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)(sin²A + cos²A - sinAcosA)/(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) - (1/3√3)] / [1 - 3(1/3)]

=  [(9 - 1)/3√3)] / (1 - 1)

=  [(9 - 1)/3√3)] / 0

=  undefined -----(2)

From (1) and (2), we get 

L.H.S  =  R.H.S

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