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M.N.RAO

SRICHAITANYA

EDUCATIONAL INSTITUTIONS

EAMCET exam is for entering in to the statewide Engineering colleges.

In this exam questions are given in Objective Type only. In EAMCET objective questions

occurs based on basic concepts of Intermediate books

In EAMCET, both speed and strike-rate matter. You need to be quick as well as accurate

to achieve high scores.

Master the fundamentals, practice a lot, and manage your time to get good rank

PREVIOUS PAPER ANALYSIS (TRIGONOMETRY)

2005 2006 2007 2008 2009 2010

TRIGONOMETRY

Up to transformations ,

periodicity and extreme

5 5 4 4 4 3

Trigonometric equations 1 1 1 1 - 1

Inverse trigonometric functions 1 - 1 1 1 1

Hyperbolic functions 1 1 1 1 1 -

Properties of triangles 3 3 3 3 2 2

Heights and distances 1 1 1 1 1 1

Complex numbers 1 2 2 3 1 2

Demoivres theorem 1 1 1 - 1 1

Trigonometric expansions - - - - - -Important Formulae and shortcut Formulae(TRIGONOMETRY)

01.1

. (60 ). (60 ) 34

Sin Sin Sin Sin + =

02.0 0 1

. (60 ). (60 ) 34

Cos Cos Cos Cos + =

03. 0 0. (60 ). (60 ) 3Tan Tan Tan Tan + =

04.2 2

( ). ( )Sin A B Sin A B Sin A Sin B+ =

05.2 2

( ). ( )Cos A B Cos A B Cos A Sin B+ =

06. If A + B =450

or 2250

then(1 )(1 ) 2

(1 )(1 cot ) 2

TanA TanB

CotA B

+ + =

=

07. If A + B = 1350

or 3150

then(1 )(1 ) 2

(1 )(1 t ) 2

TanA TanB

CotA Co B

=

+ + =

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08. If A + B + C = 1800

then TanA TanA= , 1CotACotB =

09. If A + B + C = 900

then 1TanATanB = , CotA CotA=

10. 2 sec 2CotA TanA Co A+ = 16. 2 2CotA TanA Cot A =

11. If0

60A B = 2 2 3

4

Cos A Cos B CosACosB+ =

12. If0

60A B = 2 2 3

4Sin A Sin B SinASinB+ =

13.3 3 0 3 0 3

(120 ) (120 ) 34

Cos Cos Cos Cos + + =

14.0 0

(60 ) (120 ) 3 3Tan Tan Tan Tan + + + + =

15. ( )Sin A B C SinACosBCosC SinASinBSinC+ + =

16. ( )Cos A B C CosACosBCosC SinASinBCosC+ + =

17.2 2 2

osaC bSin C aSin bCos a b c + = = +

18.4 4 2 2

1 2Sin Cos Sin Cos + =

19.6 6 2 2

1 3Sin Cos Sin Cos + =

20.0 0

(120 ) (120 ) 0Cos Cos Cos + + + =

21.2 2 0 2 0 3

(120 ) (120 )2

Cos Cos Cos + + + =

22. (120 ) (120 ) 0Sin Sin Sin + =

23. Period of Sinx, Cosx, Secx and Cosecx is 2

24. Period of Tanx, Cotx is

25. Period of [ ]x x is 1 26.Period of [ ]ax ax is1

a

27. Range of Sinx are Cosx are [-1,1] 28.Range of aCosx bSinx c+ + is

[2 2

c a b + ,2 2

c a b+ + ]

29. Minimum value of2 2 2 2

seca Sin x b Co x+ is 2ab

30. Minimum value of 2 2 2 2a Tan x b Cot x+ is 2ab

31. Minimum value of2 2 2 2

a Cos x b Sec x+ is 2b

32.Minimum value of2 2 2 2

a Sec x b Cosec x+ is2

( )a b+

33. General solution of Sin is ( 1)nn + if Principal value is

34..General solution of Tan is n + if Principal value is

35.General solution of Cos is 2n if Principal value is

36. General solution of if Sin =0 is n =

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37. General solution of if Cos =0 is = (2 1)2

n

+

38..If2 2

Sin Sin = 2 2

Cos Cos = then general solution is n = 2 2Tan Tan =

39. For aCos +bSin = c then solution exists if 2 2c a b +

40..1 1

sin cos2

x x

+ = ,1 1

tan cot2

x x

+ = ,1 1

sec cos2

x ec x

+ =

41.1 1 1 1

tan1

x y z xyzTan x Tan y Tan z

xy yz zx

+ + + + =

42. If1 1 1

Tan x Tan y Tan z + + = then x y z xyz+ + =

43, If1 1 1

2Tan x Tan y Tan z

+ + = then 1xy yz xz+ + =

43. If1 1 1

sin sin sin2

x y z

+ + = then2 2 2

2 1x y z xyz+ + + =

44.If1 1 1

cos cos cosx y z + + = then 2 2 2 2 1x y z xyz+ + + =

45.

