 2cos 2 sin 2 1cos 2 sin 2 1 cos 2 1 sin 2

A AA AyA A A A




IIT MATHS



2

2

2cos 2sin cos2sin 2sin cos

A A AA A A

cos cos sincot

sin cos sinA A A

AA A A

2


1 cos 2 sin 21 cos 2 sin 2

A Ay

A A

A AA A

2

2

2sin 2sin cos tan2cos 2sin cos

A A A AA A A

2



A Ay

A A

A AA A

2

2

cos sin2cos 2sin cos2cos 2sin cos cos sin

A AA A AA A A A A

1 tan tan cot1 tan 4 4

A A AA

4



A Ay

A A

A AA A

2

2

2sin 2sin cos cos sin2sin 2sin cos cos sin

A A A A AA A A A A

1 tan tan1 tan 4

A AA

.

30 cos5 cos 4cos 4 cos sin 4 sin

22cos 2 1 cos 2sin2 cos2 sin

22 2

2

2 2cos 1 1 cos 2.2cos sin

2cos 1

4 2

2 2

2 4cos 4cos 1 1

cos 4cos 2cos 1 1 cos

4 2

2 4

cos 8cos 8cos 1

4cos 3cos 2cos 1

4 2cos 16cos 20cos 5

5 316cos 20cos 5cos clearly, a =5, b = - 20, c = 16 and d = 0. Satisfy thegiven relation.

31. Clear;u tan 0 and tan 2 0 for0 / 2 . We have

tan 3 tan 2

tan 2 tantan 31 tan 2 tan

0 tan tan 2 tan 3

tan 3 1 tan 2 tan 3 tan 3

tan 3 2 tan 2 tan

tan 3 0 tan 2 tan 2or .

32. The given equation can be written as

2

2 2

1 tan / 2 2 tan / 2sin cot

1 tan / 2 1 tan / 2cos

x xx x

2tan 1 cos sin cot .2 tan2 2

1 cos 0

x x

2 2sin cot 1 costan tan 02 1 cos 2 1 cosx x

2 2tan 2 tan cot tan tan 02 2 2 2x x

2

2

1tan 2 tan . cot tan tan2 2 2 2 2 2

tan 02

x x

tan cot tan tan tan tan 02 2 2 2 2 2x x




IIT MATHS



tan / 2 tan / 2 tan / 2x

or tan / 2 tan / 2 tan / 2x

33 sin cos cos sin sin / 2 sin

cos sin / 2 1 sin 2 1/ 4

sin 2 3 / 4 34. The expression

15 7 3cos .cos .cos2 32 2 16 2 8

cos cos .cos32 16 8

2sin .cos cos .cos32 32 16 8

2sin32

and so on.TRANSFORMATIONS

35. sin sin cos cosa and b

2sin cos2 2

a

.... (i)

and 2sin cos2 2

b

.....(ii)Squaring and additing Eqs. (i) and (ii), then

2 2 24cos2

a b

2 21cos2 2

a b

and 2 22 1 cos a b

2 2

22 2

2

2cos2

1 tan22

21 tan2

a b

a b

By componedo and dividendo method

22 2

2 2

2 tan42

2a b

a b

2 2

2 2

4tan2

a ba b

36. 2 cos cosx

2 cos cosy

and similarly for 2x.Adding, 2 2 2 0x y z .

37 We find from the given relations that and are the roots of the equation

cos sin 2x y a (1)

2 2cos 2 sinx a y

2 2 2cos 4 cos 4x ax a

2 2 2 2sin 1 cosy y

2 2 2 2 2cos 4 cos 4 0x y ax a y

Which, being quadratic in cos , has two rootscos and cos , such that

2 2

4cos cos axx y

and

2 2

2 2

4cos cos a yx y

Similarly, we can write (1) as a quadratic insin , giving two values sin and sin ,suchthat

2 2

4sin sin ayx y

and

2 2

2 2

4sin sin a xx y

38 The given equation can be written as

2 32cos cos 2cos 12 2 2 2

24cos 4cos cos 1 02 2 2




IIT MATHS



222cos cos sin 0

2 2 2

sin 0 2cos cos2 2 2

and

(1)Also, since 0 , , we have .Therefore, from (1) we get 1/ 2 ,so that / 3 .

39 Since cos ,cos cosand arein H.P., we have

2 2

2cos coscos

cos cos

2 cos sin2cos cos

2 2 2cos cos cos sin

2 21 cos cos sin

22

2 2

2

sincos1 cos

4sin / 2 cos / 22sin / 2

2 2cos 2cos / 2

cos sec / 2 2

Periodicty & Extreme Values40. 6 6 2sin cosx x a

3 32 2 2

32 2 2 2

2 2 2

2 2 2

2 2

sin cos

sin cos 3sin cos

sin cos

1 3sin cos31 sin 24

x x a

x x x x

x x a

x x a

x a

or 2 24sin 2 13

x a

2

2

0 sin 2 140 1 13

x

a

2

2

2

30 14

30 14

1 14

1 11, ,12 2

a

a

a

a

41. 2 2 2sin sin sinA B C

21 11 cos 2 1 cos 2 sin2 2

A B C

211 cos 2 cos 2 sin2

A B C

211 2cos .cos sin2

A B A B C

21 cos .cos 1 cosC A B C

22 cos cos .cosC C A B

2 2 2 2sin sin sin 2 cos cosA B C C C ...(i)

22 cos cosC C

2 1 12 cos cos4 4

C C

29 1cos4 2

C

2 2 2 9sin sin sin4

A B C

Now, for positive quantities AM GM

2 2 2

3 2 2 2

sin sin sin3

sin sin sin

A B C

A B C

From eq. (i),




IIT MATHS



2 2 2

2 /3

9 / 4 sin sin sin3 3

sin sin sin

A B C

A B C

2 / 33 sin sin sin4

A B C

or

3/ 23 3 3sin sin sin , ,4 8

A B C ie

Also in ,sin 0,sin 0,sin 0ABC A B C and from eq. (i)

2 2sin sin 1 cosA B C 42. We know that

2 2 2 25 12 5cos 12sin 5 12x x

113 5cos 12sin 0 13 0x xa

and 10 5cos 12sin 13 0 13x xa

113 .13 13 13

113

a a a is positivea

a

and 1 ( ).

13a a is positive

43.2

4 2 2 1 3cos cos 1 cos .2 4

y x x x

min34

y and y is maximum when

22 1cos

2x

is the maximum .

max1 3 14 4

y

44. 2 0 2 23sin . sin 60 sin sin sin

41sin3 .4

y x x x x

x

45.

7 247cos 24sin 25. cos sin25 25

25cos ,

x x x x

x

where 7cos .25

46.

2

tan .tan3

tan 3 1 tan 0.

y A A

A y A y

As tan A is real,

20 3 1 4 0D y y

2 13 10 3 0 3.3

y y y or y

But each of tanA, tanB is less than 3 .or one

is 0 and the other is 3 .

3y is not possible. So 1 .3

y Also ,

tan .tan 0.A B

47. 2 5 3/2 cos 3 3/2 sin 5a b

2213 3 352 2

a

and 225 13/ 2 3 3 / 2b

2, 12a b .




