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Question-1

Find the radian measures corresponding to the following degree measures:
(i) 250 (ii)-47030 (iii)2400 (iv)5200

Solution:
Radians = 180 degrees
1 degree = radians

(i) 25° = 25 x = Radians.

(ii) -47° 30’ = -47° 30’ x = -0.2638Radians.

(iii) 240° = 240 x = Radians.

(iv) 520° = 520 x = Radians.

Question-2

Find the degree measures corresponding to the following radian measures (use π=).(i) (ii)-4 (iii) (iv)

Solution:
i) Radians = 180° 1 Radian = degrees

= 39.375
º

ii) 1 radian = -4radian
=
× -4 == -32.72 × 7 = -229o5'24''

iii) 1 radian =

= = 3000

iv) 1 radian =

= 210o

Question-3

A wheel makes 360 revolutions in one minute. Through how many radians does it turn is one second?

Solution:
One complete revolution = 2π360 revolutions = 360 × 2π= 720π radians / minute = 720π /60 radians/sec = 12π radians/sec.

Question-4

In a circle of diameter 40cm. The length of a chord is 20cm. Find the length of minor arc corresponding to the chord.

Solution:


.

 

Question-5

If, in two circles, arcs of the same length subtend angles of 60o and 75o at the centre, find the ratio of their radii.

Solution:
Let the radii be r1 and r2.
Let the angles subtend by the arcs in two circles be
θ1 and θ2.
l =
θ r where θ is the angle, l the length of an arc and r the radius of the circle.
θ1 = 60o = 60 × = radian
θ2 = 75o = 75 × = radian
r1 = r2
r1 / r2 = =
Therefore the required ratio is 5:4.

Question-6

Find the angle in radian through which a pendulum swings if its length is 75cm and the tip described an arcs of length
(i) 10cm
(ii) 15cm
(iii) 21cm

Solution:
(i) Length of the pendulum (r) = 75cm
Length of an arc (l) = 10cm

θ = l/r = 10/75radians = 2/15radians

(ii) Length of the pendulum (r) = 75cm
Length of an arc (l) = 15cm

θ = l/r = 15/75 radians = 1/5radians

(iii) Length of the pendulum (r) = 75cm
Length of an arc (l) = 21cm

θ = l/r = 21/75 radians = 7/25radians

Question-7

Find the values of other five trigonometric functions
1. Cos x = , x lies in third quadrant.

Solution:
sin x = , cosec x = -, tanx = , cotx = secx = -2

Question-8

Find the values of other five trigonometric functions
(i) Sin x = , x lies in second quadrant.

Solution:
cos x = - , secx = -, tanx = -, cotx = - ,Cosecx =

Question-9

Cot x = , x lies in third quadrant.

Solution:
Tanx = , sinx = -, Cosec x = -, cosx = - , secx = -

Question-10

Find the values of the other five trigonometric functions in the following problem: sec θ = 13/5, θ lies in fourth quadrant.

Solution:
sin θ , tan θ , cosec θ and cot θ are negative in IV th quadrant sec θ and cos θ are positive in IVth quadrant.
sec
θ = 13/5
cos
θ = 1/sec θ = 5/13
sin2 θ = 1-
sin θ =
cosec
θ =    
tan θ = =  
cot
θ = =

Question-11

Tan x = , x lies in second quadrant.

Solution:
cotx = -, sinx = , cosecx = , cosx = -, secx = -

Question-12

Find the values of the trigonometric functions sin 765o

Solution:
sin 765o = sin= sin= sin=

Question-13

Find the value of the following trigonometric function: cosec(-1410o)

Solution:
cosec(-1410o) = -cosec 1410o = -cosec (8π – 30o) = cosec 30o = 2

Question-14

Find the value of the following trigonometric function:sin ()

Solution:
Sin = -Sin= -Sin(660)= - Sin (2x360 – 60) = -(-sin60) = sin60 =

Question-15

2sin2 +2cos2 +2sec2 = 10

Solution:
L.H.S = 2sin2 +2cos2 +2sec2
        
= 2sin2 +2cos2 +2sec2
        
= 2sin2 + 2cos2 + 2sec2
        
= 2
× + 2× + 2× 4
         = 1 + 1+ 8
         = 10
         = R.H.S

Question-16

Find the value of:(i) sin 750 (ii) tan150

Solution:
i)Sin (30+45) 
= sin30cos45+cos30sin45 Sin(A+B)
=SinACosB+CosAsinB
=
=

(ii)tan15 = tan(45 – 30)
             =
             =
             =

Question-17

Prove the following:Cos cos - sinSin = sin(x+y)

Solution:
This is of the form CosACosB-SinASinB = Cos(A+B) L H S = Cos= Cos
            = Sin(x+y)
Hence proved.

