Relation Between Trigonometric Ratios Class 10th

Introduction

We know that the reciprocal of sin ϴ, cos ϴ, and tan ϴ is cosec ϴ, sec ϴ, and cot ϴ respectively. Therefore, some formulae are here which show the Relation Between Trigonometric Ratios.

Derivation of Formula and Explanation

If we assume △PQR, 

RELATION BETWEEN TRIGONOMETRIC RATIOS

sin ϴ = PQ/PR    and cosec ϴ = PR/PQ

sin ϴ⨯cosec ϴ = PQ/PR⨯PR/PQ

sin ϴ⨯cosec ϴ = 1

Similarly,

1) sin ϴ×cosec ϴ = 1  ⇒      sin ϴ = 1/cosec ϴ and       cosec ϴ = 1/sin ϴ 

2) cos ϴ×sec ϴ = 1   ⇒      cos ϴ = 1/sec ϴ     and     sec ϴ = 1/cos ϴ 

3) tan ϴ×cot ϴ = 1   ⇒    tan ϴ = 1/cot ϴ     and     cot ϴ = 1/ tan ϴ 

4) tan ϴ = sin ϴ/cos ϴ

∵ sin ϴ = PQ/PR         and          cos ϴ = QR/PR

Now, sin ϴ/cos ϴ = PQ/PR ∕ QR/PR = PQ/PR×PR/QR = PQ/QR = Perpendicular/Base = tan ϴ

Similarly,

5) cot ϴ = cos ϴ/sin ϴ    

Trigonometric Identities

(1) sin2 ϴ + cos2 ϴ = 1

∵ sin ϴ = PQ/PR  and  cos ϴ = QR/PR [from the above figure]

LHS       

sin2 ϴ + cos2 ϴ

(PQ/PR)2 + (QR/PR)2 

PQ2/PR2 + QR2/PR2

PQ2 + QR2/PR2

PR2/PR2  = 1 = RHS           (by Pythagoras theorem PR2 = PQ2 + QR2)

(2) 1 + tan2 ϴ = sec2 ϴ

∵ tan ϴ = PQ/QR [from the above figure]

LHS  

1 + tan2 ϴ

1 + (PQ/QR)2

1 + PQ2/QR2

QR2 + PQ2/QR2 

PR2/QR2 = (PR/QR)2 = sec2 ϴ = RHS   

Alternate Method –

We know that,  sin2 ϴ + cos2 ϴ = 1

Divide both side by cos2 ϴ

sin2 ϴ/cos2 ϴ + cos2 ϴ/cos2 ϴ = 1/cos2 ϴ               ( ∵ tan ϴ = sin ϴ/cos ϴ)

tan2 ϴ + 1 = sec2 ϴ   

(3) 1 + cot2 ϴ = cosec2 ϴ

∵ cot ϴ = QR/PQ [from the above figure]

LHS    

1 + cot2 ϴ

1 + (QR/PQ)2

1 + QR2/PQ2 

PQ2 + QR2/PQ2 

PR2/PQ2 = (PR/PQ)2 = cosec2 ϴ = RHS   

Alternate Method –

We know that,  sin2 ϴ + cos2 ϴ = 1

Divide both side by sin2 ϴ

sin2 ϴ/sin2 ϴ + cos2 ϴ/sin2 ϴ = 1/sin2 ϴ            (∵ cot ϴ = cos ϴ/sin ϴ)  

1 + cot2 ϴ  = cosec2 ϴ

Note – 1) (sin ϴ)2 = sin2 ϴ ≠ sin ϴ2

It means (sin ϴ)2 can be written as sin2 ϴ but cannot be written as sin ϴ2. The same applies to other trigonometric functions.

2) Trigonometric identities can be written as –

A) sin2 ϴ + cos2 ϴ = 1                     sin2 ϴ = 1 – cos2 ϴ                 cos2 ϴ = 1 – sin2 ϴ

B) 1 + tan2 ϴ = sec2 ϴ                     sec2 ϴ – tan2 ϴ = 1                 tan2 ϴ = sec2 ϴ – 1

C) 1 + cot2 ϴ  = cosec2 ϴ               cosec2 ϴ – cot2 ϴ = 1              cot2 ϴ = cosec2 ϴ – 1

Example – If sin ϴ = 5/13 then find all the trigonometric functions using the relation between trigonometric ratios, where ϴ is an acute angle.

Solution – Here, sin ϴ = 5/13 

We know that   sin2 ϴ + cos2 ϴ = 1

(5/13)2 + cos2 ϴ = 1

cos2 ϴ = 1 – (25/169)

cos ϴ = √(169-25)/169 = √(144/169) = 12/13 

∵ sec ϴ = 1/cos ϴ  

sec ϴ = 1 ∕ 12/13 = 13/12  

∵ cosec ϴ = 1/sin ϴ = 1 ∕ 5/13 = 13/5  

∵ 1 + cot2 ϴ = cosec2 ϴ

cot2 ϴ = cosec2 ϴ – 1

cot ϴ = √(13/5)2 – 1= √169/25 – 1 = √(169-25)/25 = √(144/25) = 12/5  

∵ tan ϴ = 1/cot ϴ

tan ϴ = 1 ∕ 12/5 = 5/12  

Examples Based on Trigonometric Identities

Example – 1) Prove that tan ϴ + cot ϴ = sec ϴ×cosec ϴ

Solution – tan ϴ + cot ϴ = sec ϴ×cosec ϴ

LHS         tan ϴ + cot ϴ

sin ϴ/cos ϴ + cos ϴ/sin ϴ

(sin ϴ⨯sin ϴ + cos ϴ⨯cos ϴ)/cos ϴ⨯sin ϴ

(sin2 ϴ + cos2 ϴ)/cos ϴ⨯sin ϴ {∵ sin2 ϴ + cos2 ϴ = 1}

(1/cos ϴ)⨯sin ϴ = sec ϴ × cosec ϴ = RHS

Example – 2) Prove that (1 + cot2 ϴ)(1 + cos ϴ)(1 – cos ϴ) = 1

Solution – (1 + cot2 ϴ)(1 + cos ϴ)(1 – cos ϴ) = 1

LHS          (1 + cot2 ϴ)(1 + cos ϴ)(1 – cos ϴ) {∵ 1 + cot2 ϴ = cosec2 ϴ}

(cosec2 ϴ)(1 – cos2ϴ)      {(a+b)(a-b) = a2 – b2}

cosec2 ϴ×sin2 ϴ {∵ 1 – cos2 ϴ = sin2 ϴ}

(1/sin2ϴ)×sin2ϴ               {cosec ϴ = 1/sinϴ}

1 = RHS

Relation Between Trigonometric Ratios Class 10th in Hindi

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