IIT MATHS


UPTO TRANSFORMATIONSEXERCISE-II

Linked ComprehensionsComprehension - I

0 22, 1P P and

2 22 sin cos sin cosn n n n

n nP P

2 2 2 2sin 1 sin cos 1 cosn n


2 2 4 4sin cos sin cosn n

2 24sin cos nP

2 22 4sin cosn n nP P P

...(i)for n = 4.

2 24 2 0sin cosP P P

2 2

4

2 0

1 2sin cos1, 2

PP P

2 24 1 2sin cosP

(ii) for n = 6,



2 2 2 21 2 sin cos sin cos 2 2

6 1 3sin cosP (iii)

1. 6 42 3 10P P

2 2 2 22 1 3sin cos 3 1 2sin cos 10

(From eqs. (ii) and (iii))= 2 - 3 + 10 = 9

2. Let 2 2sin cos ,k then from eq. (i),

2 4n n nP P kP

From eq. (ii), 4 1 2P k

and from eq. (iii), 6 1 3P k

Put n = 10, then 10 8 6 1 3P P kP k k 2

10 8 3P P k k ....(iv)and put n = 8,then 8 6 4 1 2P P kP k k

28 6 2P P k k

21 3 2k k k

28 2 4 1P k k

From eq. (iv), 210 5 5 1P k k

10 8 66 15 10 7P P P

2 26 5 5 1 15 2 4 1 10 1 3 7k k k k k

= 83. From eq. (i), 2 2

2 4sin cosn n nP P P

4n Comprehension - II

Given 3 5cos ,cos ,cos

7 7 7

are the roots of

the equation3 28 4 4 1 0x x x

...(i)

Replacing x by 1x in eq. (i),then we get

3 24 4 1 0x x x ... (ii)

4. From eq.(i), 3 28 4 4 1x x x

3 58 cos cos cos7 7 7

x x x

Put x = 1, then

3 51 8 cos 1 cos 1 cos7 7 7

x

2 2 23 51 8 2sin 2sin 2sin14 14 14

3 5 1sin sin sin14 14 14 8

5. From eq. (i),3 28 4 4 1

3 58 cos cos cos7 7 7

x x x

x x x

Put x = -1 , then - 8 - 4 + 4 + 1

3 58 1 cos 1 cos 1 cos7 7 7

3 57 8 1 cos 1 cos 1 cos7 7 7

2 2 23 58 2cos 2cos 2cos14 14 14




IIT MATHS



3 5 7cos cos cos

14 14 14 8

6. From eq. (iii),

2 2 23 5tan tan tan 217 7 7

32

1

2 1tan 217r

r

Also, replacing x by 1x in eq. (iii), then

3 2

1 21 35 7 0x x x

3 27 35 21 1 0x x x ....(iv)

3 2 23 5cot ,cot ,cot7 7 7 are the

roots of eq. (iv)

3 2 23 5 35cot cot cot 57 7 7 7

32

1

2 1cot 57r

r

Hence,

3 32 2

1 1

2 1 2 1tan cot 21 5 1057 7r r

r r

Comprehension - III

7.cos cotcos cotcos cot

AL a xBM b yCN c z

8. sin ,sin sinz x yA B c

a b c

9. cos .a x ec etc If / 3 3 2A A OAN

sin 2 ,sinz xa a

2

22sin cos 2. 1z x xa a a

2 22az x a x

33

3sin sin3 3sin 4sin 3 4x xAa a

2 3

3

3 4xa xa

Comprehension - IV

cos 3 ,sin 3

OL rCL r

so 3 3sin cos

1sin 3 cos 3r r

3 3sin cos

sin 3 cos 3r

10 Now

4 4cos sin cos cos 3

sin sin 3r

r

cos 2 cos 3r

cos2 cos cos2 sin sin 2r

1 cos cos 2 sin sin 2r r

1 cos tan 2sinr

r

(1)11. Next,

3 3cos sin sin coscos 3 sin sin 3 cosr

sin cos sin 2r

sin2 2 sin cos2 cos sin2r

1 2 cos sin 2 2 sin cos 2r r

1 2 cos cot 22 sin

rr

(2)

12. From (1) and (2) we get1 cos 2 sin

sin 1 2 cosr r

r r

2 2

2 21 cos 2 cos 2 cos

2 sinr r r

r

22 1

cosr

r




IIT MATHS


UPTO TRANSFORMATIONSComprehension - V13. AOB OAB OBA

2 2OAB

and 2 cos 2 sin / 22 2

AB R R

14. Next, Area of the segment AMB= Area of sector AMB - Area of the .AOB

21 1 sin2 2 2 2

R AB R

21 1 2 sin / 2 cos / 22 2

R R R

21 sin2

R

15. Now area AMBECL = 2 Area AMBEOA

21 2Rn

[ Area of the sector ABE +

Area of segment AMB]

22 21 1 12 sin2 2 2 2

R AB Rn

2 2 21 12 4 sin sin2 2 2 2 2

R R

2 1 cos sinR

2 cos sinR

1 1sin cos 1 nn n

EXERCISE-IIIMATCHING

1. 0 0 0 0 0 0sin70 sin10 sin 40 30 sin 40 30

2 0 2 0 2 0 2 0sin 40 sin 30 cos 30 cos 40 0 0 0 0sin 20 sin 80 cos 70 cos10

2. Substitute the values of A and B given onR.H.S to verify the equations on on L.H.S.

3.2cos cos ,

2 2

2sin cos2 2

tan2

a b

ba

2 2 2 2

2 22 2

1 /cos

1 /

b a a ba bb a

and 2 22 2

2 / 2sin1 /

b a aba bb a

2 2 2 2 2

2

sin sin coscos 2cos cos 2sin sin

a b

2 2 2 cosa b

4. 0 0 0

0

2 3 90 18 ,3 7 909

A A A A AA

00 0

0

13 90 22 , 2 902

30

A A A A A

A

5. cos sin 1/ 2 1 2sin cos 1/ 4x x x x

2sin 2 3 / 4 cos sin1 3/ 4 7 / 4

x x x

cos sin 7 / 2x x

2cos 1/2 7/2 cos

7 1 /4cos2 1/2 7 /2 7 /4.

x x

x




IIT MATHS



6.13 14 13 14, 3 ,48 48 16 16

3 /16, 2 /16 i.e.

3 sin 3 0IIquadrant

13 14 11 102 , ,24 24 24 24

i.e. 2 cos 2 0IIquadrant

sin 3cos 2

is negative.

Similarly

14 18 2 2, 3 ,48 48 16 16

sin3 can be positive or negative.

10 612 , cos 224 24

is negative.

sosin 3cos 2

can be positive or negative.

If 18 23, , sin 348 48

is ve

and sin 3cos 2cos 2

is ve so

is positive

and if sin 323 / 4 cos 2 0cos 2

then so

is not defined.7 2 2cos 2 cos sin xy

2 2 2sin2 1 cos sin 1 1y x

2coscot2sin

x yx y

2 1 tan 2xxy

.

