SAMPLE PROBLEMS OF PROVING TRIGONOMETRIC IDENTITIES

Problem 1 :

Prove :

cos θ/(sec θ - tan θ)  =  1 + sin θ

Solution :

L.H.S :

  =  cos θ/(sec θ - tan θ)

  =  cos θ/[1/cos θ - (sin θ/cos θ)]

  =  cos θ/[(1 - sin θ)/cos θ]

  =  cos² θ/(1 - sin θ)

  =  cos² θ/(1 - sin θ)

  =  (1 - sin² θ)/(1 - sin θ)

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

  =  1 + sin θ

=  R.H.S

Problem 2 :

Prove :

√(sec²θ + cosec²θ)  =  tan θ + cot θ

L.H.S :

=  √(sec²θ + cosec²θ)

  =  √(1 + tan²θ + 1 + cot²θ)

  =  √(2 + tan²θ + cot²θ)

  =  √(tanθ)² + (cotθ)² + 2tanθ x cotθ

  =  √(tanθ + cotθ)²

  =  tanθ + cotθ

=  R.H.S

Problem 3 :

Prove :

(1 + cos θ - sin²θ)/(sin θ)(1 + cos θ)  =  cot θ

Solution :

L.H.S :

  =  (1 + cos θ - sin²θ)/(sin θ)(1+cosθ)

  =  (1 + cos θ) - (1 - cos²θ)/(sin θ)(1+cosθ)

  =  [(1 + cos θ) - (1 - cos θ)(1 + cos θ)]/(sin θ)(1+cosθ)

  =  [(1 + cos θ) (1 - (1 - cos θ))]/(sin θ)(1+cosθ)

taking (1 + cos θ) as common term

  =  [(1 + cos θ) (1 - 1 + cos θ))]/(sin θ)(1 + cosθ)

  =  [(1 + cos θ) (cos θ)]/(sin θ)(1 + cosθ)

  =  cos θ/sin θ

  =  cot θ

= R.H.S

Problem 4 :

secθ(1 - sinθ)(secθ + tanθ)  =  1

Solution :

L.H.S :

Problem 5 : 

sinθ/(cosecθ + cotθ)  =  1 - cosθ

Solution :

L.H.S :

=  sinθ/(cosecθ + cotθ)

=  sin θ/[(1/sin θ) + (cos θ/sin θ)]

=  sin θ/[(1+cosθ)/sin θ]

=  (sin θ x sin θ)/(1+cosθ)

= sin²θ/(1+cosθ)

=  1- cos²θ/(1+cosθ)

=  (1- cosθ)(1+cosθ)/(1+cosθ)

= 1 - cos θ

=  R.H.S

Problem 6 :

Prove :

[sin(90 - θ)/(1 + sinθ)] + [cosθ/(1 - (cos(90 - θ))]  =  2secθ

Solution :

L.H.S :

  =  [sin(90 - θ)/(1 + sinθ)] + [cosθ/(1 - (cos(90 - θ))]

We can write sin (90-θ) as cos θ and cos (90 - θ) as sin θ.

  =  2/cosθ

  =  2secθ

=  R.H.S

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