# Question-1

**Find the radian measures corresponding to the following degree measures:**

(i) 25

(i) 25

^{0}(ii)-47^{0}30^{â€™}(iii)240^{0}(iv)520^{0}**Solution:**

Radians = 180 degrees

1 degree = radians

(i) 25Â° = 25 x = Radians.

(ii) -47Â° 30â€™** = **-47Â° 30â€™ x = -0.2638Radians.

(iii) 240Â° = 240 x = Radians.

(iv) 520Â° = 520 x = Radians.

# Question-2

**Find the degree measures corresponding to the following radian****measures (use Ï€=).(i) (ii)-4 (iii) (iv)****Solution:**

i) Radians = 180Â° 1 Radian = degrees

= 39.375Âº

ii) 1 radian = -4radian

= Ã— -4 == -32.72 Ã— 7 = -229^{o}5'24''

iii) 1 radian =

= = 300^{0}

iv) 1 radian =

= 210^{o}

# Question-3

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

**Solution:**

One complete revolution = 2Ï€360 revolutions = 360 Ã— 2Ï€= 720Ï€ radians / minute = 720Ï€ /60 radians/sec = 12Ï€ 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 60**

^{o}and 75^{o}at the centre, find the ratio of their radii.**Solution:**

Let the radii be r

_{1}and r

_{2}.

Let the angles subtend by the arcs in two circles be Î¸

_{1}and Î¸

_{2}.

l = Î¸ r where Î¸ is the angle, l the length of an arc and r the radius of the circle.

Î¸

_{1}= 60

^{o}= 60 Ã— = radian

Î¸

_{2}= 75

^{o}= 75 Ã— = radian

r

_{1}= r

_{2}

r

_{1}/ r

_{2 }= =

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

(i) 10cm

(ii) 15cm

(iii) 21cm

**Solution:**

(i) Length of the pendulum (r) = 75cm

Length of an arc (l) = 10cm

Î¸ = l/r = 10/75radians = 2/15radians

(ii) Length of the pendulum (r) = 75cm

Length of an arc (l) = 15cm

Î¸ = l/r = 15/75 radians = 1/5radians

(iii) Length of the pendulum (r) = 75cm

Length of an arc (l) = 21cm

Î¸ = 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.

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.

(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**

**Î¸ = 13/5, Î¸ lies in fourth quadrant.**

**Solution:**

sin Î¸ , tan Î¸ , cosec Î¸ and cot Î¸ are negative in IV th quadrant sec Î¸ and cos Î¸ are positive in IVth quadrant.

sec Î¸ = 13/5

cos Î¸ = 1/sec Î¸ = 5/13

sin

^{2}Î¸ = 1-

âˆ´ sin Î¸ =

cosec Î¸ =

âˆ´ tan Î¸ = =

cot Î¸ = =

# 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 765**

^{o }**Solution:**

sin 765

^{o}= sin= sin= sin=

# Question-13

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

^{o})**Solution:**

cosec(-1410

^{o}) = -cosec 1410

^{o}= -cosec (8Ï€ â€“ 30

^{o}) = cosec 30

^{o}= 2

# Question-14

**Find the value of the following trigonometric function:**

**sin ()**

**Solution:**

Sin = -Sin= -Sin(660)= - Sin (2x360 â€“ 60) = -(-sin60) = sin60 =

# Question-15

2sin

L.H.S = 2sin

^{2 }+2cos^{2 }+2sec^{2 }= 10**Solution:**

L.H.S = 2sin

^{2 }+2cos

^{2 }+2sec

^{2 }= 2sin

^{2 }+2cos

^{2 }+2sec

^{2 }= 2sin

^{2}

^{ }+ 2cos

^{2 }+ 2sec

^{2 }= 2Ã— + 2Ã— + 2Ã— 4

^{ }= 1

^{ }+ 1+ 8

^{ }= 10

^{ }= R.H.S

# Question-16

**Find the value of:(i) sin 75**

^{0}(ii) tan15^{0 }**Solution:**

i)Sin (30+45)

= sin30cos45+cos30sin45 Sin(A+B)

=SinACosB+CosAsinB

=

=

(ii)tan15 = tan(45 â€“ 30)

=

=

=

# Question-17

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

**Solution:**

This is of the form CosACosB-SinASinB = Cos(A+B)âˆ´ L H S = Cos= Cos

= Sin(x+y)

Hence proved.

