Question

prove that (sec A -tan A)^2=1-sin A/1+sinA​

Answer 1

Answer:

1-sinA/1+sinA

Step-by-step explanation:

$(secA-tanA)^2=1-sinA/1+sinA\\\\secA=1/cosA, tanA=sinA/cosA\\=[1-sinA/cosA]^2\\=(1-sinA)^2/cos^2A\\=(1-sinA)(1-sinA)/(1-sin^2A)\\=(1-sinA)(1-sinA)/(1-sinA)(1+sinA)\\=1-sinA/1+sinA$