Prove that sec A-1sec A-1-1-cos A1-cos A

Take LHS of ( A 1)( A + 1) = ( 1 cos A)( 1 + cos A) ( A 1)( A + 1) = 1cos A 11cos A + 1 = 1 cos Acos A1 + cos Acos A = 1 cos ...

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Basics of Geometry

Prove that ...

Question

Prove that $\frac{(s e c A - 1)}{(s e c A + 1)} = \frac{(1 - c o s A)}{(1 + c o s A)}$

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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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