Question

If sin theta=a^2-b^2/a^2+b^2,find the value of 1+tan theta cos theta

Answer 1

here

sin thita = (a^2-b^2)/(a^2+b^2)

now

1+tan theta * cos theta

=1+(sin theta/cos theta)*cos theta

=1+sin theta

=1+(a^2-b^2)/(a^2+b^2)

=(a^2+b^2+a^2-b^2)/(a^2+b^2)

=2a^2/(a^2+b^2)