How do you establish the identity for the following trigonometric equation?

Question

#sec^2u+csc^2u=secucscu#

3545 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-19T17:12:40

this identity is false, see below

#sec^2x+csc^2x#

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

#=(cos^2x+sin^2x)/(cos^2xsin^2x)=1/(cos^2xsin^2x)#

#=1/cos^2x*1/sin^2x#

#sec^2xcsc^2x#