The document contains solutions to multiple trigonometry identity and concept questions. For each question, the solution uses substitution techniques to simplify the expressions and arrive at the answer. Key steps include: 1) Letting variable angles equal specific values like 0, 30, 45, 60, 90 degrees to simplify trig functions. 2) Applying identities like sin^2 x + cos^2 x = 1 to isolate variables. 3) Substituting the simplified expressions back into the original to arrive at an identity equaling the answer choices. The techniques shown provide concise solutions through strategic substitution of angle values and use of trig identities.

1. Q. The value of (1 cot cosec )(1 tan sec ) is equals to :
(a) 3 (b) 0 (c)
S. Here the value of the expression is independe
1 *(d) 2
nt of θ,therefore givenexpression is
an identity i
       

Magical Conceptual Short Tricks
0
n θ, so we can put any suitable value of θ to minimise the calculations.
here let θ 45 then value of given expression is (1
Q. Let k=4cos x cos2x cos3x cos x cos
1 2)(1 1 2) (2
2x cos3x thenk is equals to :
(a) 1 (b) 0
2)(2 2)=2.
*(

     

 
 
0 0 0 0 0 0
2
0
2
S. Again value of the expression is independent of x, so the expression is anidentity in x,
so let x 0 then k=4cos0 cos0 cos0 cos0 cos0 cos0
c) 1 (d) 2
Q. The value of the expression
4 1 1 1
cos c (
1
os
       
 


2
0
2
S. Here options are in terms of so the value of the expression is independent
) 2cos cos cos( ) is ?
*(a) sin (b
of .
so we can put any su
)cos (
itable value of to minimise the calculations.
pu
c)sin2 (d)cos2
t =0
 
     


 

 

2 2
2 2 2 2 2
then, cos cos ( ) 2cos cos cos( )
cos 0 cos (0 ) 2cos0cos cos(0 ) 1 cos 2cos cos 1 cos s
Q. If tan A tanB x and cot A cotB y then cot(A B)
1 1 1 1
(a) *(b)
in .
(c)
x y x y
          
           
    
 



   

0 0
0 0 0
1 2 1 2
S. Let A 60 &B 30 then x 3 & y 3 ,
3 3 3 3
so L.H.S. cot(60 30 ) cot30 3
Now put the values of x and y in the options then option (
x y
b) will give
(d) x y
7
Q.
s 3.
If X sin +sin s
12 12
   
     
        
   
 
 
     
0 0 0 0
3 7 3
in ,Y cos +cos cos
12 12 12 12
X Y
then
Y X
(a) 2sin2 (b) 2cos2 *(c)2tan2 (d) 2cot
3 3
S. Let 15 then X sin120 +sin0 sin60 = 0
2
2
     
           
   
         
 
  


    
0 0 0
0
3
2
1 1 3 1 2
Y cos120 +cos0 cos60 1 1, then L.H.S.
2 2 1 3 3
Now put 15 in the options then option (c) will gives 2/ 3.
Q. If cos cos cos 0 and if cos3 cos3 cos3 k cos cos cos then k
(a) 1 (b) 8 *(c) 12 (d)

         
 
               
 0 0
0 0 0 0 0 0
S. cos cos cos 0 so let 0, 120 & 120 satisfying given condition
cos3 cos3 cos3 k cos cos cos
cos0 cos360 cos360 k cos0 cos120 cos120
1 1 1 k.1.( 1/2).( 1/2) k 1
9
2
           
         
   
       

The beauty of these short tricks is that many problems of this kind can be solved easily.

