Question

cot x+cosecx-1÷cotx-cosecx+1=1+cosx÷sinx

Answer 1

hope this answer is helpful

Answer 2

Answer:

To show: $\frac{cot\,x+cosec\,x-1}{cot\,x-cosec\,x+1}=\frac{1+cos\,x}{sin\,x}$

Consider,

LHS

$=\frac{cot\,x+cosec\,x-1}{cot\,x-cosec\,x+1}$

$=\frac{(cot\,x+cosec\,x)-(cosec^2\,x-cot^2\,x)}{cot\,x-cosec\,x+1}$

$=\frac{(cot\,x+cosec\,x)-(cosec\,x-cot\,x)(cosec\,x+cot\,x)}{cot\,x-cosec\,x+1}$

$=\frac{(cot\,x+cosec\,x)(1-(cosec\,x-cot\,x))}{cot\,x-cosec\,x+1}$

$=\frac{(cot\,x+cosec\,x)(1-cosec\,x+cot\,x)}{cot\,x-cosec\,x+1}$

$=cot\,x+cosec\,x$

$=\frac{cos\,x}{sin\,x}+\frac{1}{sin\,x}$

$=\frac{1+cos\,x}{sin\,x}$

$=RHS$

Hence Proved