Question

prove that sec theta +tan theta /sec theta -tan theta =cos theta / 1-sin theta whole square​

Answer 1

Answer:

I am using '$'as theta

=(sec$+tan$)/(sec$-tan$)

=(sec$+tan$)/(sec$-tan$)*(sec$-tan$)/(sec$-tan$)

=(sec^2$-tan^2$)/(sec$-tan$)^2

=1/(1/cos$+sin$/cos$)^2

=1/(1+sin$/cos$)^2

=(cos$/1+sin$)^2