2. :S.
Let = 15º then LHS = tan15º+2tan30º+4 tan60º + 8cot1
Q. The valu
20º
1 1
=2 3+2 × +4 3 +8
3 3
e of tan +2tan2 +4tan4 +8cot 8 is :
(a) tan (b) tan2 *(c) cot (d) cot2
 




 
   
   
Magical Method of Substitution and Balancing
 
 
2 2 2 2 2 2 2
0
2
p 1
Q. If tan A and if =6 is acute angle then (pcosec2 q sec2 )
q 2
(a) p +q *(b) 2 p +q (c) 2 p q (d)
=2+ 3 = cot 15º =cot
S. : Let A 45 then q p
2 2 2 2
and LHS pcosec15 psec15 p 2 2 p
3
p q
1 3 1

 
 
 
       
 

  
    
 
Magical Method of Substitution
 
2 2 2 2
0
2 2
8ab
Q. If a sin x sin y, b cos x cos y, c tan x tan y then
(a b ) 4a
*(a) c (b) c (c) 2c (d)
now put q p in options then b wil
2c
l match with LHS.
S. : let x y 45 ,
    





 

Magical Method of Substitution
       
2 2 2 2
0 0
Q. If cos 2cos then tan tan
2 2
1 1 1 1
a b c d
3 3 3 3
therefore a 2, b 2, c 2
8 2 2
then 2 c
( 2 2 ) 4 2
S. : cos 2cos so let 60 & 0
tan
   
      
  
 
 
      
 

 
   
  


 

Magical Method of Substitution
   
 n n1 2 3
0
1 2 3
n1 2 3
n/ 2
0
2 n/
Q. If 0 , , ,................. and if tan tan tan ...........................tan 1
2
then
1
ta
cos cos cos ............
n tan 30 tan 30
...............cos
(a) 2 *(b) 2 (c)
2 2 3

   
      
  
   

          
    
0
n1 2 3
n/4 n/
0
4
0 0
:S.
Let ............. 45 (Satisfying both given conditions)
Required value cos 45 cos 4
2 (d
5 cos 45 ............n terms
1 1 1
.....................
2 2 2
) 2

       
  
  
Magical Method of Substitution and Balancing
k k
4 6
n/2
n/2n
k k 4 4 6 6
k
k
4 6
1
Q. Let f (x)= (sin x cos x),x R and k 1,then f (x) f (x)
k
1 1 1 1
(a) *(b) (c) (
1 1
.....n tertms 2
2( 2)
1 1 1
S. f (x)= (sin x cos x) f (x) f (x) (sin x cos x) (sin x cos x)
k 4 6
As va
d
lue of above expr
)
4
ession
12 6 3
is i


  
      
   

4 4 6 6
4 6
ndependent of x soput x 0 in above expression
1 1 1 1
f (x) f (x) (sin 0 cos 0) (sin 0 cos 0)
4 6 4
.
26
1
1

        
The beauty of these short tricks is that many problems of this kind can be solved easily.

3. 0 0 0
0
2
:
sin5 sin2 sin
Q. The value of the expression is equals to :
cos5 2cos3 2cos cos
*(A) tan (B) cos (C) c
sin150 sin60 sin30
Let 30 then LH
ot (
S =
cos
D si
.
n
S
)


    


 
     
   
Magical Method of Substitution and Balancing
0 0 2 0 0
0
0 0
2 2 2
:
Q. If (cos cos
1
150 2c
) (sin
os90 2cos 30 cos30 3
Now
sin ) k sin
put 30 in options then (a) w
, then k isequals to:
2
(A) 0 (B) 1 (C) 3 *(D) 4
ill match with L.H.S.
S.
Let 90 , 0 then (0
 
  

  
    

  
 
 
  
  
Concept of Identity
2 2
4 4 2
2
2 2
2 2 2 2 2 2 2 2
4 4 2
Q. Which of the followings is not equals to unity :
(A) cos sin 2sin
sin cos
(B) (1 cot ) (1 tan )
2 2
(C) sin cos cos sin s
1
1) (1 0) k k 4
2
in si
S.
n cos cos
*(D)sin cos 2sin
    
 
    
          
 
  
 


 
Concept of Identit
0 0 0 0 0
4 4 2
2
2 2
2 2 2 2
the value of theabove expressions should be 1.
put 45 ( sin45 cos 45 and also tan45 cot 45 ).
1
(A) cos sin 2sin 0 2 1
2
sin cos 1 1 1 1
(B) (1 cot ) (1 tan ) .2 .2 1
2 2 4 4 2 2
(C) sin cos cos sin si
:
n
      
        
 
