Magical Short Tricks for JEE(Main).

Infinite many problems of this kind can be solved easily by these techniques.
for more short tricks mail me your mail ID at [email protected]

1. Q. The value of (1 cot cosec )(1 tan sec ) is equ als 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
nθ, so we can put any suitable value of θ to minimise the calculations.
here letθ 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

     






22
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
          
           
    




   

00
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
XY
then
YX
(a)  2sin2               (b)  2cos2                 *(c) 2tan2               (d) 2cot
33
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 33
Now put 15 in the options then option (c) will gives 2/ 3.
Q. If cos cos cos 0 and if cos3 cos3 cos3 kcos cos c os then k
(a) 1 (b) 8 *(c) 12 (d)

         

               
 
00
0 0 0 0 0 0
S. cos cos cos 0 so let 0, 120 & 120 satisfying give n condition
cos3 cos3 cos3 kcos cos cos
cos0 cos360 cos360 kcos0 cos120 cos120
1 1 1 k.1.( 1/2).( 1/2) k 1
9
2

           
         
   
       
 Thebeauty of these short tricksis that many problemsof thiskind canbe solved easily.

2. :S. 
Let = 15º then LHS = tan15º+2tan30º+4 tan60º + 8cot1
Q.  The valu
20º
11
=2 3+2 × +4 3 +8
33
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
p1
Q. If tan A and if =6 is acute angle then (pcosec2 q sec2 )
q2
(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
pq
1 3 1




       


  
    

Magical Method of Substitution

2 2 2 2
0
22
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
00
Q. If cos 2cos then tan tan
22
1 1 1 1
a               b           c               d    
33 33
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
   
 nn1 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
00
:S.
Let ............. 45 (Satisfying both given co nditions)
Required value cos 45 cos 4
2 (d
5 cos 45 ............n terms
111
.....................
222
)2


       
  
  
Magical Method of Substitution and Balancing
kk
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) (
11
.....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

         Thebeauty of these short tricksis that many problemsof thiskind canbe 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
00
2 2 2
:
Q. If (cos cos
1
150 2c
) (sin
os90 2cos 30 cos30 3
Now
sin ) ksin
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
22
4 4 2
2
22
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 )
22
(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
22
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
33
2
3
4 4 2
12
1111
sin cos cos 1
4444
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
  

    
Thebeauty of these short tricksis that many problemsof thiskind canbe 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 3
11
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 sin4 4 ) (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 ?
Thebeauty of these short tricksis that many problemsof thiskind canbe solved easily.