1. (n – 2)2 = n (n – 1) – 4n + 10
n2 – 4n + 4 = n2 – 5n + 10
n = 6 Ans ]
Q.138107/bin The sum of the series aC0 + (a + b)C1 + (a + 2b)C2 + ..... + (a + nb)Cn is
where Cr's denotes combinatorialcoefficient in the expansion of (1 + x)n, n N
(A) (a + 2nb)2n (B) (2a + nb)2n (C) (a +nb)2n – 1 (D*) (2a + nb)2n – 1
Q.139109/bin The coefficient ofthe middle term in the binomialexpansion in powers of x of(1 + x)4 and of
(1 – x)6 is the same if  equals
(A) –
3
5
(B)
3
10
(C*) –
10
3
(D)
5
3
Q.14029/bin (2n+1) (2n+3) (2n+5) ....... (4n 1) is equal to :
(A)
( ) !
. ( ) ! ( )!
4
2 2 2
n
n n
n (B*)
( ) ! !
. ( ) ! ( )!
4
2 2 2
n n
n n
n (C)
( ) ! !
( ) ! ( ) !
4
2 2
n n
n n
(D)
( ) ! !
! ( ) !
4
2 2
n n
n
n
[Hint: E = (2 n + 1) (2 n + 3) (2 n + 5) ......(4 n  1)
Multiply numerator and denominator by (2n + 2) (2n + 4) ...... (4n) & also by (2n )!.
E =
( ) ! ( ) ( ) ( ) ........ ( ) .
( ) ! ( ) ( ) ........ ( )
2 2 1 2 2 2 3 4 1 4
2 2 2 2 4 2 2
n n n n n n
n n n n n
   
  
=
 
( ) ! ( ) !
( ) ! ( ) ( ) ...... ( ) !
4
2 2 1 2 2
n n
n n n n n
n

 
=
 
( !) . ( )!
. ( ) !
n n
n
n
4
2 2
2
 B ]
Q.141110/bin If Sn = 

n
0
r r
n
C
1
and Tn = 

n
0
r r
n
C
r
then
n
n
S
T
is equal to
(A*)
2
n
(B) 1
2
n
 (C) n – 1 (D)
2
1
n
2 
Q.14211/bin The coefficient of xr (0  r  n1) in the expression :
(x+ 2)n1 + (x+ 2)n2. (x+ 1) + (x+ 2)n3 . (x+ 1)² + ...... + (x+ 1)n1 is :
(A) nCr (2r  1) (B*) nCr (2nr  1) (C) nCr (2r + 1) (D) nCr (2nr + 1)
[Hint: E = (x + 2)n  1


































1
n
2
2
x
1
x
......
2
x
1
x
2
x
1
x
1
= (x + 2)n  1
1
1
2
1
1
2





 

















x
x
x
x
n
= (x + 2)n ( ) ( )
( )
x x
x
n n
n
  







2 1
2
= (x + 2)n  (x + 1)n
Now co-efficient of xr in  
( ) ( )
2 1
  
x x
n n
= nCr 2n  r  nCr = nCr (2n  r
 1) ]

2. Q.12663/bin In the expansionof
x
x x
x
x x

 









1
1
1
2 3 1 3 1 2
10
/ / / , the term which does not contain x is :
(A) 10C0 (B) 10C7 (C*) 10C4 (D) none
[Hint:
 
x
x x
1 3
3
2 3 1 3
1
1
/
/ /

 
= x1/3 + 1  (x1/3  x1/2)10 ]
Q.12765/bin Ifthe 6th termin the expansion of the binomial
1
8 3
2
10
8
x
x x
/
log






 is 5600, then xequals to
(A) 5 (B) 8 (C*) 10 (D) 100
[Hint: T6 = 8C5
1
8 3
8 5
x /







. (x2 log10 x)5 = 100  x = 10 ]
Q.12868/bin Co-efficient of t in the expansion of,
( + p)m  1 + ( + p)m  2 ( + q) + ( + p)m  3 ( + q)2 + ...... ( + q)m  1
where  q and p  q is :
(A)
 
m
t
t t
C p q
p q


(B*)
 
m
t
m t m t
C p q
p q
 


(C)
 
m
t
t t
C p q
p q


(D)
 
m
t
m t m t
C p q
p q
 


[Hint: E = ( + p)m  1 1
2 1











  























q
p
q
p
q
p
m
......
 co-efficient of t =
   
 
  

