The correct option is 0.

Question Analysis from the Image:
The provided image asks us to evaluate the following definite integral:

$$I={\int }_0^{\pi /2}\frac{1-cotx}{cscx+cosx}dx$$

Step-by-Step Derivation:

Step 1: Simplify the integrand using basic trigonometric identities
We can express the cotangent and cosecant terms in terms of sine and cosine:

$$cotx=\frac{cosx}{sinx}$$

$$cscx=\frac{1}{sinx}$$

Now, let us rewrite the numerator of the integrand:

$$1-cotx=1-\frac{cosx}{sinx}=\frac{sinx-cosx}{sinx}$$

Next, let us rewrite the denominator of the integrand:

$$cscx+cosx=\frac{1}{sinx}+cosx=\frac{1+sinxcos}{sinx}$$

Substitute these expressions back into the integrand:

$$\frac{1-cotx}{cscx+cosx}=\frac{(sinx-cosx)/sinx}{(1+sinxcosx)/sinx}=\frac{sinx-cosx}{1+sinxcosx}$$

Thus, our integral can be written as:

$$I={\int }_0^{\pi /2}\frac{sinx-cosx}{1+sinxcosx}dx$$

Step 2: Apply the definite integral reflection property
Recall the property of definite integrals:

$${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$$

Applying this property to our integral, where $a=0$ and $b=\pi /2$, we replace $x$ with $\pi /2-x$:

$$I={\int }_0^{\pi /2}\frac{sin(\pi /2-x)-cos(\pi /2-x)}{1+sin(\pi /2-x)cos(\pi /2-x)}dx$$

Using the co-function identities:

$$sin(\pi /2-x)=cosx$$

$$cos(\pi /2-x)=sinx$$

The integral simplifies to:

$$I={\int }_0^{\pi /2}\frac{cosx-sinx}{1+cosxsinx}dx$$

Factoring out a negative sign from the numerator:

$$I=-{\int }_0^{\pi /2}\frac{sinx-cosx}{1+sinxcosx}dx$$

This gives us:

$$I=-I$$

Adding $I$ to both sides:

$$2I=0\Rightarrow I=0$$

Therefore, the value of the given definite integral is 0.