## Trigonometry Exercise 13.1 – Questions:

- Find sin θ and cos θ for the following:

(i)

(ii)

(iii)

2.Find the following:

- If sin x =
^{3}/_{5}, cosec x = ______________ - If cos x =
^{24}/_{25}, sec x = _______________ - If tan x =
^{7}/_{24}, cot x = _______________ - If cosec x =
^{25}/_{15}, sin x = ____________ - If sin A =
^{3}/_{5}and cos A =^{4}/_{5}, then, tan A = __________ - If cot A =
^{8}/_{15}and sin A =^{15}/_{17}, then, cos A = _________

3.Solve:

- Given tan A =
^{3}/_{4}, find the value of sin A and cos A. - Given cot θ =
^{20}/_{21}, determine cos θ and cosec θ - Given tan A =
^{7}/_{24}, find the other trigonometric ratios of angle A. - If 2 sin θ = √3, find cos θ, tan θ and cot θ + cosec θ
- If 3 tan θ = 1, find sin θ, cos θ and cot θ
- If sec x = 2, then, find sin x, tan x, cot x and cot x + cosec x
- If 4 sin A – 3 cos A = 0, find sin A, cos A, sec A and cosec A
- If 13 sin A = 5 and A is acute, find the va;ue of
^{5sin A – 2 cos A}/_{tan }_{A} - If cos θ =
^{5}/_{13}and θ is acute, find the value of^{5tanθ + 12cot θ}/_{5tanθ – 12cot θ} - IF 13 cos θ – 5 = 0, find
^{sin θ + cos θ}/_{sin θ – cos θ}

## Trigonometry Exercise 13.1 – Solutions:

**Find sinθ and cosθ for the following:**

**(i)**

**(ii)**

**(iii)**

Solution:

(i)

sin θ = ^{opposite side}/_{hypotensuse} = ^{24}/_{25}

cos θ = ^{adjacent side}/_{hypotensuse} = ^{7}/_{25}

(ii)

sin θ = ^{opposite side}/_{hypotensuse} = ^{15}/_{25}

cos θ = ^{adjacent side}/_{hypotensuse} = ^{20}/_{25}

(iii)

sin θ = ^{opposite side}/_{hypotensuse} = ^{10}/_{25}

cos θ = ^{adjacent side}/_{hypotensuse} = ^{24}/_{25}

**2.Find the following:**

**If sin x =**^{3}/_{5}, cosec x = ______________**If cos x =**^{24}/_{25}, sec x = _______________**If tan x =**^{7}/_{24}, cot x = _______________**If cosec x =**^{25}/_{15}, sin x = ____________**If sin A =**^{3}/_{5}and cos A =^{4}/_{5}, then, tan A = __________**If cot A =**^{8}/_{15}and sin A =^{15}/_{17}, then, cos A = _________

Solution:

- If sin x =
^{3}/_{5}, cosec x =^{5}/_{3} - If cos x =
^{24}/_{25}, sec x =^{25}/_{24} - If tan x =
^{7}/_{24}, cot x =^{24}/_{7} - If cosec x =
^{25}/_{15}, sin x =^{15}/_{25} - If sin A =
^{3}/_{5}and cos A =^{4}/_{5}, then, tan A =^{3}/_{4} - If cot A =
^{8}/_{15}and sin A =^{15}/_{17}, then, cos A =^{8}/_{17}

**Solve:**

**1.Given tan A = ^{3}/_{4}, find the value of sin A and cos A.**

Solution:

We know tan A = ^{sin A}/_{cos A} = (^{opposite side}/_{hypotenuse})/ (^{adjacent side}/_{hypotenuse})