p q
p q
m m
or
   
p q
p q
m m
  

 
=
 
m
t
m t m t
C p q
p q
 


]
Q.12970/bin (1 + x) (1 + x + x2) (1 + x + x2 + x3) ...... (1 + x + x2 + ...... + x100) when written in the ascending
power of x then the highest exponent of x is ______ .
(A) 4950 (B*) 5050 (C) 5150 (D) none
[Hint: Highest exponent inthe product offirst two is 3 = 1 + 2
Highest exponent inthe product offirst three is 6 = 1 + 2 + 3
Similarly, Highest exponent inthe product offirst hundred
= 1 + 2 + ...... + 100 = 5050 ]
Q.13073/bin Let  
5 2 6

n
= p + f where n  N and p  N and 0 < f < 1 then the value of, f2  f + pf 
p is
(A) a naturalnumber (B*) a negative integer
(C) a prime number (D) are irrationalnumber
[Hint: value =  1 ]
Q.13175/bin Number ofrational terms inthe expansion of  
2 3
4
100
 is :
(A) 25 (B*) 26 (C) 27 (D) 28

3. Q.11219/bin Let n
)
3
4
7
(  = p +  when n and p are positive integers and   (0, 1) then (1 – ) (p + ) is
(A) rationalwhichis not aninteger (B) a prime
(C) a composite (D*) none ofthese
[Hint: (1 – ) (p + ) = 1  (D) ]
Q.11324/bin If(11)27 + (21)27 when divided by 16 leaves the remainder
(A*) 0 (B) 1 (C) 2 (D) 14
[Hint: an + bn = (a + b) (Q (a, b) ) if n is odd
alternatively : interpret from (16 – 5)27 + (16 + 5)27 ]
Q.11425/bin Last three digits ofthe number N = 7100 – 3100 are
(A) 100 (B) 300 (C) 500 (D*) 000
[Hint: consider (5 + 2)100 – (5 – 2)100
= 2 [100C1 599 · 2 + 100C3 597 · 23 + ....... + 100C995 · 299]
= 2 [1000 · 598 + 100C3 594 + ....... + 1000 · 298]
 minimum 000 as last three digits.  (D) ]
Q.11527/bin The last two digits of the number 3400 are :
(A) 81 (B) 43 (C) 29 (D*) 01
[Hint : 3400 = 81100 = (1 + 80)100 = 100C0 + 100C1 80 + ....... + 100C100 80100
 Last two digits are 01 ]
Q.11628/bin If (1 + x + x²)25 = a0 + a1x + a2x² + ..... + a50 . x50 then a0 + a2 + a4 + ..... + a50 is :
(A*) even (B) odd & ofthe form 3n
(C) odd & of the form (3n1) (D) odd & ofthe form (3n+1)
[Hint: putting x = 1 and 1 and adding
a0 + a2 + ...... + a50 =
3 1
2
25

=
 
1 2 1
2
25
 
(
2
even
2
1
odd


)
=
25 2 2 2 1
2
0
25
1
25
2
2 25
25
25
C C C C
   
. . .
=
 
2 1 2 2
2
25
1
25
2
25
25
24
   
C C C
. ...... .
= 2 [13 + 25C2 + ...... + 25C25 . 223]
 even ]
Q.11731/bin The sum of the series (1² + 1).1! + (2² +1).2! + (3² +1). 3! + ..... + (n²+ 1). n! is :
(A) (n+1). (n+2)! (B*) n.(n+1)! (C) (n+1). (n+1)! (D) none of these
[Hint: Tn = [n (n + 1)  (n  1) ] n! = n. (n+1)!  (n  1). n!
Now put n = 1, 2, 3, ...... , n and add ]
Q.11835/bin Let Pm stand for nPm . Then the expression 1 . P1 + 2 . P2 + 3 . P3 + ..... + n. Pn =
(A*) (n+ 1) !  1 (B) (n+ 1) ! + 1 (C) (n+ 1)! (D) none of these
[Hint: Tn = n . n ! = n! [(n + 1)  1 ] = (n + 1) !  n!
Now put n = 1, 2, 3 , ....... and add ]

