Prove That:- secA-1/secA+1 = 1-cosA/1+cosA

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\frac{s e c A - 1}{s e c A + 1} = \frac{1 - c o s A}{1 + c o s A}

\frac{1 + s e c A}{s e c A} = \frac{s i n^{2} A}{1 - c o s A}

\frac{c o t θ + c o s e c θ - 1}{c o t θ - c o s e c θ + 1} = c o s e c θ + c o t θ = \frac{1 + c o s θ}{s i n θ}

\frac{1}{s e c A - t a n A} - \frac{1}{c o s A} = \frac{1}{c o s A} - \frac{1}{s e c A + t a n A}

secA - t a n A = \frac{cos A}{1 + sin A}

prove that inA-cosA+1÷sinA+cosA-1=1÷secA-tanA using identity sec2A = 1+tan2A

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