Prove ((cscx-cotx)(cscx+cotx))/secx=cosx

Question

How to prove that this identity is true?

Answer 1

Remeber the formula (a-b)(a+b) = a^2-b^2

Left handside identity =

= (Cscx-Cotx)(Cscx+Cotx)/Secx

= (Csc^2x-Cot^2x)/Secx

= 1/Secx

= Cosx

= Right hand side identity