82 2 2

cos 2 cot 2 cos 21 cos 2 2 sin 2 2

sin 2 sin 2

eck

k

21 tan 1 sin 2 11 tan 1 sin 2 1

kk

21sin 2 1 cos 4 sin 2 cos 22

2 2sin 2 1 sin 2 1k k 3 3sin 6 3sin 2 4sin 2 3 4 .k k

9

2

2 2 2 2

1 2sin /2 cos /2 cos /2 sin /2cos /2 sin /2 cos /2 sin /2

x x x xx x x x

cos / 2 sin / 2cos / 2 sin / 2

x xx x

1 tan / 21 tan / 2

xx

1 tan / 21 sincos 1 tan / 2

xxx x

2

22

2sin / 21 cos tan / 21 cos 2cos / 2

xx xx x

2 2 2

2 2 2

cos /2 sin /2 1 tan /2cos

cos /2 sin /2 1 tan /2x x x

xx x x

.10. A, B, C are in A. P.

0 0 060 ; 90 30B C A

sin sin sin 3 3 / 2A B C

cos cos cos 3 1 / 2A B C

sin 2 sin 2 sin 2 4sin sin sin

3

A B C A B C

cot cot cot cot cot cot 1B C C A A B .

11 13 / 6 2 / 6,25 / 64 / 6,37 / 6 6 / 6

So for all values in p,q,r and s

sin sin /6 1/2sin2 sin /3 3/2

tan 2 3 cot 3,cot 2 1/ 3

12 (A) sin sin sin2 2 2

1 sin sin sin sin2 2 2 2

A B C

A B C

= 1 2sin cos4 4

2cos sin4 4

A B A B

C C




IIT MATHS



1 2sin cos cos4 4 4C A B C

A B C

1 2sin4

C

2sin sin8 8

C A B C B A

1 4sin sin sin4 4 4

C B A

1 4sin sin sin4 4 4

A B C R

1 4cos cos2 4 2 4

sin4

A B

C

1 4cos cos sin4 4 4

A B C T

(B) sin sin sin2 2 2A B C

1 sin sin sin sin2 2 2 2A B C

1 2sin cos4 4

2cos sin4 4

A B A B

C C

1 2sin4

cos cos4 4

C

A B C

A B C

1 2sin4

2cos cos8 8

C

C A B C B A

1 4sin cos cos4 4 4

C B A

1 4cos cos sin4 4 4

A B C s

1 4sin sin2 4 2 4

cos2 4

A B

C

1 4sin sin cos4 4 4

A B C p

(C) cos cos cos2 2 2A B C

cos cos cos cos2 2 2 2A B C

2cos cos4 4

2cos cos4 4

A B A B

C C

2cos cos cos4 4 4

C A B C

A B C

2cos4

2sin sin8 8

C

C A B C B A

4cos sin sin4 4 4

C B A




IIT MATHS



4cos cos4 2 4

cos2 4

C B

A

4 cos cos cos4 4 4

A B C Q

13 (A) If M is mid point of PQ, then

sin sin,2 2

M

Also ,sin2 2

N

It is clear from the figure. ML NL

sin sin sin2 2

sin sin sin 2sin 22 4

sin sin 2 and

sin sin sin 14

(P,Q,R,S,T)

(B)

2 22 2 sin sin cos cosa b

2 2cos

24cos 4( , )2

S T

(C) 3sin 5cos 5 3sin 5 1 cos

Squaring both sides, then

229sin 25 1 cos

29 1 cos 1 cos 25 1 cos

9 1 cos 25 1 cos 1 cos 0

34 cos 16

8 15cos , sin17 17

then

75 245sin 3cos 317 17

R

Hence, 5sin 3cos 3

14. (A) Let 2

2

7 6 tan tan1 tan

y

2 27 cos 6sin cos sin

1 cos 2 1 cos 27 3sin 22 2

3sin 2 4cos 2 3

2 2

2 2

3 4 3 3sin 2 4cos 2

3 3 4 3

2 8y

8, 2

6, 10 ,R S

(B)Let 5cos 3cos / 2 3y

1 35cos 3 cos sin 32 2

13 3 3cos sin 32 2

22

22

13 3 3 13 3 33 cos sin2 2 2 2

13 3 33 32 2

3 7 3 7y

4 10y

10, 4

6, 14 ,R T

(C) Let 1 sin 2cos4 4

y

1 cos 2cos2 4 4

1 cos 2cos4 4

1 3cos4




IIT MATHS



1 cos 14

3 3cos 34

1 3 1 3cos 1 34

2 4y

4, 2

2, 6 ,P Q

15 (A) Let tan , tan , tanx A y B z C

tan tan tan tan tan tanA B C A B C

tan tan tan 1 tan tanA B C A B

tan tan tan1 tan tan

A B CA B

tan tanA B C

A B C

A B C (i)

2 2

tan 1 sin 21 1 tan 2

x A Ax A

1 4sin sin sin2

A B C

2 2 2

2

1 1 1

xyz

x y z

22

1

x Sx

(ii)

2 2

tan 1 tan 21 1 tan 2

x A Ax A

1 tan 2 .tan 2 . tan 22

A B C

2 2 2

1 2 2 2. . .2 1 1 1

x y zx y z

2

41

x Qx

(iii)2 2

2 2

1 1 tan cos 21 1 tan

x A Ax A

cos 2 cos2 cos2A B C 1 4cos cos cosA B C

1 4 cos A

2

11 41 tan A

2

11 41

Rx

(iv)

22 2

1 1 cos1 1 tan

Ax A

1 2cos cos cosA B C

11 2sec A

2

11 21

Tx

(B) 1xy yz zx

Let cot , cot , cotx A y B z C

cot cot cot cot cot cot 1A B B C C A

A B C (i)

2 2

cot 1 sin21 1 cot 2

x A Ax A

1 4sin sin sin2

A B C

2 2 2

2

1 cot 1 cot 1 cotA B C




IIT MATHS



2

121

Px

(ii)

2 2 2

cot tan1 1 cot 1 tan

x A Ax A A

1 tan 22

A

1 tan 2 tan 2 tan 22

A B C

2 2 2

4 tan tan tan1 tan 1 tan 1 tan

A B CA B C

2 2 2

4cot cot cotcot 1 cot 1 cot 1

A B CA B C

241

xx

241

x Qx

(iii)

2 2 2

2 2 2

1 1 cot tan 11 1 cot tan 1

x A Ax A A

= cos 2A

1 4cos cos cosA B C

1 4 cos A

2

1 4 , , , ,1

x P Q R S Tx

(iv)

22 2

1 1 sin1 1 cot

Ax A

2 2cos cos cosA B C

2 2 cos A

22 2 , , ,

1

x P Q R Sx

(C) 2 3x y z

2 2 2 2 3x y z xy yz zx

1 2 3xy yz zx

1xy yz zx

Now, 2 2 2x y y z z x

2 2 22 x y z xy yz zx

= 2 (1 -1)= 0

Which is possible only when0, 0, 0x y y z z x x y z

2 2 23 1x y z andx y z

13

x y z

(i)

2 2 22 1 1 11x x y z

x y zx

2

133 3 33

11 413

xx

3

2 2

1 12 2.1 1x x

3

31 32. 2

2113

2.3 38

3 3

4

2 2

121 1

x Px x

(ii)




IIT MATHS



2 2

13 333 3 11 1 21

3

x xx x

and

3

2 24 41 1

x xx x

31 13 3 34 4.1 813 27

4 278 3 3

3 32

2 241 1

x x Qx x

(iii)

2 2

2 2

111 1 333 11 1 213

x xx x

and

3

2 2

1 11 4 1 41 1x x

3

11 4113

3 3 31 4 , , , ,8 2

P Q R S T

(iv)

22

1 1 1 93 3 11 41 13

xx

and

3

22

1 11 2 1 211 xx

1 3 3 91 2 1 2 , , , ,8 411

3

P Q R S T

16.

