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

1 Answer
Oct 12, 2016

see below

Explanation:

#(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