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Proving Trigonometric Identities Examples


Instance 1 :

Show the next identities.

(i) sec4θ (1 – sin4θ) – 2tan2θ  =  1

Answer :

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

Therefore proved.

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

Answer :

(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)

Therefore proved.

Instance 2 :

Show the next identities.

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

Answer :

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

  =  0

Therefore proved.

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

Answer :

[(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

Therefore proved.

Instance 3 :

(i) If sin θ + cos θ  =  3 , then show that

tan θ + cot θ  =  1

Answer :

Given : sin θ + cos θ  =  3. 

(sin θ + cos θ)2  =  3

sin2θ + cos2θ + 2sin θ cos θ  =  3

2sin θ cos θ  =  3 – 1

sin θ cos θ  =  1

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

Therefore proved.

(ii) If 3 sin θ − cos θ  =  0, then present that

tan 3θ  =  (3 tan θ – tan3θ)/(1 – 3tan2θ)

Answer :

Given : √3 sin θ − cos θ = 0.

Then,

√3 sin θ = cos θ

sin θ/cos θ  =  1/√3

tan θ  =  1/√3

θ  =  30°

tan 3θ :

=  tan 3(30°)

=  tan 90°

=  undefined —–(1)

(3 tan θ – tan3θ)/(1 – 3tan2θ) :

=  (3tan 30° – tan330°) / (1 – 3tan230°)

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

=  [√3 – (1/3√3)] / [1 – 3(1/3)]

=  [√3 – (1/3√3)] / [1 – 1]

=  [√3 – (1/3√3)] / 0

=  undefined —–(2)

From (1) and (2), we get

tan 3θ  =  (3 tan θ – tan3θ)/(1 – 3tan2θ)

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