         
    




y
2 2 2
3 3
2
3
4 4 2
1 2
1 1 1 1
sin cos cos 1
4 4 4 4
1
(D)sin cos 2sin 0 2 1(not equals to unity).
2
S.
The value o
4
Q. sin sin sin is equals to :
sin3 3 3
4 3 3
(A) (B) *(C) (D) no
f
ne
3 4 4
:
    
      
         
       
 
      






Concept of Identity
 
0
3 0 3 0 3 0
0
above expressions is independent of so put 30 .
sin 30 sin 150 sin 270 1 1 3
then value of given expression 1 .
8 8 4sin90
  
 
    
The beauty of these short tricks is that many problems of this kind can be solved easily.

4. cot 1
S. 3cot cot cot3
cot 1 tan
Q. If then is equals to :
cot cot3
2cot cot3 2tan3 tan
3 tan tan3
1 2 2 1
(A) *(B) (C) (D)
3 3 3 3
Q. If 13 cos 12cos( 2 ), then va
cot cot3 3
tan 1 1 2
tan3 1tan tan3 31 1
tan 2

              
 

 
  

   

   
 
    
    


cos( 2 ) 13 cos( 2 ) cos
lue of cot( ) 25tan is :
*(A) 0 (B) 1 (C) 3 (D) 4
Q. I
25 2cos( ).cos
S. 25
cos 12 cos( 2 ) cos 1 2sin( ).sin( )
cot( ).cot 25 cot( ) 25tan cot ( ) 25t
f x y 3 cos 4 a
0
n
a
d
n
           
    
         
        
    
 
      

  

 

0
4 4 3 3 2 2
:S.
The value of the expression is independent of , so put 0 .
x y 3 1 x y 2...
x y 4sin2 the
(i) and
n:
(A) x y 9 (B) x y 9 (C)x y 2(x y ) *(D) x 2
0
y
x y .
 
  
   
  
       
    


Magical Method of SubstitConcept of Identity ution and Balancing
0 0 0 0 0 0 0 0
0 0 0 0 0 0
b(cos1 sin1 )(cos2 sin2 )(cos3 sin3 )........
.....(ii), by (i) and (ii) x y 1
Now put x y 1 in the options the
......(cos 45 sin45 )
cos1
n optio
cos2 cos3 cos 4 .......
n (D) is sat
.....cos 44 cos
isf
5
.
4
ied
Q. If = a ,
whe
 

 
  
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0
(cos1 sin1 ) (cos2 sin2 ) (cos3 sin3 ) (cos 44 sin44 ) (cos 44 sin44 )
S. ..........
cos1 cos2 cos3 cos 44 cos 4
(A) 22 (B) 23 (C) 24 *(D) 2
5
(1+tan1°)(1+tan2°)............. .
5
.
re a and b are prime numbers and a b then a b
    
  
0
0
.......(1+tan43°)(1+tan44°)(1+ tan45°)
1 44 2 43 3 42 ......... 45
So first wefind out if 45 then (1+tan )(1+tan ) ?
let 0 and 45 then(1+tan )(1+tan ) (1) 2 2
(1+tan1°)(1+tan4
            
      
         

By method of substitution
b
22 pairs
22 k 23 b
4°)(1+tan2°)(1+tan43°).............(1+tan22°)(1+tan23°). (1+1) = a
2 2 = 2 2 = a a 2 andb 23 and hence a b 25

       

*Do you know what are the methods of Substitutions & Balancing to Solves problems of Trigonometry?
*Do you know what are Master A.P.,G.P.& H.P. and How these Solve the problems of Progressions ?
*Do you know the Master Quadratic Equations to Solves problems of Quadratic Equations ?
*Do you know the Master Triangles to Solve problems of Solutions Of Triangles ?
The beauty of these short tricks is that many problems of this kind can be solved easily.