4. y =
)
2
/
cos(
)
2
/
sin(
2
2
/
cos
2
sin
cos
1 2







= cot /2
x =
2
/
tan
1
2
/
tan
1




= 1
y
1
y
1
2
/
cot
1
2
/
cot







 (C)
Applying C/D y
1
y
1
y
1
y
1
y
1
x
1
x










 (B)
Also , y = 


x
1
1
x
y – xy – x–1 = 0  (D) ]
Q.102 If2 cos + sin = 1, then the value of 4 cos + 3sin is equalto
(A*) 3 (B) –5 (C*)
7
5
(D) –4
[Hint: (2cos)2 = (1 – sin)2  5sin2 – 2sin– 3 = 0  sin = 1 or 
3
5
, now proceed
If sin = 1  cos = 0  E = 3; if sin = –
3
5
 cos =
4
5
or –
4
5
but cos –
4
5
(think!)
hence E =
7
5
]
Q.103 Ifsint + cost =
1
5
then tan
t
2
is equal to :
(A) 1 (*B) –
1
3
(C*) 2 (D) 
1
6
[Hint: sint & cost in tangent of half the angle  (3y+ 1)(y  2) = 0  y = 2 or 1/3 where y= tan(t/2) ]
BINOMIAL
There are 39 questions in this question bank.
Q.1042/bin Given that the termofthe expansion(x1/3  x1/2)15 whichdoes not contain xis 5mwhere m
 N , then m =
(A) 1100 (B) 1010 (C*) 1001 (D) none
[Hint: Tr + 1 = 15Cr (x1/3)15  r (–x 1/2) r 
15
3
 r

r
2
= 0  r = 6
Hence T7 is independent of x and T7 = 15C6 = 5005 = 5m  m = 1001 ]
Q.1053/bin In the binomial (21/3 + 31/3)n, if the ratio of the seventh term from the beginning of the
expansion to the seventh term fromits end is 1/6, then n =
(A) 6 (B*) 9 (C) 12 (D) 15
[Hint: Tr + 1 = nCr an  r . br where a = 21/3 and b = 3 1/3
T7 from beginning = nC6 an  6 b6 and T7 from end = nC6 bn  6 a6

a
b
n
n


12
12 =
1
6
 3
12
n
3
12
n
3
.
2


= 6 1  n – 12 =  3  n = 9 ]

5. [Sol. sin = sin
 = n +(–1)n
n = 0  =  sin/3 = sin/3  (A)
n = 1  = – sin/3 = sin(/3–/3)  (B)
n = –1  = –– sin/3 = sin(–/3–/3) = –sin(/3+/3)  (D) ]
Q.95 Choose the INCORRECT statement(s).
(A)iii sin 82
1
2

. cos 37
1
2

and sin 127
1
2

. sin 97
1
2

have the same value.
(B*)v If tanA=
3
4 3

& tanB =
3
4 3

then tan(A B) must be irrational.
(C*)viii The sign ofthe product sin2 . sin3 . sin5 is positive.
(D*)xiv There exists a value of  between 0 & 2 which satisfies the equation ;
sin4  – sin2  – 1 = 0.
[Sol. (A) A=
0
0
2
1
37
cos
.
2
1
82
sin =
0
0
2
75
cos
.
2
165
sin =  
0
0
45
sin
120
sin
2
1
 =
2
4
1
6 
B=
0
0
2
1
97
sin
.
2
1
127
sin =  
0
0
225
cos
30
cos
2
1
 = 






2
1
2
3
2
1
=
2
4
2
6 
=
4
2
3 
 A= B  True ]
(B) tan(A–B) = B
tan
A
tan
1
B
tan
A
tan


=
  
3
4
3
4
3
.
3
1
3
4
3
3
4
3






=
 
3
3
16
3
4
3
4
3





= 3/8  rational ]
(C) [Sol. sin2 = + ; sin3 = + ; sin5 = – ]
(D) [Sol. sin2 =
2
5
1
 sin2 =
2
5
1
(not possible)
sin2 = 1
2
5
1


 not possible ]
Q.96 Whichofthefollowing functions have the maximumvalueunity?
(A*) sin2 x  cos2 x (B*)
sin cos
2 2
2
x x

(C*) 
sin cos
2 2
2
x x

(D*)
6
5
1
2
1
3
sin cos
x x








6. Q.85 Minimum value of 8cos2x + 18sec2x  x  R wherever it is defined, is :
(A) 24 (B) 25 (C*) 26 (D) 18
[Sol. y = 8 cos2x + 18 sec2x
= 8 (cos2x + sec2x) + 10 sec2x
= 8 [ (cos x – sec x )2 + 2 ] + 10 sec2x
where cosx = secx  x = 0
ymin = 16 + 10 = 26 Ans ]
Q.86 In a ABC 









C
sin
c
B
sin
b
A
sin
a 2
2
2
. sin
2
A
sin
2
B
sin
2
C
simplifies to
(A) 2 (B*)  (C)
2

