Prove that sin^2A/cos^2A + cos^2A/sin^2A = 1/cos^2A*sin^2A - 2

Answer:prove that:Sin²A/Cos²A + Cos²A/Sin²A = 1/Cos²A Sin²A - 2LHS = \frac{Sin^2A}{Cos^2A} + \frac{Cos^2A}{Sin^2A} Cos 2 ASin 2 A + Sin 2 ACos 2 A = \begin{lgathered}= \frac{Sin^4A + Cos^4A}{Cos^2A . Sin^2A}\\\\Using\: a^2 + b^2 = (a+b)^2 - 2ab\\\\a = Cos^2A \: \& \:b = Sin^2A\\\\= \frac{(Sin^2A + Cos^2A)^2 - 2Sin^2A Cos^2A}{Cos^2A Sin^2A} \\\\Sin^2A + Cos^2A = 1\\\\= \frac{1 -2Sin^2A Cos^2A}{Cos^2A Sin^2A}\end{lgathered} = Cos 2 A.Sin 2 ASin 4 A+Cos 4 A Usinga 2 +b 2 =(a+b) 2 −2aba=Cos 2 A&b=Sin 2 A= Cos 2 ASin 2 A(Sin 2 A+Cos 2 A) 2 −2Sin 2 ACos 2 A Sin 2 A+Cos 2 A=1= Cos 2 ASin 2 A1−2Sin 2 ACos 2 A \begin{lgathered}= \frac{1}{Cos^2A Sin^2A} - 2\\\\= RHS\end{lgathered} = Cos 2 ASin 2 A1 −2=RHS LHS=RHS