TRIGONOMETRY PROBLEMS USING IDENTITIES

About "Trigonometry problems using identities"

Trigonometry problems using identities are much useful to the kids who would like to practice problems on trigonometric identities.   

Before we look at the problems on trigonometric identities, let us have a look on some important trigonometric identities. 

Some important trigonometric Identities

Let us see the reciprocal trigonometric identities

sin θ = 1/Cosec θ

Cosec θ = 1/sin θ

Cos θ = 1/sec θ

sec θ = 1/cos θ

tan θ = 1/cot θ

cot θ = 1/tan θ

sin² θ  + cos² θ = 1

sin² θ  = 1 - cos² θ

cos² θ = 1 - sin² θ

sec² θ - tan² θ = 1

sec² θ  = 1 +  tan² θ

tan² θ  =  sec² θ - 1

csc² θ - cot² θ = 1

csc² θ = 1 + cot² θ

cot² θ =  csc² θ - 1

Trigonometry problems using identities

Problem 1 : 

Prove : (1 - cos² θ) csc² θ  =  1

Solution :

Let A  =  (1 - cos² θ) csc² θ  and

B  =  1

A  =  (1 - cos² θ) csc² θ

Since  sin² θ + cos² θ  =  1, we have  sin² θ  = 1 - cos² θ

A  =  sin² θ x csc² θ

We know that csc² θ  =  1/sin² θ

A  =  sin² θ x 1/sin² θ

A  =  1

A  =  B  Proved

Let us look at the next problem on "Trigonometry problems using identities"

Problem 2 : 

Prove : sec θ √(1 - sin ² θ) = 1

Solution :

Let A  =  sec θ √(1 - sin ² θ)  and

B  =  1

A  =  sec θ √(1 - sin ² θ)

Since  sin² θ + cos² θ  =  1, we have  cos² θ  = 1 - sin² θ

A  =  sec θ √cos ² θ

A  =  sec θ x cos θ

A  =  sec θ x 1/sec θ

A  =  1

A  =  B  Proved

Let us look at the next problem on "Trigonometry problems using identities"

Problem 3 : 

Prove : tan θsin θ + cos θ  =  sec θ

Solution :

Let A  =  tan θsin θ + cos θ  and

B =  sec θ

A  =  tan θsin θ + cos θ

A  =  (sin θ/cos θ)sin θ + cos θ

A  =  (sin² θ/cos θ) + cos θ

A  =  (sin² θ + cos² θ) / cos θ

A  =  1 / cos θ

A  =  sec θ

A  =  B  Proved

Let us look at the next problem on "Trigonometry problems using identities"

Problem 4 : 

Prove : (1 - cos θ)(1 + cos θ)(1 + cot² θ)  =  1

Solution :

Let A  =  (1 - cos θ)(1 + cos θ)(1 + cot² θ)  =  1

B  =  1

A  =  (1 - cos θ)(1 + cos θ)(1 + cot² θ)

A  =  (1 - cos²)(1 + cot² θ)

Since  sin² θ + cos² θ  =  1, we have  sin² θ  = 1 - cos² θ

A  =  sin² θ x (1 + cot² θ)

A  =  sin² θ  + sin² θ x cot² θ

A  =  sin² θ  + sin² θ x (cos² θ/sin² θ)

A  =  sin² θ  + cos² θ

A  =  1

A  =  B  Proved

Let us look at the next problem on "Trigonometry problems using identities"

Problem 5 : 

Prove : cot θ + tan θ  =  sec θcsc θ

Solution :

Let A  =  cot θ + tan θ and

B  =  sec θ csc θ

A  =  cot θ + tan θ

A  =  (cos θ/sin θ) + (sin θ/cosθ)

A  =  (cos² θ + sin² θ) / sin θcosθ

(Because, cos² θ + sin² θ = 1 )

A  =  1 / sin θcos θ

A  =  (1/cos θ) x (1/sin θ)

A  =  sec θcscθ

A  =  B  Proved

Let us look at the next problem on "Trigonometry problems using identities"

Problem 6 : 

Prove : cos θ/(1 - tan θ) + sin θ/(1 - cot θ) = sin θ + cos θ 

Solution :

Let A  =  cos θ/(1 - tan θ) + sin θ/(1 - cot θ)  and

B  =  sin θ + cos θ

A  =  cos θ/{1 - (sin θ/cos θ)} + sin θ/{1 - (cos θ/sin θ)} 

A = cos θ/{(cos θ - sin θ)/cos θ} + sin θ/{(sin θ - cos θ/sin θ)} 

A  =  cos² θ/(cos θ - sin θ) + sin² θ/(sin θ - cos θ) 

