To study the vital trigonometric identities,

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]

= cos^{2} θ/(1 – sin θ)

= cos^{2} θ/(1 – sin θ)

= (1 – sin^{2} θ)/(1 – sin θ)

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

= 1 + sin θ

= R.H.S

Downside 2 :

Show :

√(sec^{2}θ + cosec^{2}θ) = tan θ + cot θ

L.H.S :

= √(sec^{2}θ + cosec^{2}θ)

= √(1 + tan^{2}θ + 1 + cot^{2}θ)

= √(2 + tan^{2}θ + cot^{2}θ)

= √(tanθ)^{2} + (cotθ)^{2} + 2tanθ x cotθ

= √(tanθ + cotθ)^{2}

= tanθ + cotθ

= R.H.S

Downside 3 :

Show :

(1 + cos θ – sin^{2}θ)/(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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