Question

prove that: cos²A+tan²A-1/sin²A=tan²A

Answer 1

($(Cos ^{2} A + tan^{2}A - 1 ) / Sin ^{2}A =\ \textgreater \ Cos ^{2}A/ Sin^{2}A + Tan ^{2}A/ Sin^{2}A$$-1/Sin^{2}A$$=\ \textgreater \ Cot^{2}A + Sec^[tex]=\ \textgreater \ 1 + Sec^{2}A =\ \textgreater \ Tan^{2}A${2}A - Cosec^{2}A =\ \textgreater \ { Cot^{2}A - Cosec^{2}A } + Sec^{2}A [/tex]



Hence proved that
LHS = RHS

Answer 2

put value of 1 = sin2 + cos2 and then solve the problem