[tex]\frac{cosec^2\ x + 2cosec\ x - 3}{cosec^2x - 1} = \frac{cosec\ x + 3}{cosec\ x + 1}[/tex]
Solution:
Given that, we have to verify
[tex]\frac{cosec^2\ x + 2cosec\ x - 3}{cosec^2x - 1} = \frac{cosec\ x + 3}{cosec\ x + 1}[/tex]
Take the LHS
[tex]\frac{cosec^2\ x + 2cosec\ x - 3}{cosec^2x - 1}[/tex]
Use the following identity, for denominator
[tex]a^2 - b^2 = (a+b)(a-b)[/tex]
Therefore,
[tex]\frac{cosec^2\ x + 2cosec\ x - 3}{(cosecx + 1)(cosec\ x - 1)}[/tex]
Numerator can be rewritten as:
[tex]\frac{(cosec\ x +3)(cosec\ x - 1)}{(cosecx + 1)(cosec\ x - 1)}[/tex]
Cancel the common terms
[tex]\frac{(cosec\ x +3)}{(cosecx + 1)}[/tex]
Thus, LHS = RHS
Thus proved
