# Trigonometry

## Exercise 8.1 Part 1

##### Important Formulae

In the right angle triangle ABC, the right angle is ∠ B text(sin A)=text(side opposite to ∠A)/text(hypotenuse)=p/h

text(cos A)=text(side adjacent to ∠A)/text(hypotenuse)=b/h

text(tan A)=text(side opposite to ∠A)/text(side adjacent to ∠A)=p/b

If sin, cos or tan ratio are calculated for angle C, then values for p, b and h shall change accordingly.

text(cosecθ)=(1)/text(sinθ)

text(secθ)=(1)/text(cosθ)

text(cotθ)=(1)/text(tanθ)

text(tanθ)=text(sinθ)/text(cosθ)

∠ A00300450600900
Sin A01/21/sqrt2sqrt3/21
Cos A1sqrt3/21/sqrt21/20
Tan A01/sqrt31sqrt3Undefined
Cosec AUndefined2sqrt22/sqrt31
Sec A12/sqrt3sqrt22Undefined
Cot AUndefinedsqrt311/sqrt30

The value of sin or cos never exceeds 1, but the value of sec and cosec is always greater than or equal to 1.

## Trigonometric Identities:

text(cos)^2A + text(sin)^2A = 1

1 – text(sin)^2A = text(cos)^2A

1 – text(cos)^2A = text(sin)^2A

1 + text(tan)^2A = text(sec)^2A

text(sec)^2A – text(tan)^2A = 1

text(sec)^2A – 1 = text(tan)^2A

text(cot)^2A + 1 = text(cosec)^2A

text(cosec)^2A – text(cot)^2A = 1

text(cosec)^2A – 1 = text(cot)^2A

## Exercise 8.1 Part 1

Question – 1 - In triangle ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (a) Sin A, Cos A

Solution: AB = 24 cm, BC = 7 cm, AC = ?

The value of AC can be calculated by using Pythagoras Theorem:

AC^2 = AB^2 + BC^2

Or, AC^2 = 24^2+ 7^2

= 576 + 49 = 625

Or, AC = sqrt(625)=25

text(sin A)=(BC)/(AC)=(7)/(25)

text(cos A)=(AB)/(AC)=(24)/(25)

(b) Sin C, Cos C

Solution:

text(sin C)=(AB)/(AC)=(24)/(25)

text(cos C)=(BC)/(AC)=(7)/(25)

Question – 2 - In the given figure, find tan P – cot R. Solution: Value of QR can be calculated by using Pythagoras theorem:

QR^2 = PR^2 – PQ^2

Or, OR^2 = 13^2 – 12^2

= 169 – 144 – 25

Or, OR = 5

Now;

text(tan P)-text(cot R)

=(QR)/(PQ)-(QR)/(PQ)=0

Question – 3 - If sin A = ¾, calculate cos A and tan A.

Solution: Sin A = ¾ = p/h

We can calculate b by using Pythagoras theorem;

b^2 = h^2 – p^2

Or, b^2 = 4^2 – 3^2

= 16 – 9 = 7

Or, b = sqrt7

Now;

text(cos A)=b/h=sqrt7/4

text(tan A)=p/b=3/sqrt7

Question – 4 - Given 15 cot A = 8, find sin A and sec A.

Solution: 15 cot A = 8

Or, cotA=(8)/(15)=b/p

This means, b = 8 and p = 15

We can calculate h by using Pythagoras theorem;

h^2 = p^2 + b^2

Or, h^2 = 15^2 + 8^2

= 225 + 64 = 289

Or, h = 17

Now;

text(sin A)=p/h=(15)/(17)

text(sec A)=h/b=(17)/(8)

Question – 5 - Given text(sec θ) = (13)/(12), calculate all other trigonometric ratios.

Solution:

text(sec θ)=(13)/(12)=h/b

This means, h = 13 and b = 12.

We can calculate p by using Pythagoras theorem;

p^2 = h^2 – b^2

Or, p^2 = 13^2 – 12^2

= 169 – 144 = 25

Or, p = 5

Other trigonometric ratios can be calculated as follows:

text(sin θ)=p/h=(5)/(13)

text(cos θ)=b/h=(12)/(13)

text(tan θ)=p/b=(5)/(12)

text(cosec θ)=h/p=(13)/(5)

text(cot θ)=b/p=(12)/(5)

Question – 6 - If ∠ A and ∠ B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B

Solution: For no two different angles the cos ratio is same. (Ref: Table of trigonometric ratio).