PROVING TRIGONOMETRIC IDENTITIES PRACTICE PROBLEMS

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/(1²- sin²θ)

=  √(1 - sinθ)²/(cos²θ)

=  √[(1 - sinθ)/(cosθ)]²

=  [(1 - sinθ)/(cosθ)]

=  [(1/cosθ) - (sinθ/cosθ)]

=  secθ - tanθ

=  R.H.S

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