(A) Let

,sec , ,sec ,secP A A Q B B andR C C

are there points on, sec ,Y X then centroidof PQR is

sec sec sec,3 3

A B C A B CG

or sec sec sec,

3 3A B CG

Draw a line GL perpendicular to x-axis , whichcuts 2secY X at M.

,sec3 3

M

or , 23

M

It is clear from the figure LG LM

sec sec sec 23

A B C

sec sec sec 6 sec 6A B C or A 6

Again, let

2 2 2,tan , ,tan ,tanP A A Q B B andR C C

are three points, on 2tan ,Y X then cen-

troid of PQR is

2 2 2tan tan tan,3 3

A B C A B CG

or 2 2 2tan tan tan,

3 3A B CG

Draw a line GL perpendicular to x-axis, whichcuts 2tanY X at M.

2, tan3 3

M

or ,33

M

It is clear from the figure LG LM2 2 2tan tan tan

33

A B C




IIT MATHS


UPTO TRANSFORMATIONS2 2 2 2tan tan tan 9 tan 9A B C or A

9

3 3 3 0 ,and Q S (B) Let

,cos , , ,cos2 2 2A B CP A ec Q B and R C ec

are three points on cos2XY ec

,then

centroid of PQR is

cos cos cos2 2 2,

3 3

A B Cec ec ecA B CG

or

cos cos cos2 2 2,

3 3

A B Cec ec ecG

Draws a line GL perpendicular to x-axis, which

cuts cos2XY ec at M.

, cos3 6

M ec

or , 23

M

It is clear from the figure LG LM

cos cos cos2 2 2 2

3

A B Cec ec ec

or cos cos cos 62 2 2A B Cec ec ec

or cos 62Aec

6

Again, let 2 2,sec , , sec

2 2A BP A Q B

and 2,sec

2CR C are three points on

2sec ,2XY then centroid of PQR is

2 2 2sec sec sec2 2 2,

3 3

A B CA B CG

or

2 2 2sec sec sec2 2 2,

3 3

A B C

G

Draw a line GL perpendicular to x-axis, which

cuts 2sec

2XY at M.

2,sec3 6

M

or 4,

3 3M

It is clear from the figure GL ML

2 2 2sec sec42 2 2

3 3

A B Csce

or 2 2 2sec sec 4

2 2 2A B Csce

or 2 4

2Asce

4

2 2 3 0 ,and P T




IIT MATHS


UPTO TRANSFORMATIONS(C) Let

,cos , ,cos ,cosP A A Q B B and R C C

are three points on cos ,Y X then centroidof PQR is

cos cos cos,3 3

A B C A B CG

or cos cos cos,

3 3A B CG

Draw a line GL perpendicular to x-axis, whichcuts, cosY X at M.

1,3 2

M

It is clear from the figure. GL MLcos cos cos 1

3 2A B C

3cos cos cos2

A B C

31 4sin / 2 sin / 2 sin / 22

A B C

(from Identity)

1sin / 2 sin / 2 sin / 28

A B C

or

cos / 2 cos /2 cos /2 8ec A ec B ec C

or cos / 2 8ec A 8

Again, let

2 2 2,cot ,cot ,cotP A A Q B B and R C C

are three pointson 2cotY X

Then, centroid of PQR is2 2 2cot cot cot,

3 3A B C A B CG

or2 2 2cot cot cot,

3 3A B CG

Draw a line GL perpendicular to x-axis which

cuts 3cotY X at M.

2, cot3 3

M

or 1,

3 3M

It is clear from the figure GL ML2 2 2cot cot cot 1

3 3A B C

or 2 2 2cot cot cot 1A B C 2 2 2cos 1 cos 1 cos 1 1ec A ec B ec C

2 2 2cos cos cos 4ec A ec B ec C

or 2cos 4ec A 4

4 R EEEEEEEEE4IIIIIII IIII IIIIIIIII1 . 2 2 2

2 2 2

2 3cos cos cos7 7 7

1 1 12 3sin sin sin

7 7 7

ec ec ec

= 2 2 2

2 4 61 cos 1 cos 1 cos7 7 7

=




IIT MATHS



4 6 2 6 2 42 1 cos 1 cos 2 1 cos 1 cos 2 1 cos 1 cos7 7 7 7 7 7

2 4 61 cos 1 cos 1 cos7 7 7

NNN NNNNNNNNNNN =2 4 61 cos 1 cos 1 cos7 7 7

= 2 4 61 cos cos cos7 7 7

+ 2 4 4 6 6 2cos cos cos cos cos cos7 7 7 7 7 7

- 2 4 6cos cos cos7 7 7

=

1 112 2

2 4 4 6 6 22cos cos 2cos cos 2cos cos7 7 7 7 7 7

18

= 1 112 2

6 2 10 2 8cos cos cos cos cos7 7 7 7 7

4 2 4 6cos cos cos cos7 7 7 7

18

= 1 1 1 712 2 8 8

AAAA AAAAAAAAA

=

2 4 63 2 cos cos cos7 7 7

22 4 4 6 6 2cos cos cos cos cos cos7 7 7 7 7 7

= 1 12 3 2 7

2 2

N N N2 2 22 3 7cos cos cos 8

77 7 78

ec ec ec

2 . LLL LLLL LLLLLLL 4p aaa 4p FFF F FFFFFFFF FF FFFF FFFF

4 2 2 4p p p

sssssss sss ssssss ssssssssss, ss sss5 1,2p lllll lllllll l ll 2.

3 . 112º 5 60º cos52

1cos(3 2 )2

A A A

A B

(ii) 1cos3 cos 2 sin 3 sin 22

A A A A

(ii) 3 2

3

(4cos cos )(2cos 1)(3sin 4sin )(2sin cos )12

A A AA A A A

(ii)

3 2

2 2 2

4cos 3cos 2cos 1

13 4sin 2sin cos2

A A A

A A A

(ii)

3 2

2 2

4cos 3cos 2cos 1

13 4 4cos 2cos 1 cos2

A A A

A A A

(ii)

3 2

3 2

4cos 3cos 2cos 1

18cos 2cos 1 cos2

A A A

A A A

SSSSSSSSSSSSSS SSSSS S 5th ddddddddddddddddii iii i.




IIT MATHS


UPTO TRANSFORMATIONS(AAA) 5.

4 .