(D)
4

where  is the area ofthe triangle
[Hint: 2R (a + b + c)  (sin
2
A
)
4 R s ·
R
4
r
= r s = ]
Q.87 If  is eliminated fromthe equations x = a cos( – ) and y = b cos ( – ) then
)
cos(
ab
xy
2
b
y
a
x
2
2
2
2




 is equalto
(A) cos2 (  – ) (B*) sin2 ( – ) (C) sec2 (  – ) (D) cosec2 ( – )
[Sol. ( – ) = ( – ) – ( – )
cos( – ) = cos ( – ) cos ( – ) + sin ( – ) sin( – )
cos( – ) = 2
2
2
2
b
y
1
.
a
x
1
a
x
.
b
y



 



























 2
2
2
2
2
b
y
1
a
x
1
)
cos(
ab
xy
 )
cos(
ab
xy
2
)
(
cos
b
a
y
x 2
2
2
2
2







 = 2
2
2
2
2
2
2
2
b
a
y
x
a
x
b
y
1 


 )
cos(
ab
xy
2
b
y
a
x
2
2
2
2




 = sin2 ( – ) ]
Q.88 Thegeneralsolutionofthetrigonometric equation
tan x + tan 2x + tan 3x = tan x · tan 2x · tan 3x is
(A) x = n (B) n ±
3

(C) x = 2n (D*) x =
3
n
where n  I
[Hint: tan x + tan 2x + tan 3x = tan 3x – tan 2x – tan x
 tan x + tan 2x = 0
 tan 2x = tan (– x)

7.  2b2 = a2 + c2 Þ a2, b2, c2 are in A.P. ]
Q.75 The number ofsolution ofthe equation, 

5
1
r
)
x
r
cos( = 0 lying in (0, p) is :
(A) 2 (B) 3 (C*) 5 (D) more than 5
[Hint: cos x + cos 2x + cos 3x + cos 4x + cos 5 x = 0
2 cos 3x cos 2x + 2 cos 3x cos x + cos 3x = 0
cos 3x [2 cos 2x + 2 cos x – 1] = 0
x = (2n – 1)
6


6

,
6
3
,
6
5
= 3





= 5
2nd equation gives cos x =
4
2
1
= 2
Q.76 If  = 3  and sin  =
a
a b
2 2

. The value of the expression, a cosec  b sec  is
(A)
1
2 2
a b

(B*) 2 a b
2 2
 (C) a + b (D) none
[Sol. a cosec – bsec =


 cos
b
sin
a
















sin
b
a
b
cos
b
a
a
cos
sin
b
a
2
2
2
2
2
2
Now sin3 = 2
2
b
a
a

gives















cos
sin
sin
3
cos
cos
3
sin
b
a 2
2
= 2
2
b
a
2  Ans ]
Q.78 The value ofcot 7
1
2
0
+ tan 67
1
2
0
– cot 67
1
2
0
– tan7
1
2
0
is :
(A) arationalnumber (B*)irrationalnumber (C) 2(3 + 2 3 ) (D) 2 (3 – 3 )
Q.79 If in a triangle ABC
2 2
cos cos cos
A
a
B
b
C
c
a
bc
b
ca
    then the value of the angle Ais :
(A)
8

(B)
4

(C)
3

(D*)
2

Q.80 The value ofthe expression (sinx + cosecx)2 + (cosx + secx)2 – ( tanx+ cotx)2 wherever defined is
equalto
(A) 0 (B*) 5 (C) 7 (D) 9

8. [Hint: C
sin
b
a
2
1
b
r
2
1
a
r
2
1





r (a + b) = 2 
r =
b
a
2


]
Q.66 For eachnatural number k , let Ck denotes the circle withradius k centimeters and centre at the origin.
Onthe circle Ck , a particle moves kcentimeters inthecounter- clockwise direction.Aftercompleting its
motion on Ck , the particle moves to Ck+1 inthe radialdirection. Themotion ofthe particle continues in
this manner .The particle starts at (1, 0).Ifthe particlecrosses the positive directionofthe x- axisfor the
first time onthe circle Cn thenn equalto
(A) 6 (B*) 7 (C) 8 (D) 9
[Hint: Totaldistance travelled = 35 cm;displacement at the instant it crosses the +ve x-axis first timeis 6cm
Angular displacement oneachcircle is 1 radian.]
Q.67 If in a ABC,
cos cos cos
A
a
B
b
C
c
  then the triangle is
(A)right angled (B) isosceles (C*) equilateral (D) obtuse
[Hint:
cos
sin
cos
sin
cos
sin
A
R A
B
R B
C
R C
2 2 2
   tanA = tanB = tanC  A = B = C  (C) ]
Q.68 If cos A+ cosB + 2cosC = 2 then the sides of the ABC are in
(A*)A.P. (B) G.P (C) H.P. (D) none
[Hint: cosA + cosB = 2(1-cosC) = 4 sin2
C
2
or 2cos
A B

