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Question-1

Simplify the following expressions:.

Solution:
==1 - sinθ cosθ.

Question-2

Prove .

Solution:

Multiply   tan θ/tan θ we get,
      
       =
cotθ + 1 + tanθ       = R.H.S.

Question-3

Prove tan2φ + cot2φ + 2 = sec2φ cosec2φ.

Solution:
L.H.S = tan2φ + cot2φ + 2
        = sec2
φ - 1 + cosec2φ - 1 + 2
[Using Identity 1 + cot2
φ = cosec2φ and 1 + tan2φ = sec2φ]
        = sec2
φ + cosec2φ         =
        =
        =      [Using Identity sin2
φ + cos2φ = 1]
        = sec2
φ cosec2φ         = R.H.S.

Question-4

 Evaluate: .

Solution:

                        
                        
= 1 + 1
                         = 2.

Question-5

Prove that: .

Solution:
L.H.S =
        =
        = 1 + 1
        = 2
        = R.H.S.

Question-6

Prove that: sec2θ - cot2 (90º - θ ) = cos2 (90º - θ ) + cos2θ.

Solution:

Question-7

Prove that: .

Solution:
L.H.S =
        =
        =                 [Using Identity sin2
θ + cos2θ = 1]
        = 1
        = R.H.S.

Question-8

Prove that: cos (81º + θ) = sin (9º - θ ).

Solution:
L.H.S = cos (81º + θ )
        = cos [90º - (9º -
θ )]
        = sin (9º -
θ )
        = R.H.S.

Question-9

Prove that: sinθ cos (90º-θ) + cos θ sin (90º - θ) = 1.

Solution:
L.H.S = sinθ cos (90º-θ ) + cos θ sin (90º - θ )
        = sin
θ sin θ + cos θ cos θ         = sin2θ + cos2 θ         [Using Identity sin2φ + cos2φ = 1]
        = 1
        = R.H.S.

Question-10

If = y, then is also y.

Solution:
=×
                  =
                  =
                  =
                  =
                  = y         (Since = y).

Question-11

Prove: = tan2 φ.

Solution:
L.H.S =.
            =
            =
            =
            = tan2
φ
            = RHS.

Question-12

Prove: (sec θ - tan θ)2 = .

Solution:
LHS = (sec θ - tan θ)2
      =
      =
      =
      =
      = RHS.

Question-13

Prove that = sin2 A cos2 A.

Solution:
L.H.S =
        =

          =
          =
          
= sin2 A cos2 A
        = R.H.S.

Question-14

If a cosθ - b sinθ = c, show that a sinθ + b cosθ = ± .

Solution:
(a cosθ - b sinθ )2 + (a sinθ + b cosθ )2
              = a2(cos2
θ + sin2θ ) + b2(sin2θ + cos2θ ) – 2ab sinθ cosθ + 2ab sinθ cosθ
              = a2 + b2 (a cosθ - b sinθ )2 + (a sinθ + b cosθ )2 = a2 + b2
(a cosθ - b sinθ )2 = a2 + b2 - (a sinθ + b cosθ )2
                             = a2 + b2 - c2
a sinθ + bcosθ = ± .

Question-15

Prove that 2(sin6θ + cos6θ ) - 3(sin4θ + cos4θ ) + 1 = 0.

Solution:
2(sin6θ +cos6θ ) – 3(sin4θ + cos4θ ) + 1
    = 2[(sin2θ )3 + (cos2θ )3] – 3[(sin2θ )2 + (cos2θ )2] + 1
    = 2[(sin2θ + cos2θ )3 – 3sin2θ cos2θ (sin2θ + cos2θ )] – 3[(sin2θ + cos2θ )2 – 2sin2θ cos2θ ] + 1
The algebraic identity
a3 + b3 = (a + b)3 – 3ab(a + b) and
a2 + b2 = (a + b)2 – 2ab
are used in the above step where
a = sin2θ and b = cos2θ.
Writing sin2θ + cos2θ = 1, we have
= 2[ 1 – 3 sin2θ cos2θ ] – 3[– 2 sin2θ cos2θ ] + 1
= 2 – 6 sin2θ cos2θ - 3 + 6 sin2θ cos2θ + 1
= - 3 + 3 = 0

Question-16

Evaluate cos(40o + θ ) - sin(50o - θ ) +  .

