Trigonometry Exercise 13.2 – Class X

Trigonometry Exercise 13.2 – Questions:

I. Answer the following questions:

  1. What trigonometric ratios of angles from 0 to 90 are equal to 0?
  2. Which trigonometric ratios of angles from 0 to 90 are equal to 1?
  3. Which trigonometric ratios of angles from 0 to 90 are equal to 1/2?
  4. Which trigonometric ratios of angles from 0 to 90 are not defined?
  5. Which trigonometric ratios of angles from 0 to 90 are equal?

II. Find θ. if 0≤θ≤90

  1. √2 cos θ = 1
  2. √3 tan θ = 1
  3. 2 sin θ = √3
  4. 5 sin θ = 0
  5. 3 tan θ = √3

III. Find the value of the following.

  1. sin 30˚ cos 60˚ – tan245˚
  2. sin 60˚ cos 30˚ + cos 60˚ sin 30˚
  3. cos 60˚ cos 30˚ – sin 60˚ sin 30˚
  4. 2sin230˚ – 3 cos2 30˚ + tan 60˚ + 3 sin2 90˚
  5. 4 sin2 60˚ + 3 tan2 30˚ – 8 sin 45˚ .cos 45˚
  6. cos 45˚/sec 30˚ + cosec 30˚
  7. (4sin260 – cos245)/(tan230 + sin20)
  8. (sin30˚ + tan45˚ – cosec60˚)/(sec30˚ + cos 60˚ +cot45˚)
  9. (5cos260˚ + 4 sec230˚ – tan245˚)/(sin230˚ + cos230˚)
  10. (5sin230˚ + cos245˚ – 4 tan230˚)/(2sin30˚ + cos30˚ + tan45˚)

IV. Prove the following equalities.

  1. sin30˚ .cos60 ˚+ cos30˚.sin60˚ = sin 90˚
  2. 2cos230˚ – 1 = 1 – 2sin230˚ = cos 60˚
  3. if θ = 30˚, prove that 4cos2θ – 3 cos θ = cos 3θ
  4. If π = 180˚ and A = π/6 prove that (1 – cosA)(1 + cos A)/(1 – sin A)(1+ sinA) = 1/3
  5. If B = 15˚, prove that 4 sin 2B.cos 4B.sin 6B = 1
  6. If A = 60˚ and B = 30˚ then prove that tan(A – B) = tanA – tanB/1 + tanAtanB

Trigonometry Exercise 13.2 – Solutions:

I. Answer the following questions:

  1. What trigonometric ratios of angles from 0 to 90 are equal to 0?

Solution:

sinθ = 0 when θ = 0˚

cosθ = 0 when θ = 90˚

tanθ = 0 when θ = 0˚

cotθ = 0 when θ = 90˚

 

  1. Which trigonometric ratios of angles from 0 to 90 are equal to 1?

Solution:

sinθ = 1 when θ = 90˚

cosθ = 1 when θ = 0˚

tanθ = 1 when θ = 45˚

cosec θ  = 1  when  θ = 90 ˚

sec θ = 1 when θ = 0 ˚

 

  1. Which trigonometric ratios of angles from 0 to 90 are equal to 1/2?

Solution:

sinθ = 1/2 when θ = 30˚

cosθ = 1/2 when θ = 60˚

 

  1. Which trigonometric ratios of angles from 0 to 90 are not defined?

Solution:

tanθ = undefined when θ = 90˚

cosecθ = undefined when θ = 0˚

secθ = undefined when θ = 90˚

cotθ = undefined when θ = 0˚


  1. Find θ. if 0≤θ≤90
  2. √2 cos θ = 1

Solution:

√2 cos θ = 1

cos θ = 1/√2

We know, cos 45˚ = 1/√2

Therefore, cos θ = cos 45˚

Thus, θ = 45˚

 

  1. √3 tan θ = 1

Solution:

√3 tan θ = 1

tan θ = 1/√3

We know, tan 30˚ = 1/√3

Therefore, cos 30˚ = cos θ

Thus, θ = 30˚

 