2
cos
A B

2
= 4sin2
C
2
or cos
A B

2
= 2sin
C
2
or 2cos
C
2
cos
A -B
2
= 4sin
C
2
cos
C
2
= 2sinC
2sin
A + B
2
cos
A -B
2
= 2sinC or sinA+ sinB = 2sinC  a, c, b are inA.P.
. ]
Q.69 IfAand B are complimentaryangles, then :
(A*) 1
2
1
2






 






tan tan
A B
= 2 (B) 1
2
1
2






 






cot cot
A B
= 2
(C) 1
2
1
2






 






sec cos
A
ec
B
= 2 (D) 1
2
1
2






 






tan tan
A B
= 2
[Sol. A = /2 – B   
A B
2 4 2

Hence 1 + tanA/2 = 1 +
1 2
1 2


tan /
tan /
B
B
=
2
1
2
 tan
B HenceAis correct ]

9. ymax =
2
625
2
252
 ]
Q.59 4 sin50 sin550 sin650 has the values equalto
(A)
3 1
2 2

(B*)
3 1
2 2

(C)
3 1
2

(D)
3 3 1
2 2

d i
[Sol. 2[2sin50 sin550] sin650
 2[cos500 – cos600]sin650
 2cos500 sin650 – sin650
 sin(1150) + sin150 – sin650 =
2
2
1
3 
Ans]
Q.60 If x, y and z are the distances of incentre from the vertices of the triangle ABC respectively then
z
y
x
c
b
a
is equalto
(A)  2
A
tan (B*)  2
A
cot (C)  2
A
tan (D)  2
A
sin
[Similar to 48]
[Sol. x = r cosec
2
A
a = r 






2
C
cot
2
B
cot
2
A
sin
.
2
C
cot
2
B
cot
x
a







 =
2
C
sin
.
2
B
sin
2
A
cos
.
2
A
sin

2
C
sin
.
2
B
sin
2
A
sin
2
C
cos
.
2
B
cos
.
2
A
cos
xyz
abc

=
2
C
cot
.
2
B
cot
.
2
A
cot
In a triangle  2
A
cot =  2
A
cot ]
Q.61 The medians ofa ABC are 9 cm, 12 cm and 15 cm respectively . Thenthe area of the triangle is
(A) 96 sq cm (B) 84 sq cm (C*) 72 sq cm (D) 60 sq cm
[Hint: Produce the medianAM to D such that
GM = MD . Join D to B and C . Now GBDC
is a parallelogram. Note that the sides ofthe
 GDC are 6, 8, 10 GDC = 90º

10. Q.49 Number of roots of the equation cos sin
2 3 1
2
3
4
1 0
x x


   which lie in the interval
[] is
(A) 2 (B*) 4 (C) 6 (D) 8
[Sol. 1 – sin2x +
2
1
3 
sinx –
4
3
– 1 = 0
sin2x –
2
1
3 
sinx +
4
3
= 0
4sin2x – 2 3 sinx – 2sinx + 3 = 0
On solvingwe get
sinx = 1/2 ;
2
3
= (/6 , 5/6 ; /3 , 2/3 ]
Q.50
sec
sec
8 1
4 1



 is equal to
(A) tan 2 cot 8 (B) tan 8 tan 2 (C) cot 8 cot 2 (D*) tan 8 cot 2
Q.51 In a ABC if b = a  
1
3  and C = 300 then the measure of the angle A is
(A) 150 (B) 450 (C) 750 (D*) 1050
[Hint: use tan
2
B
A 
=
b
a
b
a


cot
2
C
to get A– B and A+ B = 1500 (given) ]
Q.52 Number ofvalues of  
[ , ]
0 2 satisfying the equation cotx – cosx = 1 – cotx. cosx
(A) 1 (B*) 2 (C) 3 (D) 4
Q.53 The exact value of cos273º + cos247º + (cos73º . cos47º)is
(A) 1/4 (B) 1/2 (C*)3/4 (D) 1
[Sol.
2
146
cos
1 0

+
2
94
cos
1 0

+
2
26
cos
120
cos 0
0

= 1 + 2
26
cos
4
1
2
94
cos
146
cos 0
0
0







 
= 3/4 Ans ]
Q.54 In a ABC, a = a1 = 2 , b = a2 , c = a3 such that ap+1 = 




 


 p
p
p
2
p
p
2
p
a
5
2
p
4
2
a
3
5
where p = 1,2 then
(A) r1 = r2 (B) r3 = 2r1 (C) r2 = 2r1 (D*) r2 = 3r1
[Hint: put p = 1 , we get a2 = 4  b = 4
put p = 2 , we get a3 = 4  c = 4
Hence the ABC is isosceles