Solution:
Cos (40o + θ ) – sin (50o - θ ) +
    = cos [(90 - 50
° ) + θ ] – sin(50° - θ ) +
    = sin (50
° - θ ) – sin(50° - θ ) +
    =
    = 1.

Question-17

If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, prove that x2 + y2 = 1.

Solution:
x sin3 θ + y cos3 θ = sin θ cos θ  
x sin
θ (sin2θ ) + (y cos θ ) cos2θ = sin θ cos θ ⇒ x sin θ (sin2θ) + (x sin θ ) cos2θ = sin θ cos θ [ since x sin θ = y cos θ ] x sin θ (sin2θ + cos2θ ) = sin θ cos θ ⇒ x sin θ = sin θ cos θ ⇒ x = cos θ ... x sin θ = y cos θ y = sin θ Hence x2 + y2 = cos2 θ + sin2θ = 1.

Question-18

Evaluate, sec210o – cot2 80o +.

Solution:
L.H.S. = sec210o – cot2 80o +
         = sec2 (90
° - 80° ) – cot280° +
         = cosec280
° - cot280° +
         = - +
         = + 1
         = + 1
         = 1 + 1 = 2 = R.H.S.

Question-19

Prove the following identity: = 0.

Solution:
LHS =  
       =
       =
  
      == 0.

Question-20

Prove: =.

Solution:
L.H.S. =
           =

         =
         =
× + ×
           = +
         =
         =
         = R.H.S.
  = .

Question-21

Solve the following equation for 0° < θ 90°: 3 tan θ + cot θ = 5 cosec θ.

Solution:
3 tan θ + cot θ = 5 cosec θ ⇒ 3 tan θ  + = 5 cosec θ   3tan2 θ  + 1 = 5 cosec θ tan θ ⇒ 3tan2 θ + 1 = 5 sec θ ⇒ 3(sec2 θ - 1) + 1 = 5 sec θ ⇒ 3sec2 θ - 5sec θ - 2 = 0 sec θ =  =
sec
θ  = 2 or
sec
θ  (as -1 < cos θ  <1 ) θ = 60° for 0° < θ 90°.

Question-22

Solve for θ : = 1, (θ 0).

Solution:
= 1
= 1
= 1
-sin
2 θ - 3cos θ + 3 = sin2 θ -2sin2 θ - 3cos θ = - 3
-2(cos2
θ - 1) - 3cos θ = - 3
-2cos2
θ + 2 - 3cos θ = - 3
-2cos2
θ + 5 - 3cos θ = 0
2cos2
θ + 3cos θ - 5 = 0
cos
θ = === or 1 if cos θ = 1, ∴ θ = 0°.

Question-23

Using the formula tan 2θ = , obtain the value of tan 15°.

Solution:
tan 2θ =    
Putting  
θ= 15°,
tan 30
° =


(1 - tan215
°) = 3(2tan 15°)
tan2 15
° +23 tan 15° - 1 = 0
tan15
° =   =  = .

Question-24

If a cos θ – b sin θ = x and a sinθ + b cosθ = y, prove that a² + b² = x² + y².

Solution:
a cos θ – b sin θ = x and a sinθ + b cosθ = y
R.H.S.  = x² + y²
          = (a cos
θ – b sin θ)2 + (a sin θ + b cos θ)2
          = a2cos2
θ – 2ab cos θ sin θ + b2sin2 θ + a2sin2 θ + 2absin θ cos θ + b2cos2 θ           = (a2 + b2) cos2 θ + (b2 + a2)sin2 θ           = (a2 + b2) cos2 θ + (a2 + b2)sin2 θ           = (a2 + b2) (cos2 θ + sin2 θ )
          = (a2 + b2)              [
 cos2 θ + sin2 θ = 1]
          = L.H.S.
a² + b² = x² + y².