5. A triangle has three sides and three angles.
The sides opposite to angles A, B and C. are denoted
by a,b and c.
the sum of its angles is A+B+C =180°and
the sum of its angles is a+b+c
Properties and Solutions of Triangles
= 2s (perimeter),
In any triangle sum of any two sides is greater than third side a+b >c,c+b >a,a+c>b*
Most of the questions of this chapter can be solved veryeasily bymaster triangles,
for that we have invented three Master Triangles.
As we all know that a master key unlocks many l
Master Tri
ocks. Simil
angles
arly our Master triangles
can solve many questions belong to family of triangles.
Q In a triang
* The above master triangle can be used most widely to all the problems which
belong to a family of triangles i.e. problems belong to all triangles.
* If the data of this triangle satisfy the given conditions in the questions.
      
            
      
          

 


le ABC correct statement is : [ ,
B+CA A B C
(a) (b c) cos = 2a sin (b) (b c) cos = a sin
2 2 2 2
B+CB C A A
(c) (b c) sin =a cos (d) (b+c) sin =a cos
2 2 2 2
IITJEE 2005 1M]

     
S.
Let the triangle be an equilateral,so according to data of MT 1,
a b c 1 and A B C 60º after putting the value of a,b,cthen only option (b)
wil
–
Here above problem belongs to all triangles,so MT -1 can be applied here.
l balance and so it is the right choice.
A
F
B CD
E
P
R
-
-
  
   
   
   
   
0
1 2 3
3
Let a = b = c = 1 A= B = C = 60 Area= Δ =
4
abc 1
Radius of circumcircle R
4Δ 3
Δ 1
Radius of in circle r
s 2 3
3
Radii of ex circles are r r r
2
3
Length of perpendicular (altitutes) AD BE CF
2
 
 Equilateral Triangle
Master Triangle1 MT -1
CB
b
A
The beauty of these short tricks is that many problems of this kind can be solved easily.

6. * The above master triangle can be used most widely to all the problems which
belong to a family of triangles i.e. problems belong to all triangles.
and if the data of this triangle satisfy the given conditions in the questions.
Some of the exa
           
Q. If sides of a triangle are in ratio 1: 3: 2, then ratio between its orresponding angles is :
(
(a) 3 2 1 b 3 1 2 c 1
S
2 3 d 1 3 : 2
. The data
ΙΙΤJEE 2004,1M)
mples are given below : -
  
       
of MT–2, a 1,b 3 and c 2 satisfy the given conditions
so triangle is right angled & A 30º, B 60º, C 90
Q. If the angles A,B and C of a triangle are in arithmatic Progressions a
º. A:B:C 30 60:90 1
nd a,b and c are
: 2 3
sides
 
     
  


0 0 0
0 0
S. Angles are in A.P. so according t
a c
opposite to angles A,B and C then value of the expressi
o data of MT-2,
A 30 ,B 60 ,
on sin2C sin2A is :
c a
C 90 and a 1,b
1 3
(a) (b) (c)1 (d) 3
2
3,c 2
a c 1 2
sin2C sin2A sin180 sin60
c a 2 1
2
IITJEE2010
  

   
   
 
 
 


 




2 2 2
2 2 2
3
2 3
2
S. A : B : C 1: 2: 3, therefore we can apply a
a b c
Q. In ABC if A : B : C 1: 2: 3,then the value of is equals to :
ab bc ac
2 2 8 8
(a) (b) (c
bove master triangle here.
a b c 1 3 4
a 1,b 3,c
) (d)
3 1
2 and he
3 1 3 3 2 3
nce
b b ac
3
a
1
c

  
8
3 2 3 2 3 3 2
Many questions can be solved by this above Master Triangle
 
 Right Angled Triangle
Master Triangle 2 MT -2
   
   

 
 
 
0 0 0
A 30 ,B 60 ,C 90
a 1,b 3,c 2
3 1
r .
2
R 1.
31
Δ= . 3 .
2 2
2
1
r
A
BC
The beauty of these short tricks is that many problems of this kind can be solved easily.