11. =
9
sin
18
sin
6
cos
18
cos
6
sin
4






 




= 4 Ans ]
Q.41 In a ABC if b + c = 3a then cot
B
2
· cot
C
2
has the value equal to :
(A) 4 (B) 3 (C*) 2 (D) 1
[Hint: cot
B
2
· cot
C
2
=
 
s s b


.
 
s s c


.
 
s a
s a


=
s
s a

=
2
2 2
s
s a

but given that a + b + c = 4a  2s = 4a Hence cot
B
2
· cot
C
2
=
4
2
a
a
= 2 ]
Q.42 The set of values of ‘a’for which the equation, cos 2x + a sin x = 2a  7 possess a solution is :
(A) (, 2) (B*) [2, 6] (C) (6, ) (D) ()
[Sol. cos2x + a sinx = 2a – 7
i.e. 2sin2x – a sinx + 2a – 8 = 0
sinx =
a a a a a
  

 
2
8 2 8
4
8
4
( ) ( )
sinx =
a
or
 4
2
2
Hence –1 < (a– 4)/ 2 < 1  the range of a ]
Q.43 Inaright angled trianglethehypotenuse is2 2 timesthe perpendicular drawnfromthe oppositevertex.
Thentheother acute angles ofthe triangle are
(A)

3
&

6
(B*)

8
&
3
8

(C)

4
&

4
(D)

5
&
3
10

[Sol. p2sec2 + p2cosec2 =  2
2
2 p2
 8
cos
sin
1
2
2



sin22 = 1/2 =
2
2
1






2 = n + /4
 = n/2 + /8
for n = 0   = /8
for n=1   = 3/8 ]
Q.44 Let f, g, h bethe lengths oftheperpendiculars fromthe circumcentreoftheABC onthe sides a, band
c respectively . If
a
f
b
g
c
h
  = 
a b c
f g h
then the value of  is :
(A*) 1/4 (B) 1/2 (C) 1 (D) 2

12. [Hint: Note that thegiven expressionsimplifies to 3 cot3x ]
[Sol. cotx +
)
60
x
sin(
)
60
x
cos(
)
x
60
sin(
)
x
60
cos(





= )
60
x
sin(
)
60
x
sin(
)
x
2
sin(
x
sin
x
cos



= 3
x
sin
4
x
cos
x
sin
8
x
sin
x
cos
2

 =
x
sin
3
x
sin
4
x
cos
x
sin
8
x
cos
3
x
cos
x
sin
4
3
2
2



=
x
sin
]
x
cos
4
x
cos
3
[
3
3
3

= 3 cot3x 
x
tan
x
tan
3
]
x
tan
3
1
[
3
3
2


Ans ]
Q.34 In a  ABC, cos 3A+ cos 3B + cos 3C = 1 then :
(A) ABC is right angled
(B) ABC is acute angled
(C*) ABC is obtuse angled
(D) nothing definite can be said about the nature ofthe .
[Hint:  cos 3A = 1 + 4 sin
3
2
A
sin
3
2
B
sin
3
2
C
= 1  A =
=
2
3

or B =
2
3

or C =
2
3

]
Q.35 The value of
3 76 16
76 16
  
  
cot cot
cot cot
is :
(A*) cot44º (B) tan44º (C) tan2º (D) cot46º
[Sol. Using 0
0
0
0
0
0
0
0
16
cos
76
sin
16
sin
76
cos
16
cos
76
cos
16
sin
.
76
sin
3


= 0
0
0
0
0
0
0
92
sin
]
16
cos
76
cos
16
sin
76
[sin
16
sin
76
sin
2 

=






92
sin
60
cos
92
cos
60
cos
= 0
0
92
sin
92
cos
1
= 0
0
0
2
46
cos
46
sin
2
46
sin
2
= tan460 = cot440 Ans ]
Q.36 Ifthe incircle ofthe ABC touches its sides respectivelyat L, M and N and ifx, y, zbe the circumradii
ofthe triangles MIN, NIL and LIM where I is the incentre then theproduct xyz is equalto :
(A) Rr2 (B) rR2 (C*)
1
2
R r2 (D)
1
2
rR2
[Hint: note thatANIM is a cyclic quandrilateral.
 cosec
A
2
=
2x
r
 2x = r cosec
A
2
x =
r
A
2
2
sin
 xyz =
2
R
r
r
.
2
R
.
r
2
A
sin
4
.
2
r 2
3
3