Question-25

If x = p sec θ + q tan θ and y = p tan θ + q sec θ, prove that x² - y² = p² - q².

Solution:
x = p sec θ + q tan θ and y = p tan θ + q sec θ 
L.H.S. = x² - y²
           = (p sec
θ + q tan θ)2 – (p tan θ + q sec θ)2
         = p2 sec2
θ + 2pq sec θ tan θ + q2 tan2 θ - (p2 tan2 θ + 2pq tan θ sec θ + q2 sec2 θ )
         = p2sec
2 θ + 2pq sec θ tan θ + q2 tan2 θ - p2 tan2 θ - 2pq tan θ sec θ - q2 sec2 θ          = (p2 - q2) sec2 θ + (q2 - p2) tan2 θ
         = (p2 - q2) sec2 θ - (p2 - q2) tan2 θ          = (p2 - q2) (sec2 θ - tan2 θ )
         = (p2 – q2) [Since 1 + tan
2 θ = sec2 θ ]
         = R.H.S.
x² - y² = p2 – q2.

Question-26

Show that (1 + cot θ - cosec θ )(1 + tan θ + sec θ ) = 2.

Solution:
L.H.S = (1 + cot θ - cosec θ )(1 + tan θ + sec θ )
        = (1 + - )(1 + + )
        = ()()
        =
        =
        =
        =
        = 2
        = R.H.S.

Question-27

If cosec θ - sin θ = a, sec θ - cos θ = b, prove that a2b2 (a2 + b2 + 3) = 1.

Solution:
L.H.S = a2b2 (a2 + b2 + 3)
        = (cosec
θ - sin θ )2(sec θ - cos θ )2 [(cosec θ - sin θ )2 + (sec θ - cos θ )2 + 3]
        = (cosec
θ - sin θ )2(sec θ - cos θ )2 [cosec2 θ + sin2 θ - 2 cosecθ sinθ + sec2 θ + cos2 θ - 2 secθ cos θ + 3]
        = (cosec
θ - sin θ )2(sec θ - cos θ )2 [cosec2 θ + 1 - 2 + sec2 θ - 2 + 3]
        = (cosec
θ - sin θ )2(sec θ - cos θ )2 (cosec2 θ + sec2 θ )
       

Question-28

Prove that (sin θ + sec θ)2 + (cos θ + cosec θ)2 = (1 + sec θ cosec θ)2.

Solution:
L.H.S = (sin θ + sec θ )2 + (cos θ + cosec θ )2
         
= (sin
θ +)2 + (cos θ + )2
         
= sin2
θ ++ 2 sin θ + cos2 θ + + 2 cos θ
          = 1 + (+) + 2 (sin
θ + cos θ )
          = 1 + + 2
          = 1 + + 2
          = 1 + sec2
θ cosec2 θ + 2 sec θ cosec θ
          = (1 + sec θ cosec θ )2
         
= R.H.S.

Question-29

Find the value of tan2 30o + sin2 60o – 3 cos2 60o tan2 60o – 2 tan2 45o.

Solution:
tan2 30o + sin2 60o – 3 cos2 60o + tan2 60o – 2 tan2 45o
                            =
× + - 3 + - 2
                            =
× + - +× 3 - 2
                            = +
× 3 - 2
                            = +- 2
                            =
                            =.

Question-30

Find the value of 4(sin430o + cos460o) – 3(cos245o + sin290o).

Solution:
4(sin430o + cos460o) – 3(cos245o + sin290o)
                        = 4
                        = 4
                        =
                        =
                        =
                        = - 4.

Question-31

Prove that  =  .

Solution:

=
=
=.

Question-32

Find the value of tan230o + sin260o – 3 cos260o +tan260o – 2 tan245o.

Solution:

Question-33

Determine whether the following equation is identity: =.

Solution:
=
L.H.S   =
            =
            =
            =
            =
            =    

            = 
(Dividing by cosθ) 
            =
          = R.H.S

LHS = RHS.
Therefore, it is an identity.




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