7. * The above master triangle can be used most widely to all the problems which
belong to a family of triangles i.e. problems belong to all triangles.
and if the data of this triangle satisfy the given conditions in the questions.
Some of the exa
       
S. Since data of above triangle( MT–3) i.e. A
Q. In a Δ ABC if sinA + sinB + sinC =1+ 2 and cosA +cosB +cosC = 2 ,then triangle is :
a Equilateral b Isosceles c Right angled d Right angle isosceles
mples are given below : -
  
   
             
        
45º, B 45º and C 90º
when put in given conditions of the questions
sinA sinB sinC 1 2
1 1
sin45º sin45º sin90º 1 2 1 1 2 1 2 1 2(satisfy)
2 2
1 1
cos45º cos45º cos90º 2 0 2 2 2(satisfy)
2 2
therefore, it is right an
  
     2 2 2 2 2 2
0
gled isosceles.
S. Since data of above triangle( MT–3) , a 1, b 1 and c 2
satisfies given condition b c 3a i.e.1 ( 2) 3.1 3 3
therefore A=45 ,B=4
2 2 2
Q. In ΔABC if b +c = 3a then cotB +cotC - cotA =
ab
(a) 4Δ (b) Δ (c) (d) 0
4Δ

     
   
0 0
0 0 0
5 and C 90
then cotB cotC cot A cot45 cot90 cot45 0
S. Since data of above triangle( MT–3)
A 90º,B 45º and C 45º and a
0 0 0 0
2cosA cosB 2cosC a b
Q. In a ΔABC, if + + = + then the value of Ð A is:
a b c bc ca
(a) 60 (b) 90 (c) 30 (d)75
 
  
        0
2, b 1 and c 1
2.0 21 1 3 3
2 satisfies ,therefore A 90 .
12 2 2 2 2
 
 Right Angled Isoscales
Master Triangle 3 MT -3
   
   
 
  
    
0 0 0
a 1, b 1, c 2
A 45 ,B 45 ,C 90
1
R ,
2
1
r 1 ,
2
1 1
Δ 1 1
2 2
2
1
r
A
BC
1

8.             
 
   
  




3 3 3
0
a b c
Q. In a Δ ABC, if sin A sin B sin C = 3sinAsinB sinC, then b c a =
c a b
a 0 b a b c c a b c ab bc ca
S. : The data of MT–1 , A B C 60 satisfies the given condition.
d 1
MT -1
Questions which have single currect answers :
   
  
 



cosA cosB cosC
Q In triangle AB
let a=b=c, therefore thetriangle is an equ
C if and if a = 2 then area of triangle wi
ilater
ll be :
al sotake a b c 1
a b c 1 1
a b c
a 1 b 2
1
b c a 1 1 1 0
c a b 1 1 1
   




      
 
    

2
2 2
S. : The data of MT–1 satisfies the given condition. so the triangle is
3 3
an equilateral area o
3
c d 3
2
a a a
Q. In ABC, if cos , cos & cos then tan
b c
f the equilateral triangleis Δ a (2) 3
4
b c c 2
4
b
MT -1
       
 
   
  
    
        
    
    
             

 

  

  


0
2 2 2 2 0
2 2
S. : Let the ABC is an equilateral and a b c 1,
1
cos cos cos 60
2
tan +tan =
2 2
a 3 b 1 c 3 3 d 9
(a b c)(b c a) (c a b
Q. In triangle ABC,
tan tan +tan = 3tan 0 = 1
2 2 2
3
MT -1
       
 

      
2 2
2 2 2 2
S. : Let the triangle be an equilateral. Therefore, accordi
) (a b c)
=
4b c
a sin A b cos A
ng to MT–1
3
a b c 1 and A B C 60º so L.H.S.
4
after
c tan A d cot A
putting the value of A in th
MT -1
 
       
    
          

 
     
2 2 2
2 3
0
1 A 1 B 1 C
Q In triangle ABC , the value of cos cos cos is equals to ?
a 2 b 2 c 2
s s s
a s b
e options, a will give R.H.S
S. : let a b c 2 & A B C 60 ,L
c d
abc
H
c
.
abc ab
MT -1
 
 2 9
.S. 3cos 30º =
4
put values of a, b, c and s in the options,then c will give the required result.
Alart: - Here MT -1 can't be applied,as it will give more than one options same.
The beauty of these short tricks is that many problems of this kind can be solved easily.

9.        
   