12(1 ) (1 ) 2

4

n

n n ni i Cos

++ + =

46..1 2 1 2z z z z+ +

;1 2 1 2z z z z+

;1 2 1 2z z z z

47. three complex numbers 1, 2 3Z Z Z

are collinear then

1 1

2 2

3 3

0

1

1

1

z z

z z

z z

=

48.Area of triangle formed by Z, IZ,Z+Zi is

1

2 2Z

49.Area of triangle formed by Z, Z,Z+Z is

3.

4 2Z

50.If

1

2

z z

kz z

=

(k 1) represents circle with ends of diameter

2 1

1

kz z

k

If k=1 the locus of z represents a line or perpendicular bisector .

51.. 1 2 1 2,z z z z k k z z + = >

then locus of z represents Ellipse

and if 1 2k z z<

it is less, then it represents hyperbola

52. 1 2 3A(z ),B(z ),C(z ),

and is angle between AB,AC then

1 2

1 3

iz z ABe

z z AC

=

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52.

2, 1, , log2

ii i

e Cos iSin Cis e e i i i

= + = = = =

53.

2 21 11, , , (1 ) 2 , (1 ) 2

1 1

i ii i i i i i i

i i

+ = = = + = =

+

,2 2 2 2

x a x a x a x aa ib i a ib i

+ + + = + =

where2 2

x a b= +

54.

1(1 3 ) (1 3 ) 2

3

n n n ni i Cos+

+ + =

55.

12(1 ) (1 ) 2

4

n

n n ni i Cos+

+ + =

56. 1 2 1 2z z z z+ +

; 1 2 1 2z z z z+

; 1 2 1 2z z z z

PREVIOUS EAMCET QUESTIONS1. If

1 1 1

2Tan x Tan y Tan z

+ + = 1-xy-yz-zx=(EAMCET2010)

1) 1 2) 0 3) -1 4) 3

Hint: put x=y=z=1

3which satisfies given thus

1-1

3.

1

3-

1

3.

1

3-

1

3.

1

3=0

2. 11

tanh log 1

xx a x

+ =

= (EAMCET2010)

1)1

22)

1

43) 2 4) 3

Hint: Forumala =

1 1 1tanh log

2 1

xx

x

+ =

a=

1

2

3.If 1 3z i= + then Arg z Arg z (EAMCET2010)

1) 0 2)

33) 2

34)

2

Hint:1 3

arg tan1

z

= =

3

Arg z1 3

tan1

= -

3

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Thus Arg z Arg z =2

3

4. Period of4 4

Sin x Cos x+ is ? (EAMCET 2009)

1)

4

2 2)

2

2 3)

4 4)

2

Hint:4 4 4 4

2 2Sin x Cos x Cos x Sin x

+ + + = +

as it remains unaltered Thus

2

is period

5.cos( 2 )

cosx

x y=

tan(x-y)tany= (EAMCET 2009)

1)1

1

+

2)

1

1

+3)

1+

4)

1

Hint:Apply componendo& divedendocos cos( 2 ) 1

cos cos( 2 ) 1

x x y

x x y

+ + =

2cos( ) cos 1

2sin( )sin 1

x y y

x y y

+=

tan(x-y)tany=

1

1

+

6. If0 0 0

35 , 15 40A B andC= = = then tanAtanB+tanBtanC+tanCtanA= (EAMCET2008)

1)1 2)2 3)3 4)4

Hint; as0 0 0 0

35 15 40 90+ + = tanAtanB+tanBtanC+tanCtanA=1

7.In any ABC a(bcosC-c cosB)= (EAMCET2009)

1)2 2b c+ 2) 2 2b c 3)

1 1

b c+ 4)

2 2

1 1

b c

Hint : ab Cosc-c CosB

=

ab2 2 2

2

a b c

ab

+ a

2 2 2

2

a c bc

ac

+

=

2a

2 2 2b c a+

2 22 2

2

c bb c

+=

8. In ABC

( )( )( )( )2 2

4

a b c b c a c a b a b c

b c

+ + + + +

= (EAMCET2009)

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1)2

cos A 2)2

cos B 3)2

sin A 4)2

sin B

Hint: as a + b + c=2s

( ) ( ) ( ) ( ) ( )2 2 2 2

2 .2 2. 2. ( )

