# Prove that (cosx-sinx+1)/(cosx+sinx-1)=cosecx+cotx

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### 1 Answer

We'll manage the left side and we'll factorize both, numerator and denominator by sin x:

sin x(cos x/sin x - 1 + 1/sin x)/sin x(cos x/sin x + 1 - 1/sin x)

We'll simplify and we'll replace the fraction cos x/sin x by cot x:

(cot x - 1 + 1/sin x)/(cot x + 1 - 1/sin x)

We'll manage the right side:

cosec x = 1/sin x

cosec x + cot x = 1/sin x + cot x

Therefore, the left side, can be re-written:

(cot x - 1 + cosec x)/(cot x + 1 - cosec x)

We'll have to prove:

(cot x - 1 + cosec x)/(cot x + 1 - cosec x) = cosec x + cot x

We'll multiply both numerator and denominator by (cot x + 1 - cosec x):

(cot x + 1 - cosec x)(cot x - 1 + cosec x)/(cot x + 1 - cosec x)^2= cosec x + cot x

[(cot x)^2 - (1-cosec x)^2]/(cot x + 1 - cosec x)^2= cosec x + cot x

But (cot x)^2 = (cosec x)^2 - 1

[(cosec x)^2 - 1 - 1 + 2cosec x - (cosec x)^2]/[(cot x + 1)^2 - 2cosec x*(cot x + 1) + (cosec x)^2]

(2cosec x - 2)/[2(cosec x)^2 + 2cot x - 2cosec x*cot x - 2cosec x]

2(cosec x - 1)/2[cosec x*(cosec x - 1) - cot x(cosec x - 1) ]

(cosec x - 1)/(cosec x - 1)(cosec x- cot x)

Now, we'll re-write the identity to be demonstrated

1/(cosec x- cot x) = cosecx+cotx

1 = (cosec x- cot x) (cosecx+cotx)

1 = (cosec x)^2 - (cot x)^2

But (cosec x)^2 = 1 + (cot x)^2

**Therefore, the given idenity is verified using the formula (cosec x)^2 = 1 + (cot x)^2.**