Question

$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\csc A+\cot A$, using the identity ${\csc }^{2}A=1+{\cot }^{2}A$

Answer 1

LHS

$=\frac{\cos A-\sin A+1}{\cos A+\sin A-1}$

$=\frac{\cot A-(1-\csc A)}{\cot A+(1-\csc A)}$

$=\frac{\cot A-(1-\csc A)}{\cot A+(1-\csc A)}\times \frac{\cot A-(1-\csc A)}{\cot A-(1-\csc A)}$

$=\frac{{[\cot A-(1-\csc A)]}^2}{{\cot }^{2}A-{(1-\csc A)}^2}$

$=\frac{{\cot }^{2}A+{(1-\csc A)}^2-2\cot A(1-\csc A)}{{\cot }^{2}A-(1+{\csc }^{2}A-2\csc A)}$

$=\frac{{\cot }^{2}A+1+{\csc }^{2}A-2\csc A-2\cot A+2\cot A\csc A}{{\cot }^{2}A-1-{\csc }^{2}A+2\csc A}$

$=\frac{{\csc }^{2}A+{\csc }^{2}A-2\csc A-2\cot A+2\cot A\csc A}{-1-1+2\csc A}$

$=\frac{2{\csc }^{2}A-2\csc A-2\cot A+2\cot A\csc A}{2\csc A-2}$

$=\frac{2\csc A(\csc A-1)+2\cot A(\csc A-1)}{2(\csc A-1)}$

$=\frac{2(\csc A-1)(\csc A+\cot A)}{2(\csc A-1)}$

$=\csc A+\cot A$

=RHS

Hence, proved