Confirm the Identity: log(1-sin x) + log(1+sinx) = 2 log(cosx)

Confirm the Identity: log(1-sin x) + log(1+sinx) = 2 log(cosx)

log(1-sinx) + log(1 + sinx)

log((1-sinx)(1 + sinx))

log(1 -sin^2x)

log(cos^2x)

log(cosx)^2

2log(cosx)

By the rule log(a) + log(b) = log(ab), we see that the left side becomes:
LHS = log[1 - sin(x)] + log[1 + sin(x)]
= log{[1 - sin(x)][1 + sin(x)]}
= log[1 - sin^2(x)], via difference of squares
= log[cos^2(x)], since sin^2(x) + cos^2(x) = 1
= 2log|cos(x)|, since ln(a^b) = b*ln(a)
= RHS.

I hope this helps!