# Trigonometric Ratios:-

### Today in this article I will explain on what are trigonometric ratios, trigonometric identities, ans trigonometric angles, hope these will help a lot for on going school children.

## **What is trigonometry?

### Answer:-

### The word trigonometry is a greek word in which trigonon means (objects which have three angle here it is triangle)metron means measurement. So measurement of angles of a triangle is trigonometry.

## **Trigonometry Ratios:-

### Answer:-

### Let, ABC is a right angled triangle at B, whose hypotenuse= h = AC

### perpendicular = p = AB

### base =b =BC

### and measure of unknown angle(say<C) =θ

∴ SinC = Sin θ = p/h = AB/AC [where,<B=90°]

∴ Cos C = Cos θ = b/h = BC/ AC [where,<B=90°]

∴ TanC = Tan θ = p/b =AB/BC [ where <B=90°]

### Cosec C = Cosec θ = h/p = AC/AB [where <B=90°]

Sec C = Sec θ =h/b =AC/ BC [where <B=90°]

Cot C= Cot θ = b/p = BC/ AB [where <B=90°]

Tanθ =Sinθ/Cosθ

Cotθ = Cosθ/Sinθ

## **Reciprocals of trigonometric ratios:-

### Sinθ = 1/ Cosecθ; Cosθ =1/Secθ;

Cosecθ = 1/Sinθ; Secθ =1/Cosθ;

Tanθ =1/Cotθ; Cotθ=1/Tanθ

## **Trigonometric identities:-

### (i) Sin²θ + Cos²θ =1

(ii) Sin²θ =1-Cos²θ

(iii)Cos²θ =1-Sin²θ

###

—------------------------------------------------------

(iv)Sec²θ -Tan²θ =1

(v) Sec²θ =1+Tan²θ

(vi)Tan²θ =Sec²θ-1

—------------------------------------------------------

### (vii) Cosec²θ - Cot²θ =1

(viii) Cosec²θ =1+Cot²θ

(ix) Cot²θ =Cosec²θ -1

—------------------------------------------------------

## **Prove that geometrically Sin²θ + Cos²θ =1

### Answer:-

### Let, PQR is a right angled triangle at Q, whose hypotenuse is PR, perpendicular PQ and base is QR and <R =θ(say)

### To prove:- Sin²θ + Cos²θ =1

### Proof:- ∵ PQR is a right angled Δ,

Applying Pythagorus theorem,we get

### ∴ PQ² + QR² = PR² [∵∠Q = 90°]

=> PQ²/PR² + QR²/PR² = PR²/PR²

[∵dividing both sides by PR²]

=> (PQ/PR)² +(QR/PR)² =1

=> Sin²θ + Cos²θ =1

Proved

### Note:-

### [Students are advised to do (ii) and (iii) proof themselves understanding the above method.]

## **Prove that geometrically Sec²θ -Tan²θ=1

### Answer:-

### Let, PQR is a right angled triangle at Q, whose hypotenuse is PR, perpendicular PQ and base is QR and <R =θ

Proof:-

∵ PQR is a right angled Δ,

Applying Pythagorus theorem,we get

∵ PR² = PQ² + QR²[∵∠Q = 90°]

=>PR²/PQ² =PQ²/PQ² +QR²/PQ²

=>(PR/PQ)² = 1 +(QR/PQ)²

=> Sec²θ = 1+ tan²θ

=> Sec²θ - tan²θ = 1

Proved.

### Note:-

### [Students are advised to do (v) and (vi) proof themselves understanding the above method.]

## **Find the value of Sin 30° geometrically.

### Answer:-

Let ABC is an equilateral triangle whose <A = <B = <C = 60°

Let,

AB = BC = AC = 2a

To find:- Sin 30°=?

Construction:- AD⊥BC is drawn, which bisects BC also bisects <A

Solution:- ∵ AD⊥BC

∴ BD = DC = ½ BC =½ ×2a = a

also,

<BAD = <CAD = ½ <A = ½×60°= 30°

and,

<ADB = < ADC = 90°

Now in right angled Δ ADB

∵ AD² = AB² - BD² [∵<ADB=90°]

=(2a)² - (a)²

=4a² - a²

=3a²

∴ AD = a√3

∴ Sin 30°= BD/AB = a/2a =½

∴Cos 30°= AD/AB = a√3/2a =√3/2

### [Note: Like this way students can find the values of tan 30°,cot 30°,Cosec 30° and Sec 30°]

## **Find the value of tan 60° geometrically

### Answer:-

### Let ABC is an equilateral triangle whose <A = <B = <C = 60°

Let,

AB = BC = AC = 2a

To find:- Tan 60°=?

Construction:- AD⊥BC is drawn, which bisects BC.