= ^{opposite side}/_{adjacent side}

Sin A = ^{opposite side}/_{hypotenuse}

Cos A = ^{adjacent side}/_{hypotenuse}

In right angled triangle ABC, ∟C = 90˚, A = θ

By Pythagoras theorem,

AB^{2} = BC^{2} + CA^{2}

= 3^{2} + 4^{2}

= 9 + 16

= 25

AB = 5

sin A = ^{opposite side}/_{hypotenuse } = ^{3}/_{5}

Cos A = ^{adjacent side}/_{hypotenuse} = ^{4}/_{5}

**Given cot θ =**^{20}/_{21}, determine cos θ and cosec θ

Solution:

We know that, cot θ = ^{20}/_{21}

cot θ = ^{cos} ^{θ}/_{sin θ} = (^{adjacent side}/_{hypotenuse})/(^{ opposite side}/_{hypotenuse}) = ^{adjacent side}/_{opposite side}

In right angled triangle ABC, ∟C = 90˚, A = θ

By Pythagoras theorem,

AB^{2} = BC^{2} + CA^{2}

= 20^{2} + 21^{2}

= 400 + 441

= 841

AB = 29

sin A = ^{opposite side}/_{hypotenuse } = ^{21}/_{29}

Cos A = ^{adjacent side}/_{hypotenuse} = ^{20}/_{29}

**Given tan A =**^{7}/_{24}, find the other trigonometric ratios of angle A.

Solution:

We know tan A = ^{sin A}/_{cos A} = (^{opposite side}/_{hypotenuse})/ (^{adjacent side}/_{hypotenuse})

= ^{opposite side}/_{adjacent side} = ^{7}/_{24}

Sin A = ^{opposite side}/_{hypotenuse}

Cos A = ^{adjacent side}/_{hypotenuse}

In right angled triangle ABC, ∟C = 90˚, A = θ

By Pythagoras theorem,

AB^{2} = BC^{2} + CA^{2}

= 7^{2} + 24^{2}

= 49 + 576

= 625

AB = 25

sin A = ^{opposite side}/_{hypotenuse } = ^{7}/_{25}

Cos A = ^{adjacent side}/_{hypotenuse} = ^{24}/_{25}

tan A = ^{7}/_{24}

cot A = ^{24}/_{7}

sec A = ^{25}/_{24}

cosec A = ^{25}/_{7}

**If 2 sin θ = √3, find cos θ, tan θ and cot θ + cosec θ**

Solution:

sin θ = ^{opposite side}/_{hypotenuse } = ^{√3}/_{2}

In right angled triangle ABC, ∟C = 90˚, A = θ

By Pythagoras theorem,

AB^{2} = BC^{2} + CA^{2}

2^{2} = (√3)^{2} + CA^{2}

4 = 3 + CA^{2}

4 – 3 = CA^{2}

CA = 1

cos θ = ^{1}/_{2}

tan θ = ^{√3}/_{1} = √3

cot θ + cosec θ = ^{1}/_{√3} + ^{2}/_{√3} = ^{1+2}/_{√3} = ^{3}/_{√3}

**If 3 tan θ = 1, find sin θ, cos θ and cot θ**

Solution:

3 tan θ = 1

tan θ = ^{1}/_{3} = ^{sin θ}/_{cos θ}

In right angled triangle ABC, we have ∟C = 90˚ and A = θ

AB^{2} = BC^{2} + CA^{2}

AB^{2} = (1)^{2} + 3^{2}

= 1 + 9

= 10

AB = √10

Therefore, sin θ = ^{1}/_{√10}

cos θ = ^{3}/_{√10}

cot θ = ^{3}/_{1} = 3

**If sec x = 2, then, find sin x, tan x, cot x and cot x + cosec x**

Solution:

Given, sec x = 2 = ^{hypotenuse}/_{adjacent side} = ^{2}/_{1}

In right angled triangle ABC, we have ∟C = 90˚ and A = θ

AB^{2} = BC^{2} + CA^{2}

2^{2} = BC^{2} + 1^{2}

4 – 1 = BC^{2}

3 = BC^{2}

BC = √3

sin x = ^{√3}/_{2}

tan x = ^{√3}/_{1} = √3

cot x = ^{1}/_{√3}

cot x + cosec x = ^{1}/_{√3} + ^{2}/_{√3} = ^{1+2}/_{√3} = ^{3}/_{√3}

**If 4 sin A – 3 cos A = 0, find sin A, cos A, sec A and cosec A**

Solution:

