Prove that:
1 + secA - tanA / 1 + secA + tanA = 1- sinA / cosA
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Answer:
(1+secA-tanA)/(1+secA+tanA)= (1-sinA)/cosA.
L.H.S.
={(sec^2A-tan^2A)+(secA-tanA)}/(1+secA+tanA).
={(secA-tanA).(secA+tanA)+(secA-tanA)}/(1+secA+tanA).
=(secA-tanA).(secA+tanA+1)/(1+secA+tanA).
=secA-tanA.
=1/cosA-sinA/cosA.
=(1-sinA)/cosA. Proved.
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