Question

Prove that:<br />(Cos^2A + Tan^2A - 1)/Sin^2A=Tan^2A

Answer 1

According to the Question

= $\bf\huge\frac{cos^2A + tanA\:- 1}{sin^{2}A }$

= $\bf\huge\frac{cos^2A\:tan^2A\:-sin^2A\:-\:cos^2A}{sin^2A}$

= $\bf\huge\frac{tan^2A\: - \:sin^2A}{sin^{2}A }$

= $\bf\huge Sec^2\:A-1$

= $\bf\huge tan^2 A$

LHS = RHS