4

s s a s b s c as s a s b s c

b c b c

=

=

2

2 24 4b c

=

2

2

2 2

1sin

2sin

bc A

Ab c

=

9..In ABC 2 3

2 2

B C A A BCos Cos

+ + + = (EAMCET 2004)

1)1 2)2 3)3 4)0

Hint: Put A = B = C = 600

00 0 0 060 120 180 60 60

1 1 02 2

Cos Cos+ +

+ = + =

10.0 0

0

80 10

70

Tan Tan

Tan

= (EAMCET 2007)

1)1 2)2 3)3 4)4

Hint:0 0 0

0 0

10 10 202 2

20 20

Cot Tan Cot

Cot Cot

= = (as 2 2CotA TanA Cot A = )

11. 4 1Cos Sin = then 4Sin Cos + (EAMCET 2005)

1)1 2) 2 3)3 4) 4

Hint: 2 2 2 1 16 1 4a b c + = + =

12. 2 2 2 2( sec ) ( )Sin Co Cos Sec k Tan Cot + + + = + + then k = (EAMCET 2000)

1)7 2) 2 3)3 4) 4

Put 45 =

2 21 1

2 2 1 12 2

k

+ + + = + +

=9 9

2 72 2

k k+ = + =

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13.2 0 2 0 0 076 16 76 16Cos Cos Cos Cos+ =?(EAMCET 2002)

1)3

42)

1

23)3 4) 4

as 760-16

0= 60

0

0( 60 )A B = then

2 2 3

4Cos A Cos B CosACosB + =

14. 3( os ) 3 3SinA SinB CosB C A Sin A Sin B+ = + =?(EAMCET 2007)

1)3

42)

1

23)0 4) 4

Hint: 2 2 3

2 2 2 2

A B A B B A B ASin Cos Sin Sin

+ + =

00 0 3 3 3( ) 3 0

2

A BSin Sin A B Sin A Sin B Sin B Sin B

+= = = + + =

15. Extreme value of2 2 2

43 3

Cosx Cos x Cos x

+

is? (EAMCET 2005)

1)7 2) 2 3) 1 4) 4

Hint:2 21

4. 3( ) 34

Cos x Cos x= Range 1

16.2

cos2 2cos 2x x+ = then x = (EAMCET 2008)

1)6

n

2)6

n

+2

23) 2

6n

4)

2

Hint :2 2 2 32cos 1 2cos 2 cos

4x x x + = =

6 6x n

=

17. If a, b, c are + ve then

1 1 1( ) ( ) ( )a a b c b a b c c a b cTan Tan Tan

bc ca ab

+ + + + + ++ + = (EAMCET 2002)

1) 2)

2

23)

44)

2

Hint; put a=b=c=1 then

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1 1 13 3 33 3 3

Tan Tan Tan

+ + = + + =

18.. cos 2 sin 2 7x a x a+ = has a solution if a belongs to (EAMCET 2002)

Hint;2

1 2 sin sin 2 7x a x a + = 2

2sin sin 2 7x a x a+ =

( )2 8 2 8sin

22

a a ax

= But 1 sin 1x 2 6a

19. The set of values of x, for whichtan 3 tan 2

11 tan 3 tan 2

x x

x x

=

+is (EAMCET2005)

1) 2) 3)

44)

2

Hint ; Because

0

(3 2 ) an 1 45Tan x x T x x = = =

x =0

45 does not satisfies given , thus x=

20. If ( ),x such that 21 cos cos ....y x x= + + + and 8 64y = then y = __

(EAMCET 2007)

1)1 2)2 3)3 4)4

Hint :2 1

1 cos cos ...... 2

1 cos

y x x

x

= + + + = =

(as8y=64 gives y=2)

1cos 60 or 120

2x x = =

Thus1 1

1 .... 22 4 1

ay

r= + + = = =

21. In if5 2

tan , tan2 6 2 5

A C= = then a,b,c are in (EAMCET2006)

1)AP 2)GP 3)HP 4)AGP

Hint :5 2 1

tan . tan .2 2 6 5 3

A C= =

of1

tan . tan , ,2 2 3

A Ca b c= are in A.P.

