Question

Prove a validade das seguintes identidades:

1.
$- ln| \sec( \alpha ) - \tan( \alpha ) | = ln| \sec( \alpha ) + \tan( \alpha )|$
2.
$- ln| \csc( \alpha ) + \cot( \alpha ) | = ln| \csc( \alpha ) - \cot( \alpha ) |$
​

Answer 1

1.

$-\ln\left[sec(\alpha) - tan(\alpha)\right] = -\ln\left[\dfrac{1}{cos(\alpha)} - \dfrac{sin(\alpha)}{cos(\alpha)}\right]$

$-\ln\left[sec(\alpha) - tan(\alpha)\right] = -\ln\left[\dfrac{1-sin(\alpha)}{cos(\alpha)}\right]$

Aplicando a propriedade:

$c \cdot \log_{a}[b] = \log_{a}[b]^c$

Reescrevemos como:

$-\ln\left[sec(\alpha) - tan(\alpha)\right] = \ln\left[\dfrac{1-sin(\alpha)}{cos(\alpha)}\right]^{-1}$

$-\ln\left[sec(\alpha) - tan(\alpha)\right] = \ln\left[\dfrac{cos(\alpha)}{1-sin(\alpha)}\right]$

Multiplicando numerador e denominador da fração pelo conjugado do denominador:

$-\ln\left[sec(\alpha) - tan(\alpha)\right] = \ln\left[\dfrac{cos(\alpha)}{1-sin(\alpha)} \cdot \dfrac{1+sin(\alpha)}{1+sin(\alpha)}\right]$

$-\ln\left[sec(\alpha) - tan(\alpha)\right] = \ln\left[\dfrac{cos(\alpha) + cos(\alpha) \cdot sin(\alpha)}{1-sin^2(\alpha)}\right]$

Sabendo que: $sin^2(\alpha) + cos^2(\alpha) = 1$, então: $cos^2(\alpha) = 1 - sin^2(\alpha)$. Fazendo essa substituição:

$-\ln\left[sec(\alpha) - tan(\alpha)\right] = \ln\left[\dfrac{cos(\alpha) + cos(\alpha) \cdot sin(\alpha)}{cos^2(\alpha)}\right]$

$-\ln\left[sec(\alpha) - tan(\alpha)\right] = \ln\left[\dfrac{1 + sin(\alpha)}{cos(\alpha)}\right]$

$\boxed{-\ln\left[sec(\alpha) - tan(\alpha)\right] = \ln\left[sec(\alpha) + tan(\alpha)\right]}$

2. Mesma ideia da primeira:

$-\ln\left[cossec(\alpha) + cotg(\alpha)\right] = -\ln\left[\dfrac{1}{sin(\alpha)} + \dfrac{cos(\alpha)}{sin(\alpha)}\right]$

$-\ln\left[cossec(\alpha) + cotg(\alpha)\right] = -\ln\left[\dfrac{1+cos(\alpha)}{sin(\alpha)}\right]$

$-\ln\left[cossec(\alpha) + cotg(\alpha)\right]= \ln\left[\dfrac{1+cos(\alpha)}{sin(\alpha)}\right]^{-1}$

$-\ln\left[cossec(\alpha) + cotg(\alpha)\right]= \ln\left[\dfrac{sin(\alpha)}{1+cos(\alpha)}\right]$

$-\ln\left[cossec(\alpha) + cotg(\alpha)\right]= \ln\left[\dfrac{sin(\alpha)}{1+cos(\alpha)} \cdot \dfrac{1-cos(\alpha)}{1-cos(\alpha)}\right]$

$-\ln\left[cossec(\alpha) + cotg(\alpha)\right]= \ln\left[\dfrac{sin(\alpha) - sin(\alpha) \cdot cos(\alpha)}{1-cos^2(\alpha)}\right]$

Sabendo que: $sin^2(\alpha) + cos^2(\alpha) = 1$, então: $sin^2(\alpha) = 1 - cos^2(\alpha)$. Fazendo essa substituição:

$-\ln\left[cossec(\alpha) + cotg(\alpha)\right] = \ln\left[\dfrac{sin(\alpha) - sin(\alpha) \cdot cos(\alpha)}{sin^2(\alpha)}\right]$

$-\ln\left[cossec(\alpha) + cotg(\alpha)\right]= \ln\left[\dfrac{1 - cos(\alpha)}{sin(\alpha)}\right]$

$\boxed{-\ln\left[cossec(\alpha) + cotg(\alpha)\right]= \ln\left[cossec(\alpha) - cotg(\alpha)\right]}$