## Trigonometry Exercise 13.2 – Questions:

I. Answer the following questions:

- What trigonometric ratios of angles from 0 to 90 are equal to 0?
- Which trigonometric ratios of angles from 0 to 90 are equal to 1?
- Which trigonometric ratios of angles from 0 to 90 are equal to
^{1}/_{2}? - Which trigonometric ratios of angles from 0 to 90 are not defined?
- Which trigonometric ratios of angles from 0 to 90 are equal?

II. Find θ. if 0≤θ≤90

- √2 cos θ = 1
- √3 tan θ = 1
- 2 sin θ = √3
- 5 sin θ = 0
- 3 tan θ = √3

III. Find the value of the following.

- sin 30˚ cos 60˚ – tan
^{2}45˚ - sin 60˚ cos 30˚ + cos 60˚ sin 30˚
- cos 60˚ cos 30˚ – sin 60˚ sin 30˚
- 2sin
^{2}30˚ – 3 cos^{2 }30˚ + tan 60˚ + 3 sin^{2}90˚ - 4 sin
^{2}60˚ + 3 tan^{2}30˚ – 8 sin 45˚ .cos 45˚ ^{cos 45˚}/_{sec 30˚ + cosec 30˚}- (4sin
^{2}60 – cos^{2}45)/(tan^{2}30 + sin^{2}0) ^{(sin30}^{˚}^{ + tan45}^{˚}^{ – cosec60}^{˚}^{)}/_{(sec30}_{˚}_{ + cos 60}_{˚}_{ +cot45}_{˚}_{)}- (5cos
^{2}60˚ + 4 sec^{2}30˚ – tan^{2}45˚)/(sin^{2}30˚ + cos^{2}30˚) - (5sin
^{2}30˚ + cos^{2}45˚ – 4 tan^{2}30˚)/(2sin30˚ + cos30˚ + tan45˚)

IV. Prove the following equalities.

- sin30˚ .cos60 ˚+ cos30˚.sin60˚ = sin 90˚
- 2cos
^{2}30˚ – 1 = 1 – 2sin^{2}30˚ = cos 60˚ - if θ = 30˚, prove that 4cos
^{2}θ – 3 cos θ = cos 3θ - If π = 180˚ and A =
^{π}/_{6}prove that^{(1 – cosA)(1 + cos A)}/_{(1 – sin A)(1+ sinA)}=^{1}/_{3} - If B = 15˚, prove that 4 sin 2B.cos 4B.sin 6B = 1
- 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:**

**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˚

**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 ˚

**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˚

**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˚

**Find****θ****. if 0≤****θ****≤90****√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˚

**√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˚

**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˚

**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˚

**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.**

**sin 30˚ cos 60˚ – tan**^{2}45˚

Solution:

sin 30˚ cos 60˚ – tan^{2}45˚

= ^{1}/_{2} x ^{1}/_{2} – (1)^{2}

= ^{1}/_{4} – 1

= ^{-3}/_{4}

**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

**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

**2sin**^{2}30˚ – 3 cos^{2 }30˚ + tan 60˚ + 3 sin^{2}90˚

Solution:

2sin^{2}30˚ – 3 cos^{2 }30˚ + tan 60˚ + 3 sin^{2} 90˚

= 2(^{1}/_{2})^{2} – 3(^{√3}/_{2} )^{2} + (√3) + 3(1)^{2}

= 2(^{1}/_{4}) – 3(^{3}/_{4}) + √3 + 3

= ^{1}/_{2} – ^{9}/_{4} + √3 + 3

= ^{5}/_{4} + √3

**4 sin**^{2}60˚ + 3 tan^{2}30˚ – 8 sin 45˚ .cos 45˚

Solution:

4 sin^{2} 60˚ + 3 tan^{2} 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

^{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.

^{(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}

_{ }

**(5cos**^{2}60**˚****+ 4 sec**^{2}30**˚****– tan**^{2}45**˚****)/(sin**^{2}30**˚****+ cos**^{2}30**˚****)**

Solution:

(5cos^{2}60˚ + 4 sec^{2}30˚ – tan^{2}45˚)/(sin^{2}30˚ + cos^{2}30˚)

= [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}

_{ }

**(5sin**^{2}30**˚****+ cos**^{2}45**˚****– 4 tan**^{2}30**˚****)/(2sin30****˚****+ cos30****˚****+ tan45****˚****)**

Solution:

(5sin^{2}30˚ + cos^{2}45˚ – 4 tan^{2}30˚)/(2sin30˚ + cos30˚ + tan45˚)

= [5(^{1}/_{2})^{2} + (^{1}/_{√2})^{2} – 4(^{1}/_{√3})^{2}]/[2(^{1}/_{2}) + (^{√3}/_{2}) + 1]

= [^{5}/_{4} + ^{1}/_{2} – ^{4}/_{3}]/[1 + ^{√3}/_{2} + 1]

= [^{5}/_{12}]/[^{4+√3}/_{2}]

= ^{5}/_{6(4+√3)}

_{ }

**IV.Prove the following equalities.**

**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

**2cos**^{2}30˚ – 1 = 1 – 2sin^{2}30˚ = cos 60˚

Solution:

2cos^{2}30˚ – 1 = 1 – 2sin^{2}30˚ = cos 60˚

Take, 2cos^{2}30˚ – 1 ………………..(1)

= 2(^{√3}/_{2})^{2} – 1

= 2(^{3}/_{4}) – 1

= ^{3}/_{2} – 1

= ^{3 – 2}/_{2} = ^{1}/_{2}

1 – 2 sin^{2} 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, 2cos^{2}30˚ – 1 = 1 – 2sin^{2}30˚ = cos 60˚

**If****θ = 30****˚****, prove that 4cos**^{3}θ – 3 cos θ = cos 3θ

Solution:

If θ = 30˚, we have to prove that 4cos^{3}θ – 3 cos θ = cos 3θ

LHS = 4cos^{3}θ – 3 cos θ

= 4cos^{3}30˚ – 3 cos 30˚

= 4(^{√3}/_{2})^{3} – 3(^{√3}/_{2})

= 4(^{3√3}/_{8}) – 3(^{√3}/_{2})

= ^{3√3}/_{2} – ^{3√3}/_{2}

= ^{3√3 – 3√3}/_{2}

= 0

RHS = cos 3θ

= cos 3(30˚)

= cos(90˚)

= 0

LHS = RHS

**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 – cos^{2}30˚]/[1 – sin^{2}30˚]

= [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

**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

**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