  1. 2 sin θ = √3

Solution:

2 sin θ = √3

sin θ = √3/2

We know, sin 60˚ = √3/2

Therefore, sin 60˚ = sin θ

Thus, θ = 60˚

 

  1. 5 sin θ = 0

Solution:

5 sin θ = 0

sin θ = 0/5

sin θ = 0

We know, sin 0˚ = 0

Therefore, sin 0˚ = sin θ

Thus, θ = 0˚

 

  1. 3 tan θ = √3

Solution:

3 tan θ = √3

tan θ = √3/3

i.e., tan θ = 1/√3

We know, tan 30˚ = 1/√3

Therefore, tan 30˚ = tan θ

Thus, θ = 30˚


III. Find the value of the following.

  1. sin 30˚ cos 60˚ – tan245˚

Solution:

sin 30˚ cos 60˚ – tan245˚

= 1/2 x 1/2 – (1)2

= 1/4 – 1

= -3/4

 

  1. sin 60˚ cos 30˚ + cos 60˚ sin 30˚

Solution:

sin 60˚ cos 30˚ + cos 60˚ sin 30˚

= √3/2 x √3/2 + 1/2 x 1/2

= 3/4 + 1/4

= 3+1/4

= 4/4

= 1

 

  1. cos 60˚ cos 30˚ – sin 60˚ sin 30˚

Solution:

cos 60˚ cos 30˚ – sin 60˚ sin 30˚

= 1/2 x √3/2  – √3/2 x 1/2

= √3/4√3/2

= 0

 

  1. 2sin230˚ – 3 cos2 30˚ + tan 60˚ + 3 sin2 90˚

Solution:

2sin230˚ – 3 cos2 30˚ + tan 60˚ + 3 sin2 90˚

= 2(1/2)2 – 3(√3/2 )2 + (√3) + 3(1)2

= 2(1/4) – 3(3/4) + √3 + 3

= 1/29/4 + √3 + 3

= 5/4 + √3

 

  1. 4 sin2 60˚ + 3 tan2 30˚ – 8 sin 45˚ .cos 45˚

Solution:

4 sin2 60˚ + 3 tan2 30˚ – 8 sin 45˚ .cos 45˚

= 4(√3/2)2 + 3(1/√3)2 – 8(1/√2) .( 1/√2)

= 4(3/4) + 3(1/3) – 8(1/2)

= 3 + 1 – 4

= 0

 

  1. cos 45˚/(sec 30˚ + cosec 30˚)

Solution:

cos 45˚/(sec 30˚ + cosec 30˚)

= (1/√2)/( 2/√3+ 2)

= (1/√2)/(2+2√3/√3)

√3/2√2(1+√3)

 

7.

Trigonometry Exercise 13.2

 

  1. (sin30˚ + tan45˚ – cosec60˚)/(sec30˚ + cos 60˚ +cot45˚)

Solution:

(sin30˚ + tan45˚ – cosec60˚)/(sec30˚ + cos 60˚ +cot45˚)

=(1/2  +1 –  2/√3)/(2/√3 + 1/2 + 1)

= (9  – 4√3/6)/(9 + 4√3/6)

9  – 4√3/9  – 4√3

 

  1. (5cos260˚ + 4 sec230˚ – tan245˚)/(sin230˚ + cos230˚)

Solution:

(5cos260˚ + 4 sec230˚ – tan245˚)/(sin230˚ + cos230˚)

= [5(1/2)2 + 4(2/√3)2 – (1)2]/[(1/2)2 + (√3/2)2]

= [5/4 + 16/3 – 1]/[1/4 + 3/4]

= (67/12)/(1)

= 67/12

 

  1. (5sin230˚ + cos245˚ – 4 tan230˚)/(2sin30˚ + cos30˚ + tan45˚)

Solution:

(5sin230˚ + cos245˚ – 4 tan230˚)/(2sin30˚ + cos30˚ + tan45˚)

= [5(1/2)2 + (1/√2)2 – 4(1/√3)2]/[2(1/2) + (√3/2) + 1]

= [5/4 + 1/24/3]/[1 + √3/2 + 1]

= [5/12]/[4+√3/2]

= 5/6(4+√3)

 


IV.Prove the following equalities.

