SAMPLE PROBLEMS IN TRIGONOMETRIC IDENTITIES

Abbreviations used :

L.H.S -----> Left hand side

R.H.S -----> Right hand side

Problem 1 :

Prove :

sec4θ (1 - sin4θ) - 2tan2θ  =  1  

Solution :

L.H.S :

  =  sec4θ (1 - sin4θ) - 2tan2θ

  =  (1/cos4θ) (1 - (sin2θ)2) - 2tan2θ

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

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

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

  =  [(1 + sin2θ - 2sin2θ)/cos2θ]

  =  [(1 - sin2θ)/cos2θ]

  =  [cos2θ/cos2θ]

  =  1

=  R.H.S

Problem 2 :

Prove :

(cotθ - cosθ)/(cotθ + cosθ)  =  (cosecθ - 1)/(cosecθ + 1)

Solution :

L.H.S :

=  (cotθ - cosθ)/(cotθ + cosθ)

=  [(cosθ/sinθ) - cosθ] / [(cosθ/sinθ) + cosθ]

=  [cosθ (1/sinθ) - 1) / cosθ(1/sinθ) + 1)]

=  [(1/sinθ) - 1)/(1/sinθ) + 1)]

=  (cosecθ - 1)/(cosecθ + 1)

=  R.H.S

Problem 3 :

Prove :

[(sinA - sinB)/(cosA + cosB)] + [(cosA - cosB)/sinA + sin B)] 

=  0

Solution :

L.H.S :

=  [(sinA - sinB)/(cosA + cosB)]

+ [(cosA - cosB)/sinA + sin B)]

=  [(sinA - sinB)(sinA + sinB) + (cosA - cosB)(cosA - cosB)]

(cosA + cosB)(sinA + sinB)

=  [(sin2A - sin2B) + (cos2A - cos2B)]

(cosA + cosB)(sinA + sinB) -----(1)

Numerator :

  =  (sin2A - sin2B) + (1 - sin2A - (1-sin2B)

  =  sin2A - sin2B + 1 - sin2A - 1 + sin2B

  =  0

Substitute the value of numerator in (1).

(1)----->  =  0/(cos A + cos B)(sin A + sin B)

  =  0

=  R.H.S

Problem 4 : 

Prove :

[(sin3A + cos3A)/(sinA + cosA)] + [(sin3A - cos3A)/(sinA - cosA)] 

=  2

Solution :

L.H.S :

Part 1 :

=  (sin3A + cos3A)/(sinA + cosA)

=  (sinA + cosA)(sin2cos2sinAcosA)/(sinA + cosA]

=  (1 - sinAcosA) ----(1)

Part 2 :

=  [(sin3A - cos3A)/(sinA - cosA)]

 =  (sinA - cosA)(sin2A + cos2A + sinAcosA)/(sinA + cosA)

  =  (1 + sinAcosA) ----(2)

(1) + (2) 

=   (1 - sinAcosA) +  (1 + sinAcosA)

=  1 + 1 

=  2

=  R.H.S

Problem 5 :

Prove :

If sinθ + cosθ  =  3 , then prove that

tanθ + cotθ  =  1

Solution :

sinθ + cosθ  =  3

Square both sides.

(sinθ + cosθ)2  =  32

sin2θ + cos2θ + 2sinθcosθ  =  3

1 + 2sinθcosθ  =  3

2sinθcosθ  =  2

sinθcosθ  =  1  ----(1)

tanθ + cotθ  =  (sinθ/cosθ) + (cosθ/sinθ)

  =  (sin2θ + cos2θ)/sinθ cos θ

Using (1), we get

  =  1/1

  =  1

Problem 6 :

If 3sinθ - cosθ = 0, then show that

tan3θ  =  (3tanθ - tan3θ)/(1 - 3tan2θ) 

Solution :

3sinθ − cosθ  =  0

3sinθ  =  cosθ

sinθ/cosθ  =  1/3

tanθ  =  1/3

θ  =  30°

L.H.S : 

=  tan 3θ

=  tan 3(30°)

=  tan 90°

=  undefined -----(1)

R.H.S :

=  (3tanθ - tan3θ)/(1 - 3tan2θ)

=  [3(1/3) - (1/3)3] / [1 - 3(1/3)2]

=  [(3/3) - (1/33)] / [1 - 3(1/3)]

=  [(9 - 1)/33)] / (1 - 1)

=  [(9 - 1)/33)] / 0

=  undefined -----(2)

From (1) and (2), we get 

L.H.S  =  R.H.S

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