Question #c20a4
2 Answers
Apr 4, 2017
I tried this:
Explanation:
Have a look:
Apr 4, 2017
#(tanx+cotx)/(tanx-cotx)=sec^2x/(tan^2x-1)#
Multiply the left-hand side by the conjugate of the denominator:
#(tanx+cotx)/(tanx-cotx)=(tanx+cotx)/(tanx-cotx)*(tanx+cotx)/(tanx+cotx)#
#=(tan^2x+2tanxcotx+cot^2x)/(tan^2x-cot^2x)#
Rewriting
#=(tan^2x+2tanx(1/tanx)+1/tan^2x)/(tan^2x-1/tan^2x)#
Multiplying through by
#=(tan^4x+2tan^2x+1)/(tan^4x-1)#
Factoring:
#=(tan^2x+1)^2/((tan^2x+1)(tan^2x-1))=(tan^2x+1)/(tan^2x-1)#
The numerator is a form of the Pythagorean identity:
#=sec^2x/(tan^2x-1)#
This is the right-hand side of the original equation, so the identity has been verified.