4 sin A – 3 cos A = 0

^{sinA}/_{cos A} = ^{3}/_{4}

tan A = ^{3}/_{4} = ^{opposite side}/_{adjacent side}

In right angled triangle ABC, we have ∟C = 90˚ and A = θ

AB^{2} = BC^{2} + CA^{2}

AB^{2} = 3^{2} + 4^{2}

= 9 + 16

= 25

AB = 5 = hypotenuse

sin A = ^{3}/_{5}

cos A = ^{4}/_{5}

sec A = ^{5}/_{4}

cosec A = ^{5}/_{3}

**If 13 sin A = 5 and A is acute, find the value of**^{5sin A – 2 cos A}/_{tan }_{A}

Solution:

13 sin A = 5

sin A = ^{5}/_{13} = ^{opposite side}/_{hypotenuse}

In right angled triangle ABC, we have ∟C = 90˚ and A = acute angle

AB^{2} = BC^{2} + CA^{2}

13^{2} = 5^{2} + CA^{2}

169 – 25 = CA^{2}

144 = CA^{2}

12 = CA = adjacent side

We have to find , ^{5sin A – 2 cos A}/_{tan }_{A}

sin A = ^{5}/_{13}

cos A = ^{12}/_{13}

tan A = ^{5}/_{12}

^{5sin A – 2 cos A}/_{tan }_{A} = (5. ^{5}/_{13} – 2 . ^{12}/_{13})/(^{5}/_{12})

^{5sin A – 2 cos A}/_{tan }_{A} = (^{25}/_{13 }– ^{24}/_{13})/(^{5}/_{12})

^{5sin A – 2 cos A}/_{tan }_{A} =( ^{1}/_{13})/(^{5}/_{12})

^{5sin A – 2 cos A}/_{tan }_{A} = ^{12}/_{65}

_{ }

**If cos θ =**^{5}/_{13}and θ is acute, find the value of^{5tanθ + 12cot θ}/_{5tanθ – 12cot θ}

Solution:

cos θ = ^{5}/_{13} = ^{Adjacent side}/_{hypotenuse}

In right angled triangle ABC, we have ∟C = 90˚ and A = θ = acute angle

AB^{2} = BC^{2} + CA^{2}

13^{2} = BC^{2} + 5^{2}

169 – 25 =BC^{2}

144 = BC^{2}

12 = BC = opposite side

We have to find, ^{5tanθ + 12cot θ}/_{5tanθ – 12cot θ}

tan θ = ^{12}/_{5}

cot θ = ^{5}/_{12}

^{5tanθ + 12cot θ}/_{5tanθ – 12cot θ} = (5.^{12}/_{5} + 12.^{5}/_{12})/(5.^{12}/_{5} – 12.^{5}/_{12})

= ^{(12 + 5)}/_{(12 – 5)}

= ^{17}/_{5}

**If 13 cos θ – 5 = 0, find**^{sin θ + cos θ}/_{sin θ – cos θ}

Solution:

13 cos θ – 5 = 0

cos θ = ^{5}/_{13} = ^{adjacent side}/_{hypotenuse}

In right angled triangle ABC, we have ∟C = 90˚ and A = θ = acute angle

AB^{2} = BC^{2} + CA^{2}

13^{2} = BC^{2} + 5^{2}

169 – 25 =BC^{2}

144 = BC^{2}

12 = BC = opposite side

Therefore, sin θ = ^{12}/_{13}

We have to find, ^{sin θ + cos θ}/_{sin θ – cos θ}

^{sin θ + cos θ}/_{sin θ – cos θ} = (^{12}/_{13} + ^{5}/_{13})/(^{12}/_{13} – ^{5}/_{13})

= (^{17}/_{13})/(^{7}/_{13})

= ^{17}/_{7}