Question-18

= cot2x

Solution:
=
= -cos2x/-sin2x
= cot2x

Question-19

cos cos(2+x) = 1

Solution:
= sinxcosx
=Sinx.cosx
=sinx.cosx
=sin2x+cos2x = 1
 (Hence Proved)

Question-20

sin (n + 1)x sin (n + 2)x + cos(n + 1)xcos(n + 2)x = cosx.

Solution:
L.H.S = sin (n + 1)x sin (n + 2)x + cos(n + 1)xcos(n + 2)x
         = cos[(n + 1)x - (n + 2)x]
         = cos (-x)
         = cosx
         = RH.S

Question-21

cos = - sinx

Solution:
coscosx – sinsinx-

= coscosx – sinsinx -
= -2sinsinx
= -

Question-22

sin2 6x – sin2 4x = sin2x sin10x

Solution:
L.H.S = sin2 6x – sin2 4x
[Using the formulae 2sin2x = 1 - cos2x]
        =
        = - (cos12x – cos8x)
        = - [cos(10x + 2x) – cos(10x - 2x)]
[Using the formulae -2sin
θ sinφ = cos(θ + φ ) - cos(θ - φ )]
        =
× 2 sin10x sin2x
        = sin10x sin2x

Question-23

cos22x – cos26x = sin4x sin8x

Solution:
L.H.S = cos22x – cos26x
[Using the formulae 2cos2x = 1+ cos2x]
         =
         = - (cos12x – cos4x)
         = - [cos(8x + 4x) – cos(8x - 4x)]
[Using the formulae -2sin
θ sinφ = cos(θ + φ ) - cos(θ - φ )]
         =
× 2 sin8x sin4x
         = sin4x sin8x

Question-24

sin2x + 2sin4x + sin6x = 4cos2x sin4x

Solution:
L.H.S = sin2x + 2sin4x + sin6x
         = 2sincos + 2sin4x
         = 2sin4xcos2x + 2sin4x
         = 2sin4x(cos2x + 1)
         = 4sin4xcos2x
         = 4cos2xsin4x
R.H.S

Question-25

cot4x(sin5x + sin3x) = cotx(sin5x – sin3x)

Solution:
L.H.S = cot4x(sin5x + sin3x)
[Using the formula sin5x + sin3x = 2 sin cos ]
        = 2cot4x sin cos
        = 2cot4x sin cos
        = 2cot4x sin4x cosx
        = 2cos4x cosx
R.H.S = cotx(sin5x – sin3x)
[Using the formula sin5x - sin3x = 2 cos sin ]
        = 2cotx cos sin
        = 2cotx cos sin
        = 2cos4x cosx

L.H.S = R.H.S

Question-26

Prove that =

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

Question-27

Prove that = tan4x

Solution:
L.H.S =
        =
        =
        =
        = tan4x
        = R.H.S

Question-28

Prove that = tan

Solution:
L.H.S =
        =
        =
        = tan
        = R.H.S

Question-29


Solution:
=
=
= tan2x.

Question-30

Prove that =2sinx

Solution:
L.H.S =
       =
       =
[Using the identity cos2x = cos2x – sin2x and sinx – siny = 2cos sin ]
       =
       = 2sinx
       = R.H.S

Question-31

Prove that  

Solution:
  =

  

  =   cot3x

  =   R.H.S

Hence proved

Question-32

cot x cot 2x – cot 2x cot3x – cot3x cotx =1

Solution:
Cot2x(cotx – cot 3x)- cot 3x.cotx

= cot 2x - cot3x.cotx

= cot2x

= cot 3x.cotx

=

=

=

=

=

=

=

=

=

=

=

= =
= 1.

Question-33

tan4x =

Solution:
L.H.S

= tan4x

= tan(3x+x)

=

=

=

=

=

L.H.S = R.H.S

Question-34

Prove that cos4x   =  1 - 8sin2xcos2x.

Solution:
= cos4x = 1-2sin22x
= 1-2 (2sinx cosx)2  
= 1-8 sin2x cos2x = R.H.S
Hence proved.

Question-35

Prove that cos6x = 32cos6 x- 48cos4x + 18cos2x-1

Solution:
cos6x  = 4 cos32x - 3 cos2x

          = 4(2cos2x-1)3 - 3(2cos2x-1)

          = 4[(2cos2x)3 - 3×(2cos2x)2×1 + 3 × 2 cos2x×1 - 13]-6cos2x + 3

          = 32cos6x - 48cos4x + 18cos2x -1

          = R.H.S

Hence proved




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