The document contains a series of mathematical expressions and problems involving trigonometric identities and simplifications. It presents multiple-choice questions with various options focused on evaluating trigonometric functions and relationships. The expressions demonstrate concepts such as identities and substitutions to derive values independent of the variables involved.

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 ) 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
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.