  1. sin30˚ .cos60 ˚+ cos30˚.sin60˚ = sin 90˚

Solution:

sin30˚ .cos60 ˚+ cos30˚.sin60˚ = sin 90˚

LHS = sin30˚ .cos60 ˚+ cos30˚.sin60˚

= 1/2 . 1/2 + √3/2.√3/2

= 1/4 + 3/4

= 1+3/4

= 1

RHS = sin 90˚ = 1

Therefore, LHS = RHS

 

  1. 2cos230˚ – 1 = 1 – 2sin230˚ = cos 60˚

Solution:

2cos230˚ – 1 = 1 – 2sin230˚ = cos 60˚

Take, 2cos230˚ – 1 ………………..(1)

= 2(√3/2)2 – 1

= 2(3/4) – 1

= 3/2 – 1

= 3 – 2/2 = 1/2

1 – 2 sin2 30˚ ………………(2)

= 1 – 2(1/2)2

= 1 – 2(1/4)

= 1 – 1/2

= 1/2

cos60˚ …………………(3)

we know, cos60˚ = 1/2

Therefore, (1) = (2) = (3)

Thus, 2cos230˚ – 1 = 1 – 2sin230˚ = cos 60˚

 

  1. If θ = 30˚, prove that 4cos3θ – 3 cos θ = cos 3θ

Solution:

If θ = 30˚, we have to prove that 4cos3θ – 3 cos θ = cos 3θ

LHS = 4cos3θ – 3 cos θ

= 4cos330˚ – 3 cos 30˚

= 4(√3/2)3 – 3(√3/2)

= 4(3√3/8) – 3(√3/2)

= 3√3/23√3/2

= 3√3 – 3√3/2

= 0

RHS = cos 3θ

= cos 3(30˚)

= cos(90˚)

= 0

LHS = RHS

 

  1. If π = 180˚ and A = π/6 prove that (1 – cosA)(1 + cos A)/(1 – sin A)(1+ sinA) = 1/3

Solution:

If π = 180˚ and A = π/6 = 30˚ we have to prove that (1 – cosA)(1 + cos A)/(1 – sin A)(1+ sinA) = 1/3

LHS = (1 – cosA)(1 + cos A)/(1 – sin A)(1+ sinA)

= (1 – cos 30˚)(1 + cos 30˚)/(1 – sin 30˚)(1+ sin30˚)

= [1 – cos230˚]/[1 – sin230˚]

= [1 – (√3/2)2]/[1 – (1/2)2]

= [1 – 3/4]/[1 – 1/4]

= [4 – 3/4]/[4-1/4]

=[ 1/4]/[3/4]

= 1/3

= RHS

 

  1. If B = 15˚, prove that 4 sin 2B.cos 4B.sin 6B = 1

Solution:

If B = 15˚, we have to prove that 4sin 2B.cos 4B.sin 6B = 1

LHS = 4sin 2B.cos 4B.sin 6B

= 4sin 2(15˚).cos 4(15˚).sin 6(15˚)

= 4 sin(30˚).cos(60˚).sin(90˚)

= 4(1/2).(1/2).(1)

= 4(1/4)

= 1

= RHS

 

  1. If A = 60˚ and B = 30˚ then prove that tan(A – B) = tanA – tanB/1 + tanAtanB

Solution:

If A = 60˚ and B = 30˚ then we have to prove that tan(A – B) = tanA – tanB/1 + tanAtanB

LHS = tan(A – B)

= tan(60˚ – 30˚)

= tan30˚

= 1/√3

RHS = tanA – tanB/1 + tanAtanB

= tan60˚ – tan30˚/1 + tan60˚tan30˚

= [√3 – 1/√3]/[1 + √3.(1/√3)]

= (2√3/3)/(1+1)

= √3/3

= 1/√3

Therefore, LHS = RHS