Question

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

Answer 1

Consider the left hand side,

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

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

Substituting this in the numerator of...............(1)

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

We know that, $x^2-y^2=(x+y)(x-y)$.

Applying it in the numerator,

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

Taking common,

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

Canceling like terms in numerator and denominator, we are left with,

$\csc A+\cot A$

$=\frac{1}{\sin A}+\left(\frac{\cos A}{\sin A}\right)$

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