OpenStudy (anonymous):

Verify the identity: ln|1+cscθ|+ln|1−cosθ|=2ln|sinθ|

5 years ago
OpenStudy (anonymous):

holy sh*t...

5 years ago
OpenStudy (slaaibak):

\[\ln{|(1-\cos \theta)(1+ \cos \theta) |} = 2\ln|\sin \theta|\] \[\ln|1-\cos^2 \theta| = \ln|\sin^2 \theta|\] 1-cos^2 x = sin^2 x Therefore: \[\ln|\sin^2 \theta| = \ln|\sin^2 \theta|\] LHS = RHS.

5 years ago
OpenStudy (anonymous):

where did the ln|(1−cosθ)(1+cosθ)| come from?

5 years ago
OpenStudy (slaaibak):

ln a + ln b = ln(ab) Logarithm laws

5 years ago
OpenStudy (anonymous):

aw. Thank you c:

5 years ago
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