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22. In ABC is b = 20, c = 21 and3

sin5

A = then a = (EAMCET2003)

1)7 2)12 3)13 4) 4

Hint :2 2 2

3 4sin cos

5 5 2

b c aA A

bc

+ = = =

24 400 441

5 2.20.21

a+ = 13a =

23 In ABC ( ) tan tan2 2

A B Cb c + (EAMCET2005)

1)0 2)c+b 3)bc 4) 4

Hint : ( ) . ot2 2

A b c Ab c Tan C

b c

+

+ = ( ) 0b c+ =

24 In a ABC of

1 1 3

b c c a a b c+ =

+ + + + then C= (EAMCET2008)

1).600

2).300

3).900

4).100

Hint:

By cyclic symmetry C=600

25. If 1 2 32 3r r r= =

then

a b c

b c a+ +

= (EAMCET2008)

Hint:

1 2 3

1 1/ 2 1/3

r r r= =

a:b:c=2+3 : 1+3 : 1+2

= 5 : 4 : 3

a b c

b c a+ +

=

5 4 3 191

4 3 5 60+ + =

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Other Examples with hints andshort cuts

1.1 11 1

tan tan4 2 4 2

a aCos Cos

b b

+ + =

Hint: Put 11

2

aCos

b

=x cos2a

xb

=

2 2tan( ) tan( ) 2sec 2

4 4 cos 2

bx x x

x a

+ + = = =

2. Range of1 1 1

Sin x Cos x Tan x

+ + is

Hint: as

1 1

2

1 1

Sin x Cos x

x

+ =

is valid when

Thus Range of1 1 1

Sin x Cos x Tan x

+ + is =

,2 4 4

+

=[ ]/ 4,3 / 4

3. If2 2 2 2

x y z r+ + = , then1 1 1xy yz xz

Tan Tan Tanzr xr yr

+ + =

Hint:

Put x=y=z=12

3 3r r = =

1 13tan

23

=

4. .If 1 1 2tan 2 cot3

x x

+ = find x

Hint:1 1

cot 2cot 1202

x x

+ = 1

cot 30x = 3x =

5. If1 2 1 2 2

( ) ( ) 5 / 8Tan x Cot x + = , then x =

Hint: Put x=-12

1 2 1 2 2 2 5

( 1) ( 1) ( ) (3 )4 4 8Tan Cot

+ = + =

6. If1 1

sin cos6

x x

= find x = ?

Hint :1 1

sin sin2 6

x x

=

1 1sin sin

2 6x x

+ =

12sin 120x

=

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1

sin 60x = 3

2x =

7.1

1

cos 0n

i

i

=

= then1

n

i

i

=

=

Hint :1 1

1 2cos cos ......... 0

+ =

1 1

1 2cos cos ........ 0

= = 1 2 3....... 1 = = =

i i n = =

8. 1sin (sin10) ..........

=

1 1 0 1 0sin (sin10) sin sin(10 57 ) sin sin(570 )x

= =

1 1 0 1sin sin(540 30) sin ( 30 ) sin sin(3 10 57)x

= + = = (3 10)=

9 If

2

1

5

7

z

z is purely imaginary then

1 2

1 2

2 3

2 3

z z

z z

+

Hint : Let

2 12

1

5 7

7 5

z z ii z

z= =

121

21

1

72 3.

10 21 10 2157 10 21 10 212 35

z iz

i

z i iz

2

2

++ +

= = +

= 1

10. If, 2

cos cos cos

a b cb

A B c= = =

then area of triangle

Hint : As it is equilateral triangle

Area( )

223 32 3

4 4a = =

11. Sides of triangle are 6,8,10 then circum radius is---

Hints : As it is right angled triangle

Circum Radius =( )

1 110 5

2 2hyp = =

12. If2 2

1,2 1, 1x x x x+ + + are sides of triangles then its largest angle is ______

Hints : put x = 22 2

2 2 1,2.2 1,2 1+ + + gives 7,5,3 as sides2 2 2 2 2 2

5 3 7 1cos2 2.5.3 2

b c aAbc

+ + = = =

120A =

13.

334 365

1 3 1 34 5 3

2 2 2 2

i i + + + + =

__________

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334 3654 5 5 + +

( ) ( )111 3 1213 2

4 5 3 + +

21 4 5 3 3 = + + 23 3 3 1 2 = + + + +

= 3(0) + 1 + 2 = 1 + 2

14.( )1tan log a ib

a ib

+

Use polar form ifi

a ib re

+ = where2 2

r a b= + ,

1tan

b

a =

i

a ib re

=

2tan log tan log

ii

i

rei i e

re

=

2

tan( 2 ) tan 2i = =

22 2 2

2

22 tan 2

1 tan1

b

aba

b a b

a

= =

15.1

2 2sin cos

7 7k

k ki

=

Hint

1 2

7

kcis

i

=

22 2 .2 2 .677 7 7

6

1. .....

