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 tanA 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 2S. 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

7Q.

s 3 .

 If X sin +sin s12 12

0 0 0 0

3 7 3in ,Y cos +cos cos

12 12 12 12

X Ythen

Y X

(a)  2sin2               (b)  2cos2                 * (c) 2tan2               (d) 2cot

3 3S. Let 15 then X sin120 +sin0 sin60 = 0

2

2

0 0 0

0

32

1 1 3 1 2Y 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 kcos 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 kcos cos cos

cos0 cos360 cos360 kcos0 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 1Q. If tanA 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 2and LHS pcosec15 psec15 p 2 2 p

3

p q

1 3 1

Magical Method of Substitution

2 2 2 2

0

2 2

8abQ. If a sinx sin y, b cos x cos y, c tanx 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 tan2 2

1 1 1 1a               b           c               d    

3 3 3 3

therefore a 2, b 2, c 2

8 2 2then 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 12

then

1ta

cos cos cos ............

n tan 30 tan 30

...............cos

(a) 2 * (b) 2 (c)

2 2 3

0n1 2 3

n/4 n/

0

4

0 0

:S.

Let ............. 45 (Satisfying both given conditions)

Required value cos45 cos4

2 (d

5 cos45 ............n terms

1 1 1.....................

2 2 2

) 2

Magical Method of Substitution and Balancing

k k4 6

n/2n/2n

k k 4 4 6 6k

k

4 6

1Q. Let f (x)= (sin x cos x),x R and k 1,then f (x) f (x)

k1 1 1 1

(a) * (b) (c) (

1 1.....n tertms 2

2( 2)

1 1 1S. 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 64 6

ndependent of x soput x 0 in above expression

1 1 1 1f (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.

0 0 00

2

:

sin5 sin2 sinQ. The value of the expression is equals to :

cos5 2cos3 2cos cos

* (A) tan (B) cos (C) c

sin150 sin60 sin30Let 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

22 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

11) (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

22 2

2 2 2 2

the value of theabove expressions should be 1.

put 45 ( sin45 cos45 and also tan45 cot45 ).

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 1sin cos cos 1

4 4 4 41

(D)sin cos 2sin 0 2 1(not equals to unity).2

S.

The value o

4Q. sin sin sin is equals to :

sin3 3 3

4 3 3(A) (B) * (C) (D) no

f

ne3 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 3then 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.

cot 1S. 3cot cot cot3

cot 1 tanQ. 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 1tan 2

cos( 2 ) 13 cos( 2 ) cos

lue of cot( ) 25tan is :

* (A) 0 (B) 1 (C) 3 (D) 4

Q. I

25 2cos( ).cosS. 25

cos 12 cos( 2 ) cos 1 2sin( ).sin( )

cot( ).cot 25 cot( ) 25tan cot ( ) 25t

f x y 3 cos4 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 0b(cos1 sin1 )(cos2 sin2 )(cos3 sin3 )........

.....(ii), by (i) and (ii) x y 1

Now put x y 1 in the options the

......(cos45 sin45 )

cos1

n optio

cos2 cos3 cos4 .......

n (D) is sat

.....cos44 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 ) (cos44 sin44 ) (cos44 sin44 )S. ..........

cos1 cos2 cos3 cos44 cos4

(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 we find 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.