      
S. : Let the triangle be right angled isoscales.
Therefore, according to MT–3 a
2 2 2 2
2 2 2
C C
Q. In any triangle ABC, (a - b) cos +(a + b) sin is equals to : -
2 2
a a b b c c d
3
1
MT -
   
      
     
         2 2 2 2 2
1, b 1, c 2 and A 45º, B 45º,C 90º.
C C 1
L.H.S. (a b) cos (a b) sin 0 4 2 c (since
A
Q. In a triangle ABC,
c 2)
2 2 2
2ac sin
Alart: - Here MT -1 can't be applied,as it will give more than one options same.
 
 
    
    
  

2 2 2 2 2 2 2 2 2 2 2 2
S. : Let the triangle is a right angled. Therefore, according to MT 2
a 1, b = 3 ,c 2 & A
– B+C
=
2
(a
30º,
) a b – c (b) c a b (c) b – c – a (
B 60º,C 90º.
1
then
d) c –
L.H.S. 2 1 2 sin 30º 4
a – b
MT - 2
 


    2 2 2
4
2.
2
put values of a, b, c in the options, then b will match with L.H.S.
Q. In a triangle ABC, If c a b , then 4s(s a)(s b)(s c)
(a) s (b
Alart: - Here MT -1 can't be applied,as it will give more than one options same.
 

 
  
  
          

2 2 2 2 2 2
2 2 2
2 2 2
2 2 2
S. : c a b the triangle is right angled and c is hypoteneous.
1 a b
L.H.S. 4s(s a) (s b) (s c 4Δ ×a b = 4 a b
2 4
Therefore, option d is the right o
) b c (c) c a (d
ption.
Q. In triang
) a b
MT - 2

        

 
 

2
2
2
2
S. : Let the triangle be an equilateral so according
le ABC if is the area of the triangle then a sin2C c sin2A
(a) Δ (b) 2Δ
to MT–1 ,
3 3 3
a b c 1, A B C 60°, Δ= (1) , Δ 3 4Δ
4
(
4 4
LHS
c) 3Δ (d
a
) 4
2
Δ
sin C
MT -1
        
 

       
2 2 2
2
2
2 2
8a b c
Q. In a triangle,if (a b c)(a b c)(b c a)(c a b) , then the triangle is
a b c
[Note: All symbols used have usual meaning
3
c sin2A 1 sin120° 1 sin120°= 2sin120° 2 3 4Δ .
2
in ABC.]
(a) isosceles (b) right angled (c)
   S. : The sides of MT 2 i.e. a 1, b = 3 ,c 2 satisfy the given relation of
the t
equilateral (d) ob
riangle so it is
tuse angl
right an e .
ed
gl d
MT - 2
The beauty of these short tricks is that many problems of this kind can be solved easily.

10.    
    

   
Q. In a ABC, if G is the centroid of thetriangle and GA,GB,GC makes angles , ,
with each other then cot A cotB cotC cot
S. : Let the triangle be an equilater
cot cot
(a)1 (b
al
then accor
) 0 (c
ding
)3 (d)3 3
to data o
MT -1
 
      
             
      
     
    
  

0 0
0 0 0 0 0 0
0 0
2 2 2
f MT-1.
A B C 60 then 120
cot60 cot60 cot60 cot120 cot120 cot120
3(cot60 cot120 ) 3 3 3 0
a b c A B C
Q. In a ABC, the value of the expression sin sin sin is :
sinA sinB sinC 2 2 2
(a
       
                 

       
     
 



0
2 2 2
S. : let the triangle be an equilateral, then according to data of MT-1
3
a b c 1, A B C 60 ,
4
a b c A B C 2 2 2 1 1 1
sin sin sin
sinA sinB sinC 2 2 2 2
)2 (b) (c) (d)
2 4
Q. Let f,g,h are th
2 2
e
3 3 3
MT -1

    

 


lengths of perpendiculars from circumcentre of ABC on the sides
a b c abc
a,b and c and if k then the value of k is :
S. : Let thetrianglebe anequilateral so according to data of MT-1
f g h fgh
1 1
(a)1 (b) (c) (d) 2
2 4
1
a b c 1and f g h
2 3
th
MT -1
    
        
          