A  =  cos² θ/(cos θ - sin θ) - sin² θ/(cos θ - sin θ) 

A  =  (cos² θ - sin² θ)/(cos θ - sin θ) 

A  =  [(cos θ + sin θ)(cos θ - sin θ)]/(cos θ - sin θ) 

A  =  (cos θ + sin θ) 

A  =  B  Proved

Let us look at the next problem on "Trigonometry problems using identities"

Problem 7 : 

Prove : tan θ + tan² θ = sec θ - sec² θ 

Solution :

Let A  =  tan θ + tan² θ  and 

B  =  sec θ - sec² θ 

A  =  tan² θ (tan² θ + 1) 

A  =  (sec² θ - 1) (tan² θ + 1)  [since, tan² θ = sec² θ – 1] 

A  =  (sec² θ - 1) sec² θ  [since, tan² θ + 1 = sec² θ] 

A  =  sec θ - sec² θ

A  =  B  Proved

Let us look at the next problem on "Trigonometry problems using identities"

Problem 8 : 

Prove : √{(sec θ – 1)/(sec θ + 1)}  =  cosec θ - cot θ. 

Solution :

Let A  =  √{(sec θ – 1)/(sec θ + 1)} and

B  =  cosec θ - cot θ

A  =  √{(sec θ – 1)/(sec θ + 1)} 

A  =  √[{(sec θ - 1) (sec θ - 1)}/{(sec θ + 1) (sec θ - 1)}]

[multiplying numerator and denominator by (sec θ - l) under radical sign] 

A  =  √{(sec θ - 1)²/(sec² θ - 1)} 

A  =  √{(sec θ -1)²/tan² θ}

[since, sec² θ = 1 + tan² θ ⇒ sec² θ - 1 = tan² θ] 

A  =  (sec θ – 1)/tan θ 

A  =  (sec θ/tan θ) – (1/tan θ) 

A  =  {(1/cos θ)/(sin θ/cos θ)} - cot θ 

A  =  {(1/cos θ) × (cos θ/sin θ)} - cot θ

A  =  (1/sin θ) - cot θ

A  =  cosec θ - cot θ

A  =  B  Proved

Let us look at the next problem on "Trigonometry problems using identities"

Problem 9 : 

Prove : (1 - sin A)/(1 + sin A) = (sec A - tan A)²

Solution :

Let A  =  (1 - sin A)/(1 + sin A) and 

B  =  (sec A - tan A)²

A  =  (1 - sin A)/(1 + sin A) 

A  =   (1 - sin A)²/(1 - sin A) (1 + sin A)

[Multiply both numerator and denominator by (1 - sin A) 

A  =  (1 - sin A)²/(1 - sin² A) 

A  =  (1 - sin A)²/(cos² A)

[Since sin² θ + cos² θ = 1 ⇒ cos² θ = 1 - sin² θ] 

A  =  {(1 - sin A)/cos A}²

A  =  (1/cos A - sin A/cos A)²

A  =  (sec A – tan A)²

A  =  B  Proved

Let us look at the next problem on "Trigonometry problems using identities"

Problem 10 : 

Prove :

(tan θ + sec θ - 1)/(tan θ - sec θ + 1) = (1 + sin θ)/cos θ

Solution :

Let A  =  (tan θ + sec θ - 1)/(tan θ - sec θ + 1)  and 

B  =  (1 + sin θ)/cos θ

A  =  (tan θ + sec θ - 1)/(tan θ - sec θ + 1) 

A  =  [(tan θ + sec θ) - (sec
² θ - tan² θ)]/(tan θ - sec θ + 1)

[Since, sec² θ - tan² θ  =  1] 

A = {(tan θ + sec θ) - (sec θ + tan θ) (sec θ - tan θ)} / (tan θ - sec θ + 1) 

A  =  {(tan θ + sec θ) (1 - sec θ + tan θ)}/(tan θ - sec θ + 1) 

A  =  {(tan θ + sec θ) (tan θ - sec θ + 1)}/(tan θ - sec θ + 1) 

A  =  tan θ + sec θ

A  =  (sin θ/cos θ) + (1/cos θ) 

A  =  (sin θ + 1)/cos θ

A  =  (1 + sin θ)/cos θ 

A  =  B  Proved

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