# Question-18

**= cot**

^{2}x**Solution:**

=

= -cos

^{2}x/-sin

^{2}x

= cot

^{2}x

# Question-19

**cos cos(2+x) = 1**

**Solution:**

= sinxcosx

=Sinx.cosx

=sinx.cosx

=sin

^{2}x+cos

^{2}x = 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 â€“ sinsinx-

= coscosx â€“ sinsinx -

= -2sinsinx

= -

# Question-22

**sin**

^{2 }6x â€“ sin^{2}4x = sin2x sin10x**Solution:**

L.H.S = sin

^{2 }6x â€“ sin

^{2}4x

[Using the formulae 2sin

^{2}x = 1 - cos2x]

=

= - (cos12x â€“ cos8x)

= - [cos(10x + 2x) â€“ cos(10x - 2x)]

[Using the formulae -2sinÎ¸ sinÏ† = cos(Î¸ + Ï† ) - cos(Î¸ - Ï† )]

= Ã— 2 sin10x sin2x

= sin10x sin2x

# Question-23

**cos**

^{2}2x â€“ cos^{2}6x = sin4x sin8x**Solution:**

L.H.S = cos

^{2}2x â€“ cos

^{2}6x

[Using the formulae 2cos

^{2}x = 1+ cos2x]

=

= - (cos12x â€“ cos4x)

= - [cos(8x + 4x) â€“ cos(8x - 4x)]

[Using the formulae -2sinÎ¸ sinÏ† = cos(Î¸ + Ï† ) - cos(Î¸ - Ï† )]

=Ã— 2 sin8x sin4x

= sin4x sin8x

# Question-24

**sin2x + 2sin4x + sin6x = 4cos**

^{2}x sin4x**Solution:**

L.H.S = sin2x + 2sin4x + sin6x

= 2sincos + 2sin4x

= 2sin4xcos2x + 2sin4x

= 2sin4x(cos2x + 1)

= 4sin4xcos

^{2}x

= 4cos

^{2}xsin4x

R.H.S

# Question-25

**cot4x(sin5x + sin3x) = cotx(sin5x â€“ 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 â€“ sin3x)

[Using the formula sin5x - sin3x = 2 cos sin ]

= 2cotx cos sin

= 2cotx cos sin

= 2cos4x cosx

âˆ´ 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 = cos

^{2}x â€“ sin

^{2}x and sinx â€“ siny = 2cos sin ]

=

= 2sinx

= R.H.S

# Question-31

**Prove that**

**Solution:**

=

= cot3x

= R.H.S

Hence proved

# Question-32

**cot x cot 2x â€“ cot 2x cot3x â€“ cot3x cotx =1**

**Solution:**

Cot2x(cotx â€“ 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 - 8sin**

^{2}xcos^{2}x.**Solution:**

= cos4x = 1-2sin

^{2}2x

= 1-2 (2sinx cosx)

^{2}

= 1-8 sin

^{2}x cos

^{2}x = R.H.S

Hence proved.

# Question-35

**Prove that cos6x = 32cos**

^{6}x- 48cos^{4}x + 18cos^{2}x-1**Solution:**

cos6x = 4 cos

^{3}2x - 3 cos2x

= 4(2cos

^{2}x-1)

^{3}- 3(2cos

^{2}x-1)

= 4[(2cos

^{2}x)

^{3}- 3Ã—(2cos

^{2}x)

^{2}Ã—1 + 3 Ã— 2 cos

^{2}xÃ—1 - 1

^{3}]-6cos

^{2}x + 3

= 32cos

^{6}x - 48cos

^{4}x + 18cos

^{2}x -1

= R.H.S

Hence proved