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