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

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