Q. In a triangle ABC with fixed base BC, the vertex A moves such that
B C A
cot +cot =2cot and If a,b a
a b c abc
nd c de
1
erefore k 2 3 2 3 2 3 k 2 3 2 3 2 3 k
f g
note the lengths of the sides of
2 2 2
h fgh 4
 
a triangle
opposite to the angles A, B and C, then:
(a) locus of points A is an ellipse.
(b) b + c = 2a
(c) b + c = 4a.
(d) locus of point A is a pair of st
S. : The data of M
raight l
aster Tri
ines.
angle-1 sMT -1

     
atisfy the given condition so
let the triangle be an equilateral. b+c=2a
AB AC BC 2a distance between two fixed points locus of vertex A is ellipse.
A
B C
G

11.        
   
  

    
2
1 2 3
2 2 2 2 2 2 2 2 2 2 2 2
2 3 1 1 2 3 2 3 1 2 3 1
Q. Let in a Δ ABC with side 'a',AD is a median of If AE and AF are medians of Δ ABD
a
and Δ ADC and AD m ,AE m ,AF m then
8
a m m 2m b m m 2m
S. : Let
c m m m d m m
the triangle i a
m
s n eMT -1
   
     
  
     
1 2 3
2 2
2 3
1 2 3
2
quilateral with side 'a'.
therefore according to MT-1
3 a
AD = m , AE = AF, m m
2
3 a 13 aa a
ED , AE = AF m m
4 2 4 4
After putting the values of m , m , m in the options
a
then (a) will give ,therefore (a) is the correct ch
8

Q. Let DEF is the triangle formed by joining the points of contacts of the incircle with
sides of ABC, also let R,r and Δ are radius of circumcircle, radius of incircle, and
area of the triangle ABC
oice
, the
.
n
(i)
Questions based on Paragraph
          
                    
    
          
. The sides of ΔDEF are :
A B C A B C
*(a ) 2rcos ,2rcos ,2rcos (b) r cos ,rcos , rcos
2 2 2 2 2 2
r r rA B C
(c) rsin2A,rsin2B, rsin2C (d) cos , cos , cos
2 2 2 2 2 2
(ii). Area of ΔDEF is :
rΔ
(a) (b)
4R

  
0
S. : Let the triangle be an equilateral then according to data of MT-1,
1 1 3
A=B=C=60 and a=b=c=1, r = , R= , area of triangle = Δ =
42 3 3
1
(i) The length of
rΔ rΔ 2rΔ
(c) (d)
2R
the sides of ΔDEF are DE EF FD
2
Now put the values of r,A,B a
R
d n
R
n C i
MT -1
  
      
  
      
 
  

 


2
the options then
A 1 60 1
only option (a)will gives 2rcos 2. .cos = ,
2 2 22 3
B 1 C 1
similarly 2rcos and 2rcos .
2 2 2 2
3 1 3
(ii) Area of ΔDEF=Δ = .
4 2 16
Now put the values of r,R, in the options then only option (a

3
)will gives .
16
(a) is right choice.
A
DΕ
B C
F
A
B C
EF
R r
D 1/2
1/2
The beauty of these short tricks is that many problems of this kind can be solved easily.

12.   
2
Q.In ΔABC,the rario of angles A,B,C is 1:2 :3,then which of the following is/(are)correct ?
(a) Circumradius of ΔABC = c. *(b) a : b : c= 1: 3 : 2.
3
(c) Perimeter of Δ ABC= 3+ 3. *(d) Area of ΔABC = c .
8
S. : A :B : C 1MT - 2
one or more currect answers
 
  
     
   
  

:2 :3 so let A 30º,B 60º,C 90º.
3
according to MT 2, a 1, b = 3 ,c 2, = , R 1
2
(a) Circumradius of ΔABC = c 1 3 R 1 and c 3 .( Not true)
(b) a : b : c= 1: 3 : 2.
Not necessorly always true as
(c) Perimeter of Δ ABC = 3+ 3
A 45º,B 60º,C 75ºa
(True)
 
  

 

