PROVING TRIGONOMETRIC IDENTITIES EXAMPLES

Proving Trigonometric Identities Examples :

Here we are going to see, some example problems to show proving trigonometric identities.

Proving Trigonometric Identities Examples - Questions

Question 1 :

Prove the following identities.

(i) sec4θ (1 - sin4θ) - 2tan2θ  =  1

Solution :

   =  sec4θ (1 - sin4θ) - 2tan2θ

   =   sec4θ [(12)2 - (sin2θ)2] - 2tan2θ

   =   sec4θ [(1 + sin2θ) (1 - sin2θ)] - 2tan2θ

   =   (1/cos4θ)[(1 + sin2θ) cos2θ] - 2tan2θ

   =   [(1 + sin2θ)/cos2θ] - 2tan2θ

   =   (1/cos2θ)  + (sin2θ/cos2θ) - 2tan2θ

   =   sec2θ  + tan2θ - 2tan2θ

   =   sec2θ  - tan2θ

   =  1

Hence proved.

(ii)  (cot θ-cos θ)/(cot θ+cos θ) = (cosec θ-1)/(cosec θ+1)

Solution :

L.H.S :

  =  (cot θ - cos θ)/(cot θ + cos θ) 

  =  ((cos θ/sin θ) - cos θ) / ((cos θ/sin θ) + cos θ) 

  =  ((cos θ-sin θ cos θ)/sin θ) / (cos θ + sin θ cos θ)/sin θ)

  =  ((cos θ-sin θ cos θ) / (cos θ + sin θ cos θ)

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

  =  (1 - sin θ)/(1 + sin θ)

  =  [1 - (1/cosec θ)] / [1 + (1/cosec θ)]

  =  (cosec θ - 1) / (cosec θ + 1)

Hence proved.

Question 2 :

Prove the following identities.

(i) [(sin A -  sin B)/(cos A + cos B)]  + [(cos A - cos B)/sin A + sin B)]  =  0

Solution :

L.H.S

  =  (1 - 1)/(cos A + cos B) (sin A + sin B)

  =  0

R.H.S

Hence proved.

(ii)  [(sin3A + cos3A)/(sin A + cos A)] + [(sin3A - cos3A)/(sin A - cos A)]  =  2

Solution :

L.H.S

[(sin3A + cos3A)/(sin A + cos A)] + [(sin3A - cos3A)/(sin A - cos A)] 

a3 + b3  =  (a + b)(a2 - ab + b2)

a3 - b3  =  (a - b)(a2 + ab + b2)

[(sin3A + cos3A)/(sin A + cos A)]

   =  (sin A+cos A)(sin2A+cos2A-sinAcosA)/(sinA+cosA)

=  (sin2A + cos2A - sin A cos A)

=  (1 - sin A cos A)  -------(1)

[(sin3A - cos3A)/(sin A - cos A)]

   =  (sin A-cos A)(sin2A+cos2A+sinAcosA)/(sinA-cosA)

=  (sin2A + cos2A + sin A cos A)

=  (1 + sin A cos A)  -------(2)

(1) + (2)

  =  1 - sin A cos A + 1 + sin A cos A

  =  2

R.H.S

Hence proved.

Question 2 :

(i) If sin θ + cos θ = 3 , then prove that tan θ + cot θ = 1.

Solution :

Given that sin θ + cos θ = 

(sin θ + cos θ)2  =  3

sin2θ + cos2θ + 2sin θ cos θ  =  3

2sin θ cos θ  =  3 - 1

 sin θ cos θ  =  1

L.H.S :

tan θ + cot θ  

  =  (sin θ/cos θ) + (cos θ/sin θ)

  =  (sinθ + cos2θ)/(sin θ cos θ)

  =  1/(sin θ cos θ)

  =  1/1  

  =  1 R.H.S

Hence proved.

(ii) If 3 sin θ − cos θ = 0, then show that tan 3θ = (3 tan θ - tan3θ)/(1 - 3tan2θ)

Solution :

3 sin θ − cos θ = 0

√3 sin θ = cos θ

sin θ/cos θ  =  1/√3

tan θ  =  1/√3

θ  =  30

3θ  =  90

L.H.S :

tan 3θ = tan 90

  =  undefined

R.H.S :

  =  (3 tan θ - tan3θ)/(1 - 3tan2θ)

  =  [3(1/√3) - (1/√3)3]/[1 - 3(1/√3)2)]

  =  [√3 - (1/3√3)]/[1 - 3(1/3)]

  =  [√3 - (1/3√3)]/0

  =  undefined

After having gone through the stuff given above, we hope that the students would have understood, "Proving Trigonometric Identities Examples". 

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