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Pattern Issues of Proving Trigonometric Identities


To study the vital trigonometric identities,

please click on right here

Abbreviations used within the issues : 

* L.H.S —–> Left hand facet

* R.H.S —–> Proper hand facet

Downside 1 :

Show :

cos θ/(sec θ – tan θ)  =  1 + sin θ

Answer :

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

Downside 2 :

Show :

√(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

Downside 3 :

Show :

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

Answer :

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 widespread time period

  =  [(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

Downside 4 :

secθ(1 – sinθ)(secθ + tanθ)  =  1

Answer :

L.H.S :

Downside 5 : 

sinθ/(cosecθ + cotθ)  =  1 – cosθ

Answer :

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

Downside 6 :

Show :

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

Answer :

L.H.S :

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

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

  =  2/cosθ

  =  2secθ

=  R.H.S

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