How do you evaluate #\sqrt { \frac { 1- \cos ^ { 2} 2\theta } { 1- \sin ^ { 2} 2\theta } }#?

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Answer 1 · 2017-03-23T16:24:27

#=tan 2theta#

we will use the fact that

#cos^2x+sin^2x=1#

so we have

#cos^2x=1-sin^2x#

#sin^2x=1-cos^2x#

#sqrt((1-cos^2 2theta)/(1-sin^2 2theta))=sqrt((sin^2 2theta)/(cos^2 2theta))#

#=sqrt(tan^2 2theta)#

#=tan 2theta#