Conceptual Short Tricks for JEE(Main and Advanced)

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.

: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

       
  
  
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

        

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
)


    


 
     
   
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
  
 
    

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 ?

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

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

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

            
 
   
  




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.

       
   
      
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
 
 
    
    
  

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
 

 
  
  
          

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

   
    

   
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

       
   
  

    
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

  
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)
          
                    
      
            
   
 
   
 
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

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
  
   
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

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