1-1 - cos A -tan 2 A - sec A -1- Prove it

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Implicit Differentiation

1+1 / cos A =...

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1+1/cosA= tan²A/secA-1. Prove it

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

1 + \frac{t a n^{2} A}{1 + s e c A} = s e c A .

\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}

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