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Prove ln(secθ + tanθ) + ln(secθ - tanθ) = 0?

0 votes
When secθ = 1/cosθ and tanθ = sinθ/cosθ
asked Dec 26, 2012 in TRIGONOMETRY by johnkelly Apprentice

2 Answers

0 votes

ln(secθ + tanθ) + ln(secθ - tanθ) = 0

First take left hand side

= ln(secθ + tanθ) + ln(secθ - tanθ)

= ln((secθ + tanθ)(secθ - tanθ))

= ln(sec^2θ - tan^θ)

= When secθ = 1/cosθ and tanθ = sinθ/cosθ

= ln(1/cosθ ^2θ -sin^2θ/cos^2θ)

= ln(1 -sin^2θ/ cos^2θ)

= ln(1 -sin^2θ/ cos^2θ)

Using trigonometric formula sin^2θ+cos^2θ=1

= ln(cos^2θ/ cos^2θ)

= ln(1)

= 0

ln(secθ + tanθ) + ln(secθ - tanθ) = 0

answered Dec 26, 2012 by ashokavf Scholar
0 votes

ln(secθ + tanθ) + ln(secθ - tanθ) = 0

First take left hand side

= ln(secθ + tanθ) + ln(secθ - tanθ)

= ln((secθ + tanθ)(secθ - tanθ))

= ln(sec^2θ - tan^2θ)

= When secθ = 1/cosθ and tanθ = sinθ/cosθ

= ln(1/cosθ ^2θ -sin^2θ/cos^2θ)

= ln(1 -sin^2θ)/( cos^2θ)

= ln(1 -sin^2θ/ cos^2θ)

Using trigonometric formula sin^2θ+cos^2θ=1

= ln(cos^2θ/ cos^2θ)

= ln(1)

= 0

ln(secθ + tanθ) + ln(secθ - tanθ) = 0

 

answered Dec 29, 2012 by krish Pupil

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