How do you prove #(cos^2alpha+cotalpha)/(cos^2alpha-cotalpha)=(cos^2alphatanalpha+1)/(cos^2alphatanalpha-1)#?

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Answer 1 · 2016-10-12T16:55:32

see below

#(cos^2alpha+cotalpha)/(cos^2alpha-cotalpha)=(cos^2alphatanalpha+1)/(cos^2alphatanalpha-1)#

Left Side : #=(cos^2alpha+cotalpha)/(cos^2alpha-cotalpha)#

#=(cos^2alpha+1/tanalpha)/(cos^2alpha-1/tanalpha)#

#=((cos^2alphatan alpha+1)/tanalpha)/((cos^2alphatanalpha-1)/tanalpha)#

#=(cos^2alphatan alpha+1)/tanalpha *(tanalpha)/ (cos^2alphatanalpha-1#

#=(cos^2alphatan alpha+1)/canceltanalpha *cancel(tanalpha)/ (cos^2alphatanalpha-1#

#=(cos^2alphatan alpha+1)/(cos^2alphatanalpha-1#

#:.=# Right Side