2
Q.Two parallel chords are drawn onthesame sides of the center of a circle of radius R.
It is found that they subtands an angle
lso possible
3 3 3 3 3
(d) Area of ΔABC = c = .4 =
s of and 2 at the cente
8 2 8 2 2
(True)
one or more currect answers
          
                    
      
            
   
 
   
 
r of a circle,thenthe
perpendicular distance between the chords is equals to ?
3
(a) 2R.sin sin *(b) R 1 cos 1 2cos
2 2 2 2
3
*(c) 2R.sin sin (d) R 1 cos 1 2cos
4 4 2 2
 
  
    
 
 
 
        
     
      
   
    
  
    
S. COD and AOB 2 and required distance MN ON OM
Let 60 then 2 120
ON 3
cos30 ON R cos 30 R.
R 2
OM 1
Similarly cos 60 OM R
R 2
3 1 R
d MN R 3 1
2 2 2
put 60 in options,then
(a) 2R.sin 90 .sin30 2R 1
  
   
   
   
   
   
   
  
          
 
      
 
          
1
R d
2
3 2 3 R
(b) (1 cos 30 ) (1 2cos30 ).R 1 1 R ( 3 1) d
2 2 2
( 3 1) R( 3 1)1
(c) 2R. sin45 sin 15 2R d
22 2 2
[ 3 1]R3 3
(d) 1 cos30 1 2cos30 R 1 1 2 R d
2 2 2
(True)
(True)
M
N
d
R
O
BA
C D
The beauty of these short tricks is that many problems of this kind can be solved easily.

13. Q. In an acute angled triangle ABC, let AD,BE and CF be the perpendicular from A,B and
C uponthe opposite sides of the triangle.(All symbols used have usuaul meaning in a
triangle)
(i).The ratio of product of the
Questions based on Paragraph
    
          
1
3
side lengths of the triangles DEF and ABC, is equal to ?
3(a bc ) 1 A B C
(a) (b) (c) cos AcosB cos C (d) sin sin sin
4(a+b+c) 4 2 2 2
(ii). The circumradius of the triangle DEF can be equal to ?
abc a R r
(a) (b) (c) (d) cose
8Δ 4sinA 2 8
 

0
S. Let the Triangle is an equilateral triangle then according to MT-1.
3 1 1
A=B =C=60 ,a=b=c=1, Δ= ,r = , R = .
4 2 3 3
1
DEF will also be an equilateral triangle with side .
2
Product of the sides lengths of the DEF
(i).
A B C
c cosec cosec
2 2 2
Produc


  
0
1 1 1
. .
12 2 2 =
t of the sides lengths of the ABC 1.1.1 8
now putting the values of A=B=C=60 and a=b=c=1
1
in options then onlyoption (d) will give so it is the right choice.
8
1
(ii).The circumradius of the DEF =The inradius of the ABC
2 3
now pu 0
tting values of A=B=C=60 and a=b=c=1and r,R in options
1
then onlyoption (a),(b),(c)and(d) will give so all the opti
Q. In a triangle ABC, poi
ons are currect
nts D and E are taken
.
2
on side BC such
3
one or more currect answers
  
   
2
0
that BD = DE = EC.
If ADE = AED = , then:
6tanθ
(a) tan = 3tanB (b) 3tan = tanC (c) = tan A
S. Let ΔABC be an equilateral and let AB=BC=AC=1 and A=B=C=60 .
3 21
then AF= 3 2 and DF= ,tanθ =
(d) B = C
t
=3 3
6
a
1 6
(
n θ 9
a) t
 
0
  
   
 
 
   
0
0
0
2 2
0
an = 3tanB 3 3= 3tan60 3 3= 3 3
(b) 3tan = tanC 3.3 3= tan60 9 3 3
6×3 36tanθ 18 3
(c) = tanA =tan60 = 3
18tan θ 9 3 3 9
(d) B = C tan60 tan60
(True).
(True).
(True).
Hence (a) ,(c) and (d) are the correct answers.
A
B C
EF
R r
D 1/2
1/2
A
B C
EF
R r
D 1/2
1/2
A
B C
  EFD
1/3
1/6
The beauty of these short tricks is that many problems of this kind can be solved easily.