In a $\mathrm{\Delta }ABC$, right angled at $B$ if $\tan C=\sqrt{3}$ then find $\frac{{\sin }^{2}C+{\cos }^{2}C}{1+{\cot }^{2}C}$

The correct option is:
$\frac{3}{4}$

Step-by-Step Explanation:

We are given a triangle ABC, right-angled at B. We are also given the value of the tangent of angle C:
$\tan C=\sqrt{3}$

We need to find the value of the following expression:
$\frac{{\sin }^{2}C+{\cos }^{2}C}{1+{\cot }^{2}C}$

Let us simplify the numerator and the denominator separately using fundamental trigonometric identities.

Step 1: Simplify the numerator
Recall the standard trigonometric Pythagorean identity:
${\sin }^{2}C+{\cos }^{2}C=1$

Thus, the numerator of our expression is simply equal to 1.

Step 2: Simplify the denominator
We know that the cotangent of an angle is the reciprocal of its tangent:
$\cot C=\frac{1}{\tan C}$

Substituting the given value of $\tan C=\sqrt{3}$, we get:
$\cot C=\frac{1}{\sqrt{3}}$

Now, let us calculate the denominator $1+{\cot }^{2}C$:
$1+{\cot }^{2}C=1+{\left(\frac{1}{\sqrt{3}}\right)}^2$
$1+{\cot }^{2}C=1+\frac{1}{3}$
$1+{\cot }^{2}C=\frac{4}{3}$

Step 3: Evaluate the final expression
Substitute the values we found for the numerator and the denominator back into the main expression:
$\frac{{\sin }^{2}C+{\cos }^{2}C}{1+{\cot }^{2}C}=\frac{1}{\left(\frac{4}{3}\right)}$
$\frac{{\sin }^{2}C+{\cos }^{2}C}{1+{\cot }^{2}C}=\frac{3}{4}$

Thus, the value of the given expression is $\frac{3}{4}$.