0 00

0 00 0

sin 1sin1sin sin 1 sin sin 1

x x

x x x x

=

0 00 0

00

sin 1 cos cos 1 sinsin sin 1

x x x xx x

= 00cot cot 1x x . . . . . . . . . . . . . (1 )

DDDD

0 0

0 0

0

1 1sin sin1

sin1 sin1 ......sin 45º sin 46º sin 47ºsin 48º

sin1sin133º sin134º

n

=

0

cot45º cot46º cot47º cot48º1sin1 .... cot133º cot134º

(1)

=

0

00

cot 45 cot 46º cot134º1 cot 47º cot133º ....

sin1cot89º cot 91º cot 90

= 0

1 1sin1

TTTT 0 0

1 1 1sin sin1

nn

5 .2 sin costan

2 sin cos

=

2 2cos4

2sin4

=

21 cos 2sin4 2 8

sin 2sin cos4 2 8 2 8

= tan2 8

ccccccccc ccc, ccc 8 6 . AAAAAAAA

2 3

1

1

cos cos 2 cos 2 cos 2 .......cos 2sin 22 sin

n

h

h

A A A A AAA

wwwww 2 4 8 16 1cos cos cos cos15 15 15 15

tttt 1 1 cos 1616

7 . cos sin 3 cos cos 33

n n ie x x

2 32

x n x

x xxxxxx xx ,2 2

nnnnnn 3 , ,8 4 8

nnnnnn nn nnnnnn nn nnnnnnnnnnnn nnn3.8 . 0 0180 , 45A B C B

0 0135 cot cot135A C A C

cot cot 1cot cot

1 cot cot 1cot cot

A CieC A

ie A CC A

cot cot cot cot 1

1 cot cot cot cot 1 1

1 1 cot cot 1 cot 2

1 cot 1 cot 2

ie A C A C

ie A C A C

ie A C A

ie A C

9 .0

01142 2 2852

A A

NNN 0tan 2 tan 285A

0

0 02 2

0 tan2 tan 285

2 tan 2 tantan 270 151 tan 1 tan

cot15 A

A AieA A

2

2 tan 2 31 tan

AieA




IIT MATHS



22 3 tan 2 tan 2 3 0ie A A

AAA tan 0,A ss ssssssss ssss ss ssssseeeeeeee ee

24 4 2 3

tan2 2 3

A

1 8 4 3 1 2 4 2 32 3 2 3

1 2 3 1

1 2 6 2 32 3

2 2 3 6

TTTT tan 2 2 3 6A BBB BB BBBB

01tan142 tan 2 2 32

A

tttt 6

1 0 . 0 0 0 0 0sin40 tan20 tan80 tan60 tan40

0 0

0 0 0 0

0 0

0 0 0 0

sin100 sin 20cos 20 cos80 cos 60 cos 40

sin80 sin 20cos 20 cos80 cos 60 cos 40

0 0 0 0 0 0

0 0 0 0

sin80 cos60 cos40 sin20 cos20 cos80cos20 cos40 cos60 cos80

NNNNNNNNN 0 0 0 01 1sin 80 cos 40 sin 40 cos802 2

0 0 0 0 01 2sin120 sin40 sin120 sin40 sin404 4

AAAAAAAA 0 0cos cos 60 cos 60

1 1cos3 , Denominator4 16

A A A

A

tttt 0

0

1 sin 402sin 40 8

116

WORK SHEET-1SSSSSS SSSSSS SSSSSSSSS

1 . 2 3

4sin 3sin 2sin sin2 2 2 2

= 2 sin cos2 2

= 2 1 sin 2 1 K

2 . GGGGG

=

2 2

2 2

2 1 tan 1 tan

2 1 cot 1 cot

= 8

4 44

1 tancot tantan

=

8

8

4

4

1 ab

ab

= 8 8

4 4

b aa b

3 . tan p tan tan o tan tan tan r

GGGGG= 2 2 2 2 2 21 tan tan tan tan tan tan =

2 21 tan 2 tan tan tan tan

22 tan tan tan tan tan tan tan

= 2 21 2p rp r = 21 p r

4 .10 10

3

0 0

1cos cos 3cos3 4 3r r

r rr

= 1 1 1 1 1... 1

43 2 4 10cos0 cos cos cos ... cos4 3 3 3 3




IIT MATHS



= 1 3 1 114 4 2 8

5 .

0

2 tan 20 tan 50sin 70cos50cos 20sin 50

cos 20 cos50sin 70 sec50

cos 20cos50

x

2 tan 20 cot 20 2cos 402sec20 , 2y ec

y x

6 . 2 2tan 1 tan secx x x

4 3 2tan 2 tan tan 2 tan 1x x x x

4 3 2

4 3

tan 2 tan 2 tan sectan 2 tan tan

x x x xx x x

= 22tan tan 1x x

7 . 2 22cot cos 2cot 1 cotec

= 21 cot 1 cot 1 cot

8 . 2sin 1 2 2 1 2x y x

2 22 1 0x x y

2 21 0x y 1 0x y

9 . 22

n

tan tan 2 .... tan 2 2 tan 2 1n n

tan tan 2 ..................cot 2 cot 1

1 0 . 06

sssss ss ss > sssss ss ss

,sin , sin6 6

A B

sinsin 6

6

cos / 3ec

MMMMMMMM MMMMMM MMMMMMMMM11. tan tan ,tan tan 0p q p

tan tantan1 tan tan 1

pq

1p

q

[ Alternate. (b) ]

Alernate. (a) :

22

tancos

tanLHS

p q

2

2

tan tan1 tan

p q

2 2

2

2

2

22 2

2 2

11

11

1 11

p p qqqp

q

p p q q qq p

22

22

1

1

q p q

p q

= q

Alternate. (c) : tan1

pq

22

1cos

1

q

p q

Alternate. (d) :

22

sin1

p

p q

12. cot tan sec cosxand y Alternate. (a) :

2 2

1 1 sin coscot tan cos sinx

sin cos1

1sin cosx




IIT MATHS


UPTO TRANSFORMATIONSAlternate. (b) :

2 21 cos sinsec coscos cos

y

sin tan Hence, sin tan y Alternate. (c) :

22

2 2 3

1 sin 1sin cos cos cos

x y

2 / 32

2

1 cosx y

2 /32 2secx y

... (i)

and4

2 32

1 sin tansin cos cos

xy

From eqs. (i) and (ii),

2 /3 2 / 32 2 1x y xy

13. We have,

2 2 2 4 4 2 2sin cos sin cosx y a

2 4 4 2 5 5sin cos sin cosa and xy a

42 2 2

52

sin cossin cos

p pp

q qq

x y axy a

Which is independent of , if 4p = 5qie, if p = 5 and q = 4

14. Let tan / 2 tan / 2 ,and to that

2b

Also,

2 2

2 2

1 tan / 2 1cos1 tan / 2 1

and 2 2

2 tan / 2 2sin1 tan / 2 1

Similarly, 2

2 2

1 2cos sin1 1

and

We have from the given relations

2

2 2

1 21 1

x a y a

2 2 2 0x y a x

Similarly, 2 2 2 0x y a x

We see that and are the roots of the

equation 2 2 2 0,xz yz a x so that

2 /y x and (2 ) /a x x .

Now, from 2 2 4 , weget

22 4 22 2

a xy bx x

2 2 22 1y ax b x

Also, from 2 2y and bx

we get, y yband bx x

1tan2

y bxx

and 1tan2

y bxx

15.