]

13. Q.25 With usual notation in a  ABC
1 1 1 1 1 1
1 2 2 3 3 1
r r r r r r






 





 





 =
K R
a b c
3
2 2 2
where K has the value
equalto :
(A) 1 (B) 16 (C*) 64 (D) 128
[Hint: 1st term = )
b
s
a
s
(
1




=
c

 L H S =
a bc
3 . Use  =
a bc
R
4
to get the result ]
Q.26 If
5
2
3


 
x , then the value ofthe expression
1 1
1 1
  
  
sin sin
sin sin
x x
x x
is
(A) –cot
x
2
(B) cot
x
2
(C) tan
x
2
(D*) –tan
x
2
[Hint: Onrationalizing;we get
x
sin
1
x
sin
1
|
x
cos
|
2
x
sin
1
x
sin
1







=
 
)
x
(sin
2
|
x
cos
|
1
2


=
)
x
(sin
x
cos
1


 (D) ]
Q.27 If x sin = y sin 








2
3
= z sin 








4
3
then :
(A) x + y + z = 0 (B*) xy + yz + zx = 0 (C) xyz + x + y + z = 1 (D) none
[Hint:







sin
sin
.
3
/
2
cos
cos
.
3
/
2
sin
y
x
= 









sin
sin
cos
3
2
1
=
2
1
cot
2
3

 ....(1)
|||ly







sin
3
/
4
sin
.
cos
3
/
4
cos
.
sin
z
x
= 

 cot
2
3
2
1
....(2)
1
z
x
y
x


  xz + xy + yz = 0 ]
Q.28 In a ABC, the value of
a A b B c C
a b c
cos cos cos
 
 
is equal to :
(A*)
r
R
(B)
R
r
2
(C)
R
r
(D)
2r
R
[Hint: LHS
 
 
R A B C
R A B C
sin sin sin
sin sin sin
2 2 2
2
 
 
=
4
2 4 2 2 2
sin sin sin
. cos cos cos
A B C
A B C
= 4 sin
A
2
sin
B
2
sin
C
2
=
r
R
]
Q.29 The value of cos

10
cos
2
10

cos
4
10

cos
8
10

cos
16
10

is :
(A)
1
32
(B)
1
16
(C)
 
cos /
 10
16
(D*) 
10 2 5
64


14. Q.17 With usualnotations, in a triangleABC, a cos(B – C) + b cos(C –A) + c cos(A– B) is equal to
(A*) 2
R
abc
(B) 2
R
4
abc
(C) 2
R
abc
4
(D) 2
R
2
abc
[Sol. Here a(cosB cosC + sinB sinC) + ........
using
C
sin
c
B
sin
b
A
sin
a

 = 2R
a (cosB cosC + 2
R
4
bc
) + ......
= 2
R
4
abc
3
+ a cosB cosC + b cosC cosA + c cosA cosB
= 2
R
4
abc
3
+ c cosC + c cosA cosB
= 2
R
4
abc
3
+ c [cosA cosB – cos(A + B)]
= 2
R
4
abc
3
+ c sinA sinB
= 2
R
4
abc
3
+ 2
R
4
abc
= 2
R
abc
 (A) ]
Q.18
sin cos
sin cos
3 3
 
 



cos
cot


1 2

 2 tan cot  =  1 if :
(A)  0
2
,






 (B*) 


2
,





 (C)  

,
3
2





 (D) 
3
2
2


,






[Hint: simplifies to –cos |sin| + sin cos = 0 provided sin  cos 
Q.19 With usual notations in a triangleABC, ( I I1 ) · ( I I2 ) · ( I I3 ) has the value equalto
(A) R2r (B) 2R2r (C) 4R2r (D*) 16R2r
[Hint: BICI1 is a cyclic quadrilateral with I I1 as the diameter
also  BI1C =
2
A
2


applying sine law in BCI1
1
I
I
2
A
cos
a

 I I1 =
2
A
cos
2
A
cos
·
2
A
sin
2
·
R
2
= 4R sin
2
A
  1
I
I = 64R3 sin
2
A
sin
2
B
sin
2
C
= 16R2 r ]

15. Q.8
     
   
tan . cos sin
cos . tan
x x x
x x
   
 
  
 
2
3
2
7
2
2
3
2
3
whensimplified reduces to :
(A) sinx cosx (B) sin2 x (C) sinx cosx (D*) sin2x
[Sol.
     
   
tan . cos sin
cos . tan
x x x
x x
   
 
  
