PROVING TRIGONOMETRIC IDENTITIES PRACTICE PROBLEMS

To learn the important trigonometric identities,

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Abbreviations used in the problems : 

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

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

Problem 1 :

Determine whether each of the following is an identity or not.

(i) cos2θ + sec2θ  =  2 + sinθ

Solution :

L.H.S :

=  cos2θ + sec2θ

  =  1 - sin2θ + 1 + tan2θ

  =  2 + tan2θ - sin2θ

≠  R.H.S

Since L.H.S and R.H.S are not equal, the given statement is not an identity.

(ii) cot2θ + cos θ  =  sin2θ

L.H.S :

=  cot2θ + cosθ

=  cosec2θ - 1 + cosθ  ≠  R.H.S

Since L.H.S and R.H.S are not equal, the given statement is not an identity.

Problem 2 :

Prove the following identities

(i) sec2θ + cosec2θ  =  sec2θ  cosec2θ

Solution :

L.H.S :

=  sec2θ + cosec2θ

=  (1/cos2θ)+ (1/sin2θ)

=  (sin2θ + cos2θ)/(cos²θsin2θ)

=  1/(cos2θ ssin2θ)

=  (1/cos2θ)(1/sin2θ)

=  sec2θ cosec²θ

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

Solution :

L.H.S :

=  sinθ/(1 - cosθ)

Multiply both numerator and denominator by the conjugate of denominator. 

=   [sin θ(1 + cosθ)] x [(1 - cosθ)(1 + cosθ)]

Instead of (1 + cosθ)(1 + cosθ), we can write 1 - cos2θ by using the algebraic formula 

=  [sinθ (1 + cosθ)]/(1 - cos2θ)

=  [sinθ (1 + cosθ)]/sin2θ

=  (1 + cosθ)]/sinθ

=  (1/sinθ) + (cosθ/sinθ)

=  cosecθ + cotθ

(iii)  √(1 - sinθ)/(1 + sinθ)  =  secθ - tanθ

Solution :

L.H.S :

=  √(1 - sinθ)/(1 + sinθ)

=  (1 - sinθ)/(1 + sinθ) x (1 - sinθ)/(1 - sinθ)

=  (1 - sinθ)2/[(1 + sinθ) x (1 - sinθ)]

=  (1 - sinθ)2/(12sin2θ)

=  (1 - sinθ)2/(cos2θ)

=  [(1 - sinθ)/(cosθ)]2

=  [(1 - sinθ)/(cosθ)]

=  [(1/cosθ) - (sinθ/cosθ)]

=  secθ - tanθ

=  R.H.S

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