Question

tan2a +sin2a= 4tana÷1- tan4a​

Answer 1

Answer:

sin2A+tan2A=sin2A+sin2A/cos2A=sin(2A)(cos(2A)+1)/cos(2A)

=2sin(A)cos(A)(2cos2(A))/(cos2(A)-sin2(A))

=4sin(A)cos(A)/(1-tan2(A))

=4sin(A)cos(A)(1+tan2(A))/(1-tan4(A))

=4sin(A)cos(A)sec2(A)/(1-tan4(A))

=4tan(A)/(1-tan4(A))