k

ee e e

i i

=

2 4 6 8 10 12

76

1e

i

+ + + + +

=

( )62 31 1

cos6 sin 6( ) 1

e ii

= = +

1=

16..3

1 2sinh 2

= ____________-

Hint : As

1 2sinh log 1x x x

= + +

( ) ( )3

1 2sinh 2 log 2 2 8 1 log 3 8 = + + = +

17..

log cot4

x

= + then sinh x = ___________

Hint :

log cot4

ex

= +

( )0cot tan 45 04

x xe e

= + + =

( )0tan 45 0 xe =sub

1 tan 1 tan

1 tan 1 tan

x xe e

+

= +

sinh tan 2x = cosh sec 2x =

18.. If 1 2 1 2z z z z+ =

then 1 2Arg z Arg z

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Hint : given implies

1 1

1 2

x y

y x

=

Then

1 11 21 2

1 2

tan tany y

Arg z Arg zx x

=

1 11 1

1 1

tan tany x

x y

=

1 11 1

1 1

tan cot

2

y y

x x

= + =

19.. 2i +-2i =

Hint :(1+i)2+(1-i)2 =( ) ( )1 1 2i i i+ =

13. 0 0(1 an1 )(1 44 ) 2T Tan+ + = because 10+44

0=45

0

20.0 0 0 0 0 0

40 60 80 40 60 80Tan Tan Tan Tan Tan Tan+ + = because400+60

0+80

0=180

0

( 2 ) (2 ) 120 150Tan Tan Tan Tan + + = 1

( 3) 13

= =

21.2 0 2 0 2 0 2 0 180

5

15 10 15 ...... 180 ( ) 18

2Sin Sin Sin Sin+ + + + = =

22.2 0 2 0 2 0 81

9

1 9cos 9 cos 18 ...... cos 81 ( )

2 2+ + + = =

23. If cos tan ,cos tan ,cos tanx y y z z x= = = then0

sin sin sin 2sin18x y z= = =

24.2 2 2

sin cos 3 sin cos 3 2 7ec ec + = + = =

25.2 2 2

tan cot 8 tan cot 8 2 62 + = + = =

26. If3 2

3

sin costan tan tan

cos

x xa x b x c x d

x

+

= + + + then a + b + c + d =

Hint: Put x=0

45

1 1

2 2.1 .1 .1 .1 4

1

2 2

a b c d a b c d

+

= + + + + + + =

27. Minimum value ofcos2 sin 2

27 .81x x

Hint:3cos2 4sin 2 25 5

3 3 3x x+

= = 1

243=

28. 9 27 63 81Tan Tan Tan Tan +

Hint: ( 9 9) ( 27 27)Tan Cot Tan Cot + +

4 42 sec18 2 sec54 2 4

5 1 5 1Co Co

= = =

+

29. If2

1, , are cube roots of unity then roots of

( )3

1 8 0x + =are

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Hint :( )

31 8x =

2

1 2, 2 , 2x =

21,1 2 ,1 2x =

30. Value of

2008 2008

1 3 1 3

2 2

i i + +

is _____________

( ) ( )20082008 2

1 + =

31. If

11

1

z

z

=

+ then locus of z is

Hint :

11

1

z

z

=

+

1 1z z + = (Real part differ)

0x = locus is y - axis

32. Locus of z of

1z zi

z zi

=

+

Hint :z zi z zi = +

imaginary part differ

y = 0 Locus is x - axis

33. If 1 2,z z be two roots of

20z az b+ + = ,origin and

1 2,z z from equilateral triangle then

Hint : 1 2z z a+ =

1 2z z b= and 30z =

condition ofequilateral triangle

2 2 21 2 3 1 2 2 3 3 1z z z z z z z z z+ + = + + 2 2

1 2 1 20 0 0z z z z+ + = + +

( )2

1 2 1 2 1 22z z z z z z+ = 2

2a b b = ,2

3a b=

34. 0

2

3

n

n

i

=

=

---------------value

Hint :

0 1 22 2 2

.............3 3 3

i i i

+ + +

GP

( )3 3 21 321 3 2 131

3

iair i

+= =

35 A value of n such that

31

2 2

n

i + =

is

Hint :

cos sin 16 6 6

nn

i cis

+ = = 12n =

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36. is a non real root6

1x = of then

5 3

2

1

1

+ + +

+

Hint : is root6

1 = 6

1 0 =

( )( )6 2 4 21 1 1 0 = + + =

Thus

( )4 25 32 2

11

1 1

+ ++ + +=

+ +

=

2

2 2 4 2

.0 1 1

1 1

+= =

+ + +

22

1

= =