22 2 2

2 2 2

cos /2 sin /2 cos /2 sin /2

cos / 2 sin /2 cos / 2 sin /2

2

2

sin / 2 1 coscos / 2 1 cos

16. 02

2 2 4

0

cos 1 cos cos ....n

n

x

2

11 cos

2

1sin

2 1sinx

...(i)

and 2 2 4

0

sin 1 sin sin ...n

n

y

2

11 sin




IIT MATHS



2

1cos

2 1cosy

...(ii)From eqs. (i) and (ii)

2 2sin cos 1

1 1 1x y

x y xy ...(iii)

and2 2 2 2

4 40

cos sin 1 cos sincos sin ...

n n

n

z

2 2

11 sin cos

11 11 .x y

[ form eqs. (i) and (ii)

1xyz

xy

xyz xy z [ Alternate. (b) ]xyz x y z [ from eq. (iii) ]

[ Alternate. (c) ]Linker Comprehension sComprehension- I17. Let the angles of the triangle be

0 00, .a b a and a d

Then , 180a b a a d

3 180a

60a Thus, the angles are

0 0060 ,60 60d and d Number of grades in the least angle

10609

d

10 609

d

and number of radians in the greatest angle

60180

d

60180

d

According to question,

10 60 409

60180

d

d

10 4060 609 180

d d

40d Hence, the angles of the triangle are

0 00 0 0 060 40 ,60 , 60 40 , 20 ,60 ,100 .ie 18. Let the angles of the triangle be

0 00, .a d a and a d

Then, 180a d a a d

3 180a

60a Thus, the angles are

0 0060 ,60 60 .d and d Number of

degrees in the least angle 60 d andnumber of radians in the greatest angle

60180

d

60180

d

According to question,

60 60

60180

d

d

160 603

d d

180 3 60d d

30d Hence, the angles of the triangle are

0 00 0 0 060 30 ,60 , 60 30 ,30 ,60 ,90 .ie

19. If the number of sides be 5 and 4 , then we




IIT MATHS



have 2 25 4 20

1 124 5 20

1220 20

2 Hence, number of sides be 10, 8

Comprehension- II20. 3 3sin cos sin cosx y

and sin cos 0x y

From eq. (ii), tan yx

2 2 2 2sin cosy xand

x y x y

From eq. (i),

3 3

3/ 2 3/ 2 2 22 2 2 2

y x xyx yx yx y x y

or

2 2

3/ 2 2 22 2

1x y

x yx y

1/ 22 2 1x y

or 2 2 1x y which is a circle

21. 2 tan , 2sinm n m n ...(i)

and 2 2 2 2tan sin sin sec 1mn

2 2sin tan 2 2

2 2m n m n

[from eq. (i)]

22 2 16m n mn

22. sin cos a ...(i)3 3sin cos b ...(ii)

From eq. (i),2 2 2sin cos 2sin cos a

or 2 1

sin cos2

a ....(iii)

From eq. (ii),

3sin cos 3sin cos sin cos b

23

3 12

aa a b

[from eqs. (i)

and (iii) ]3 32 3 3 2a a a b

3 2 3 0a b a

On comparing, we get 1, 2, 3v

0v 3 3 3 3v v

3 1 2 3 18 Match the following

23. 2 2 2 22 2

1 1sin cosa b x xk k

22 2 2 2

2 2 2 2 4

1 1 sec 1tan xa b c xk k k b k

2

1 sin / 2bc x ak

1 sincot1 1 cos

ak xxck bk x

2cos 1 cos sin 1 1 .sin 1 cos sinx x x

x x x ak

24.1 5sin sin sin2 6 6

1cos cos cos 22 3 3

tan 3 tan tan3 3




IIT MATHS



1 3sin cos 2 .2 2 6 6

and or

Integer Answer Questions

25. cos cos 2 , sin sin 2x ya b

2 22

2 2

2 2

cos cos 2 2cos

sin sin 20 2sin sin 2

x ya b

2 2cos

2 cos cos2 2 2cos 2 2cos k

2 22cos 4cos 2 4 2 4cos 8 2 cosk k

4 2 2k

6 2 3k k

26 . 2 2cos sinnU

6 6 2 26 cos sin 1 3sin cosU

4 4 2 24 cos sin 1 2sin cosU

6 42 2 2 2

2 1

2 6sin cos 2 sin cos 10

U KUk k

2 22 6 sin cos 3 3k k o k

27.2sin 2 4sin 4 6sin 6 .....

90sin 90 ..... 178sin178ºx

= 02 178º sin2 4 176 sin4 ...... 90sin90º

= 180 sin 2 sin 4 ..... sin88 90

0sin190 2sin 2sin1 2sin 4sin1 .... 9sin88sin190sin1

x

= 90 cos1 cos3 cos3 cos5..... cos87 cos8990sin1

0sin1 90 cos1 sin1 sin1x

90ºx cot1º

28. 2 2tan 2 0a x a

2ta n 2 0a and 0a

2a x n

2nx

29.3tan4

2

47 5.7 5cot 133 189 4 sec 1 9 4.4

30. 72 2sin 21 cos 21 1

WORK SHEET-2SSSSSS SSSSSS SSSSSSSSS

1 . 4 2 25 15 3 AAAAA AAA AA AAAA AAAAA

4 2 4 2tan 3tan tan 315 15 5 15

tan

2 . 2 21 1z Sin x y y x

2 2 2 2 2 2 22 2 1 1z x y x y xy x y

2 21 1Cos x y xy

= 2 2 2 2 2 2 2 22

2 2z x y x y xy z x y

xy xy

3 .6 4 2

4 2 3 , sin 4 sin 364sin 112sin 56sin 7 0

n

LLL 2sin t

3 264 112 36 7 0t t t

2 2 22 4 8 112 7sin sin sin7 7 7 64 4

2 4 8 7sin sin sin7 7 7 2

4 . sin sin . cos cos .a c




IIT MATHS



2sin cos2 2

2cos cos2 2

a

C

2 2

2tan sin2

a acc a c

5 . 2

1 tantan tantan1 tan tan 1 tan

1cot tan

nn

nn

2

22 2 2

1tan

cot tan 2x

n n

2tan ii iiiiiii iiii iiiiiiiiiii iimmmmmmm

2 2 2 2cot tan 2 2 1. 2n n n n

2 2 4n n n

mmm mmmmm mm 2

2 1tan

4n

n

6. 24 16 15 0x x 24 10 6 15 0x x x

2 2 5 3 2 5 0x x x

3 5 22 2

x x

tan 2 cos 1

7 . 1 tan 1 tan 2A B

TTTT 4 14

A B n

2 ,5 2

94

A B

92 34

920

8 . 22 2cos 2 cos 2cos 1y x x x

= cos 2 2 cos 2 1 cos(2 2) cos 2 1x x

2sin 1y

9. 22 2cos 2 cos 2cos 1y x x x

= cos 2 2 cos 2 1 cos(2 2) cos 2 1x x

2sin 1y

10.

sin 3. 47 .cos7sin

7

Sum

( See the hint to

Q.No. 52)

sin sin 17 .22sin

7

MMMMMMMM MMMMMM MMMMMMMMM

11 .3cos5

4sin5

and 5cos

13

12sin13

Alternate. (a) :

cos cos cos sin sin

3 5 4 12. .5 13 5 13

3365

Alternate. (b) :

sin sin cos cos sin

4 5 3 12. .5 13 5 13

5665




IIT MATHS



Alternate : 2 1 cos

sin2 2

1 cos cos sin sin

2

3 5 4 121 . .5 13 5 13

2

63165

2

165

12.cos cos sin sinsin sin cos cos

n nA B A BA B A B

cos cot2 2

n nA B B A

If n even, 2cot2

n A B , if n odd, 0

13. In a triangletan tan tan tan tan tanA B C A B C or 6 = 2 tan C

tan 3C

tan tan 3, tan tan 2A B A B

tan 1 2A or and tan B = 2 or 1.

