# Prove: ln|secA+tanA|=-ln|secA-tanA|

Asked on by tobydator

lemjay | High School Teacher | (Level 2) Senior Educator

Posted on

`ln|secA+tanA| = -ln|secA-tanA|`

`ln|secA+tanA|=(-1)ln|secA-tanA|`

Re-write right side by applying the property of logarithm which is `mlna = lna^m` .

`ln|secA+tanA|=ln|secA-tanA|^(-1)`

`ln|secA+tanA|=ln|1/(secA-tanA)|`

On both sides of the equation, use the property `e^(lna) = a` .

`e^(ln|secA+tanA|) = e^(ln|1/(secA-tanA)|)`

`secA+tanA=1/(secA-tanA)`

Note that `secA= 1/(cosA) `  and  `tanA= (sinA)/(cosA)` .

`1/(cosA) + (sinA)/(cosA) = 1/(1/(cosA)-(sinA)/(cosA))`

`(1+sinA)/(cosA) = 1/((1-sinA)/(cosA))`

`(1+sinA)/(cosA) = (cosA)/(1-sinA)`

Then, cross multiply.

`(1+sinA)*(1-sinA)=cosA*(cosA)`

`1-sin^2A = cos^2A`

From the Pythagorean identity `sin^2A + cos^2A=1` , replace left side with `cos^2A` .

`cos^2A = cos^2A`    (True)

Since the resulting condition is True, this proves that `ln|sec A+tan A|=-ln|sec A-tan A|` .

We’ve answered 317,443 questions. We can answer yours, too.