Question

express 1-cosA+sinA÷1+cos A+sin A in terms of tan a÷2​

Answer 1

Answer:

question is wrong. denominator should be cosA +sinA -1

(1-cosA +siA)/(-1+cos A+sinA)=divide numerator and denominator by cosA

= secA -1 +tanA/(-secA +1 + tanA)=

(secA +tanA) -1/(-secA+1 +tanA)=

{(secA +tanA) - (sec^2A - tan^2A)}/(-sdcA +1 + tanA) = {(secA+ tanA)(1 - secA +tanA)}/(1-secA + tanA) =

secA + tanA = (1+sinA)/cos A = (sinA/2 + cosA/2)^2/(cos^2A/2 - sin^2A/2)=

(sinA/2 + cosA/2)/(cosA/2 - sinA/2)

Divide numerator and denominator by cosA/2

=(tanA/2+1)/(tanA/2 - 1) Ans