14. 2cos cos 12 2

A B A B

1cos2 3

A B

2 12cos 12 3

A B

1cos3

A B

cos cos 2sin sin2 2

A B B AA B

1 1 22 12 3 3

15. From the first relation, we have

sin sin

sin sin

a

b

2 sin cos 2 sin cosa b

tan tana b

2 2

2 tan 2 tan2 2

1 tan 1 tan2 2

a b

From the second relation replacing

1tan tan2 2

b ca

we have,

2

2

tan tan / 2211 tan 1 tan / 22

a b b c

a b ca

22tan tan

2 2a b c

2tan 1 tan2 2

b b c

2 2 2 2tan 1 tan2 2

a b c bc

2 2 22

2 tan 221 tan

2

bca b c

2 2 2

2sin bca b c

Now, replacing by , b by a and c by -c,we get

2 2 2 2 2 2

2 2sin ac acb a c a b c

16 From the given equation, we have

tan tan 4A B A B (1)

tan tan 1A B A B (2)From (1) and (2) we get

tan tan / 2A B A B




IIT MATHS


UPTO TRANSFORMATIONS2 / 2 / 4A A

and from (1) we get

2 2

2

1 tan 1 tan 41 tan 1 tan

1 tan 1 tan4

1 tan

B BB B

B BB

2 2

2 2

2 1 tan 1 tan 141 tan 1 tan 2

B BB B

cos2 1/2 2 /3 /6B B B COMPREHENSIONS

Comprehension -I17. sin sin 2 /3 sin 4 /3x y z

sin sin 4 / 3 sin 2 / 3

2sin 2 /3 cos 2 /3 sin 2 / 3

2sin 2 /3 1/ 2 sin 2 /3 0

Similarly 0p q r

18.

sin sin 2 / 3sin 2 / 3 sin 4 / 3sin 4 / 3 sin

yz zx xy

1[cos 2 /3 cos 2 2 /3 cos 2 /3 cos

2cos 4 /3 cos 2 4 /3 ]

1 3 32cos2 cos /3 cos22 2 4

19.x y z x y z y zp q r p q r q rxz zx xy yz zx xy zx xy

1 1 2 3C C C C

00 3/ 4

3 / 4

y zq r yr qzzx xy

3/ 4 [sin 2 / 3 cos 4 / 3cos 2 / 3 sin 4 / 3

3/ 4 sin 2 / 3 3 / 4 3 / 23 3 .

8

Comprehension -IILet tan t

3

2

3 3tan tan1 tan1 tan 1 3tan

34 2

2

1 33 3 6 8 1 01 1 3

t t t t t tt t

20. 1 2 3 4 0t t t t

1 2 6 / 3 2t t 21. 1 2 3 8 / 3t t t 1 2 3 4 1/ 3t t t t

22.1 2 3

1 2 3 4 1 2 3 4

1 1 1 1 8 / 3 81/ 3

t t tt t t t t t t t

Match the following23. sin sin cos cos 0A B B A

2cos sin2 2

2sin sin2 2

A B A B

A B A B

sin cos sin 02 2 2

A B A B A B

sin 0 tan 12 2

A B A BEither or

sin 2 sin 2 2sin cos 2sin cosA B A A B B 2sin cos 2cos sinA B A B

2sin A B

cos cos cos 2sin sinA B A B A B 2 2cos sin cos 2A A A .

24 tan tan tantan tan tan

A B C A B CA B C

and / 2tan tan tan tan

tan tan 1

A B CB C C A

A B

So ( ) ( ), ( ), ( ) ( ), ( )a p s b q r

/ 4 tan tan 1 tan tanA B A B A B

tan tan tan

tan tan tan tan 1A B C

A B C as C




IIT MATHS



Next 0 0 0

tan tan tantan15 tan 45 tan120

A B C

2 3 1 3 3 2 3 Integer Answer Questions

252 1cos20º 2sin 55º 1 cos20º 1 cos110º x

130º 90º2sin sin

2 2

0 12sin 65

2

02 sin 65 165K

26. 23º + 22º = 45º

cot 23º 22º cot 45º

cot 23cot 22 1 1cot 23 cot 22

cot 23cot 22 1 cot 23cot 22

cot 23cot 22 cot 23 cot 22 1

1 cot 23 1 cot 221 cot 23 cot 22 cot 23cot cot 22

1 1 2

27. 0 01 tan1 1 tan44 1 tan2 1 tan43º

1 tan 22 1 tan 23 01 tan 45

2.2...........2.2 = 23228. 20º + 40º = 60º

tan 20º 40º tan60º

tan 20 tan 40 31 tan 20 tan 40

0tan 20 tan 40 3 3 tan 20 tan 40

2

tan 20 tan 40 3 tan 20 tan 40 3

29. 0 01 tan1 1 tan44 1 tan2 1 tan43º

1 tan 22 1 tan 23 01 tan 45

2.2...........2.2 = 232

30. 2

2 2 2

p q r

p q r pq qr rp

p q r p q q p r p

3 p q r

3 15 tan tan 3 15 1 48

48 4 3p q r

WORK SHEET-3SINGLE ANSWER QUESTIONS

1 .3 0

cos 2Sin

fff 2 3, ,

4 3 3 4

2 . sec cot2

co cot

cos cot cot2

ec

= cot32

3 .sin 2sin cos sin 2

cos3 2cos cos3 2cos cos3x x x xx x x x x

= sin 32cos cos3

x xx x

= sin 3 cos3 cos3 sin2 cos cos3

x x x xx x

= 1 tan 3 tan2

x x

SSSSSSSSS, SSSSSS SSSS SS SSSSSSSSSS SS2cos3x , ttttt tttt tt ttttttttt tt 2cos9x