 
2
3
2
7
2
2
3
2
3
=
x
cot
.
x
sin
x
cos
x
sin
.
x
cot 3



=
x
sin
x
cos
.
x
sin
x
sin
x
cos
.
x
sin
x
cos3


= sin2x Ans ]
Q.9 If in a ABC, sin3A+ sin3B + sin3C = 3 sinA· sinB · sinC then
(A) ABC may be a scalene triangle (B) ABC isa right triangle
(C) ABC is anobtuse angled triangle (D*) ABC isan equilateraltriangle
[Hint: Use : a3+b3+c3–3abc = (1/2) (a+b+c) [(a–b)2+(b–c)2+(c–a)2 ]
[Hint: either sinA + sinB + sin C = 0 (which isnot possible)
or sinA = sinB = sinC  equilateral ]
Q.10 In a triangleABC, CH and CM are the lengths of the altitude and median to the base AB. If a = 10,
b = 26, c = 32 then length (HM)
(A) 5 (B) 7 (C*) 9 (D) none
[Hint: cos B =
32 10 26
2 3210
7
10
7
2 2 2
 
  
. .
BH
 MH = BM  BH = 16  7 = 9 ]
Q.11 The value of
1
tan
cos
sin
cos
sin
sin
2
2










for allpermissible vlaues of
(A) is less than – 1 (B) is greater than 1
(C) lies between – 1 and 1 including both (D*) lies between – 2 and 2
Q.12 sin 3 = 4 sin  sin 2 sin 4 in 0  has :
(A) 2 realsolutions (B) 4 realsolutions
(C) 6 realsolutions (D*) 8 realsolutions.
[Sol. given equation canbe written as
3 sin  - 4sin3 = 4sin sin 2 sin4
hence either sin  = 0   = n
or 3 – 4sin2 = 4 sin 2 sin 4
3 – 2 (1 – cos 2) = 2 (cos 2 – cos 6)
or 1 = – 2 cos 6
cos 6 = –
2
1
= cos
3
2
6 = 2n ±
3
2

16. Question bank on Compound angles, Trigonometric eqn and ineqn, Solutions of Triangle & Binomial
There are 142 questions in this question bank.
Select the correct alternative : (Only one is correct)
Q.1 If x + y = 3 – cos4 and x – y = 4 sin2 then
(A) x4 + y4 = 9 (B) 16
y
x 

(C) x3 + y3 = 2(x2 + y2) (D*) 2
y
x 

[Sol. On addingand subtracting
x =
2
2
sin
4
4
cos
3 



; y =
2
2
sin
4
4
cos
3 



x =
2
)
4
cos
1
(
)
2
sin
1
(
4 




; y =
2
)
4
cos
1
(
)
2
sin
1
(
4 




x = 2 (1 + sin2 ) – cos22 ; y = 2 (1 – sin2) – cos22
x = 1 + 2 sin2 + sin22 ; y = 1 – 2 sin2 + sin22
x = (1 + sin2)2 ; y = (1 – sin2)2  2
y
x 
 ]
[Alternate : Or put  =
4

and verify ]
Q.2 If in a triangle ABC, b cos2
A
2
+ a cos2
B
2
=
3
2
c then a, b, c are :
(A) inA.P. (B) in G.P. (C) in H.P. (D*) None
[Hint:
b
2
(1 + cos A) +
a
2
(1 + cos B) =
3
2
c
 a + b + c = 3c  a + b = 2c  a,c,b are in A.P. ]
Q.3 If tanB =
A
cos
n
1
A
cos
A
sin
n
2

then tan(A + B) equals
(A*)
A
cos
)
n
1
(
A
sin

(B)
A
sin
A
cos
)
1
n
( 
(C)
A
cos
)
1
n
(
A
sin

(D)
A
cos
)
1
n
(
A
sin

[Sol. tan(A + B) =
B
tan
A
tan
1
B
tan
A
tan


=
A
cos
n
1
A
cos
A
sin
n
·
A
tan
1
A
cos
n
1
A
cos
A
sin
n
A
tan
2
2




=
A
cos
A
sin
n
)
A
cos
n
1
(
A
cos
A
cos
A
sin
n
)
A
cos
n
1
(
A
sin
2
2
2
2




=
)
A
sin
n
A
cos
n
1
(
A
cos
0
A
sin
2
2



=
A
cos
)
n
1
(
A
sin

]
Q.4 Given a2 + 2a + cosec2

2
( )
a x

F
H
G I
K
J= 0 then, which ofthe following holds good?
(A) a = 1 ;
x
I
2
 (B*) a = –1 ;
x
I
2

(C) a  R ; x  (D) a , x are finite but not possible to find