Solution:- ∵ AD⊥BC

∴ BD = DC = ½ BC =½ ×2a = a

Now in right angled Δ ADB

∵ AD² = AB² - BD² [∵<ADB=90°]

=(2a)² - (a)²

=4a² - a²

=3a²

∴ AD = a√3

∴ TanB = Tan 60°= AD/BD = a√3/a = √3

[Note: Like this way students can find the values of Sin 60°,Cos 60°,cot 60°,Cosec 60°,

Cot 60° and Sec 60°]

## **Find the value of Sec 45°geometrically.

### Answer:-

### Let ABC is an isosceles right angled triangle at B, whose AB = BC = a (say)

also <A = <B = 45°

To find :- Sec 45°=?

Solution:-∵ In right angled Δ ABC,

∵ AC² = AB²+BC² [∵ <B=90°]

=(a)² +(a)²[ putting value]

= a² + a²

=2a²

∴ AC = a√2

###

Sec C= Sec 45°=AC/BC = a√2/a =√2

### Note:- If you love my article and my blog please share it and support me.

## FREQUENTLY ASKED QUESTIONS:-
##
What are the three basic ratios of trigonometry ?

Answer :- The three basic ratios of trigometry are :- (i) Sinθ = perpendicular/hypotenuse(ii) Cosθ = base/hypotenuse (iii) tanθ = perpendicular/base. for any assumed angle θ

## What are the six ratios of trigonometry?

Answer :- The six ratios of trigonometry are :- (i) Sinθ = perpendicular/hypotenuse (ii) Cosθ = base/hypotenuse (iii) tanθ = perpendicular/base. (iv) Cosecθ = hypotenuse/perpendicular(v) Secθ = hypotenuse/base (vi) Cotθ = base/perpendicular for any assumed angle θ

## Find the value of tan120°

Answer :- The value of tan120°
= tan (180°- 60°)
= tan (90°× 2 - 60°)
= - tan ( 90° × 2 - 60°) [ ∵ 120° is in 2nd quadrant so tan will be negative]
= - tan 60° [ ∵ 90 × 2 even multiple of 90°, so there will be no change, tan will remain same]
= - √3

## Find the value of sin150°

Answer:- sin150°
= sin(180°- 30°)
= sin (90°×2 - 30°)[ ∵ 150° is in 2nd quadrant so sin will be positive]
= sin30°[ ∵ 90 × 2 even multiple of 90°, so there will be no change, sin will remain same]
= ½

##
What are the basic three formulas of trigonometry?

Answer:- The basic three formulas of trigonometry are :- (i)sin²θ + cos²θ =1
(ii)sec²θ - tan²θ =1
(iii)cosec²θ - cot²θ =1
for any assumed angle θ

##
What is Pythagorus law ?

.
Answer:- "The sum of the squares of base and perpendicular of a right triangle is equal to the square of its hypotenuse" which is known as Pythagorus law.

##
Which triangle is associated with trigonometry?

.
Answer:- Right triangle.

## What are the three basic ratios of trigonometry ?

Answer :- The three basic ratios of trigometry are :- (i) Sinθ = perpendicular/hypotenuse(ii) Cosθ = base/hypotenuse (iii) tanθ = perpendicular/base. for any assumed angle θ

## What are the six ratios of trigonometry?

Answer :- The six ratios of trigonometry are :- (i) Sinθ = perpendicular/hypotenuse (ii) Cosθ = base/hypotenuse (iii) tanθ = perpendicular/base. (iv) Cosecθ = hypotenuse/perpendicular(v) Secθ = hypotenuse/base (vi) Cotθ = base/perpendicular for any assumed angle θ

## Find the value of tan120°

Answer :- The value of tan120° = tan (180°- 60°) = tan (90°× 2 - 60°) = - tan ( 90° × 2 - 60°) [ ∵ 120° is in 2nd quadrant so tan will be negative] = - tan 60° [ ∵ 90 × 2 even multiple of 90°, so there will be no change, tan will remain same] = - √3

## Find the value of sin150°

Answer:- sin150° = sin(180°- 30°) = sin (90°×2 - 30°)[ ∵ 150° is in 2nd quadrant so sin will be positive] = sin30°[ ∵ 90 × 2 even multiple of 90°, so there will be no change, sin will remain same] = ½

## What are the basic three formulas of trigonometry?

Answer:- The basic three formulas of trigonometry are :- (i)sin²θ + cos²θ =1 (ii)sec²θ - tan²θ =1 (iii)cosec²θ - cot²θ =1 for any assumed angle θ

## What is Pythagorus law ?

. Answer:- "The sum of the squares of base and perpendicular of a right triangle is equal to the square of its hypotenuse" which is known as Pythagorus law.

## Which triangle is associated with trigonometry?

. Answer:- Right triangle.

Leave your comment: