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

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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.

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