4 . 1 1 2 1 2tan sin2 tan tan cos2s S 3 1 2 3tan tan tan cosS v

4 1 2 3 4tan tan tan tan sinS




IIT MATHS



1 31 2 3 4

2 4

tan1

S SS S

5 . 23sin 3O y 4 4cot tan 2x x 4 4cot tan 1 3x x

2tan 1x 2sin 1y 4

x

2y

ttt tttttt ttt ,4 2

4 2

,4 2

4 2

6 . 2 sin cossin cos

=

2 2 cos4

2 sin4

= tan2 8

7 . , ,a b c aaa aaaaa aaaa aaaaaaa. aaaa aaaa

. cosa b x . cosb c y

. cosa c x y

,a b c o

2 2 2 2 . . .a b c a b b c c a o

3 2 cos cos cosx y x y O

8 . AA GG sin cos sin cos2 2 2 2x x x x

sin cos1sin cos 22 2 2x x

x x

sin cos 2x x

11sin cos 22 2 2x x

G G G G G 11sin cos 22 2 2x x

9. 2 2 2 2 2 2 2 24 sin 2U a b a b a b

II 0 22 2 2

min 2U a b ab a b

2 2 2 2 2 2 2max 2U a b a b a b

1 0

2 32cos cos 2cos 12 2 2 2

24cos 4cos cos 1 02 2 2

222cos cos cos 1

2 2 2

TTT TTTT TTTTTTTTTTT TT 2cos 1 02

1cos2

3

3sin sin

2

MUTIPLE ANSWERS

11.0 0

0 0

3 cot 76 cot16cot 76 cot16

0 0 0 0

0 0

3sin 76 sin16 cos 76 cos16sin 76 16

0 0 0 0

0

2sin 76 sin16 cos 76 16

sin 92

0 0 0

0

cos 60 cos92 cos60sin 92

0

0

1 cos92sin 92

0tan 46[Alternate. (c) ]




IIT MATHS



0 0 0tan 90 44 cot 44

[Alternate. (d) ].

12.2

sin tansin tan

sin sin sin cos. .sin sin sin cos

sin cossin cos

sin 3 0sin

sin 2 sin 2

2 2

2 tan 2 tan1 tan 1 tan

2 2

6 tan 2 tan1 9 tan 1 tan

[ tan 3 tan ]given

2 1tan3

1tan3

and tan 313 The given relation can be written as

2 21 sin costan / 2

sin sinx xx

x x

22 2 22sin / 2 cos / 2 sin / 2x x x

2

22 2

2 tan / 2

1 tan / 2 / 1 tan / 2

x

x x

22 1 1y y y

[ where 2tan / 2y x ]2 4 1 0y y

4 16 4 2 52

y

Since 0y , we get

25 2 2 55 2 .5 2 2 5

y

9 4 5 2 5 .

14 From the given conditions

2 22 1 sin 3 1 sin 3 1 sinA B A

22sin 3sin 1 0A A

2sin 1 sin 1 0A A

sin 1 sin 1/ 2A or A

/ 2 / 6A or But since A is acute, we have / 6A .

2sin sin / 6 1/ 2B

sin 1/ 2 / 4B B

15 2 2 2sin sin 2sin sinl and2 2 2cos cos 2cos cosm

2 22cos 2l m ( by adding)and

2 2cos 2 cos 2 2cos m l

(by subtracting)

2 2

2cos cos 2cos

m l

2 2

2 2cos m lm l

.

Next,

cos / 2 1 cos11 1 coscos / 2

nn

= 2 2

2l m

m

16 sin A = - 7/25, cosA = 24/25336sin 2625

A




IIT MATHS



2 2cos / 2 49 / 50,sin / 2 1/ 50,

sin / 2

A A

A

0 0

2 /10, tan / 2

1/ 7, 135 / 2 180 .

A

as A

COMPREHENSIONSPASSAGE -1

17.2 4 6cos cos cos7 7 7

2 4 8cos cos cos 27 7 7

2 4 8cos cos cos7 7 7

3

3

2 2 2sin 2. sin 2 sin17 7 7

2 2 2 82 sin 8sin 8sin7 7 7

18

6

1cos cos cos2 cos3 cos4 cos5 cos6 ....

rr i

13

13

or 5 8

cos5 cos8

and 12

cos cos12 Then from Eq. (i),

6

1cos cos 2 cos 3 cos 4 cos5 cos 6

rr

cos12 cos2 cos3 cos4 cos8 cos6

cos 2 cos 4 cos8 cos3 cos 6 cos12

3 3 3

3 3

sin 2 .2 sin 2 .32 sin 2 2 .sin 3

sin16 sin 248sin 2 8sin 3

sin 3 sin 2 213

8sin 2 8sin 3

sin 3 sin 2 18sin 2 8sin 3 64

19.5 7sin sin sin

18 18 18

5 7cos cos cos2 18 2 18 2 18

8 4 2cos cos cos18 18 18

3

2 4cos cos cos9 9 9

2 1here

3

1 12 8

PASSAGE-IILet BD, CE and AF be of lengths y-1, y and y+ 1 respectively. Since the lengths of thetangent, from an external point to the cirecleare equal we get BF=BD=y - 1, CD = CE =y and AE = AF = y +1. Let r be the inradiusso that r = 4

2 1, 2 1 2BC y CA y and AB y ,3s y

20. Now tan tan2 2 1C r B rand

y y

so that

2

1tan2 2 1

1

r rB C y y

ry y




IIT MATHS



2 22cot

2A ry r

y y r

2 21 2y ry r

r y y r

3 23 0y r y y 3 48 0y y y

[ 4 ]as r given20 49 7y or y y

( 1)as y Hence, the sides of the triangle ABC are 13, 14and 15 units.

21. From quadrilateral ,IEAF EIF A

42

EDF .

. From triangle , 4EIF IF IE

/ 2 2 4cos / 2IEF IFE A EF A

22. cot / 2 cot / 2 cot / 2A B C

1 1 3 21.4

y y y yr r r r

MATCH THE FOLLOWING23.

0 0 0 00 0 85 55 85 55sin85 sin 55 2sin cos

2 2

0 0 0 02sin 70 cos15 2 cos 20 cos15 0 0 0 0sin 85 sin 55 2cos 70 sin15 0 0 0 0cos85 cos55 2cos 70 cos15

0 02cos 70 sin 75

24 0 0 5 1 5 1 1cos36 cos724 4 2

0 0cos36 cos 72 1/ 4

0 0 5 2 5tan 36 tan18 5 2 55

5 2 55

0 0 10 2 5 10 2 5sin 36 cos184 4

54

Integer answer Questions

25.4 4

2 2 2 2

cos sin24 24

cos sin cos sin24 24 24 24

3 1 6 2cos , 6

12 42 2K

26 . 3 2

22

3 tan tan 3 tan1 3 tan1 3tan tan

A A AKAA A

2 23 tan 3 tanK K A A

23 3 1 tanK K A

2 3tan 03 1KAK

3 3 1 0K K

1 , 33

K K

27. 2 2 04 cos 73 cos 47 cos 47 cos 73

2 0 2 0 014 cos 73 1 sin 47 2cos47cos732

14 1 cos120cos 26 cos120 cos 262

1 1 1 34 1 cos 26 cos16 4 32 2 4 4

.

28. 0 020 , 3 60 tan3 tan60 3,9

3

2

3 tan tan 31 3 tan

6 4 2tan 33tan 27 tan 3

29. 2

3

4 3sin0sin

f

2

3

4 3sin 4cossin

ec

2sin7




IIT MATHS



40cos 2 cos 449

30.

6

1

sin 14 41 4 2

sin sin 1 .sin4 4 4

m

m m

mm

,

6

12 cot 1 cot 4 2

4 4m

mm

cot tan 4 5,

12 12




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