Question

Solve:
$\frac{\cot A+\csc A-1}{\cot A-\csc A+1}=\frac{1+\cos A}{\sin A}$

Answer 1

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

$LHS$

$\frac{\cot A+\csc -1}{\cot A-\csc A+1}$

we know that $1={\csc }^2-{\cot }^{2}A$

put $1={\csc }^2-{\cot }^{2}A$ in numerator

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

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

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

$=\csc A+\cot A$

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

$=RHS$

Hence, proved.