\text{If}\mathrm{ 6}\sin Y+\cos Y=\sqrt{5}\sin Y\text{, then find the value of}\tan Y.

The correct option is:
$$\frac{-6-\sqrt{5}}{31}$$

Let's find the value of $\tan Y$ step-by-step from the given equation:
$$6\sin Y+\cos Y=\sqrt{5}\sin Y$$

Step 1: Divide the entire equation by $\cos Y$
To express the equation in terms of $\tan Y$, we can divide each term by $\cos Y$ (assuming $\cos Y\ne 0$):

Since $\frac{\sin Y}{\cos Y}=\tan Y$, the equation simplifies to:
$$6\tan Y+1=\sqrt{5}\tan Y$$

Step 2: Isolate the terms containing $\tan Y$
Rearrange the terms to bring all terms with $\tan Y$ to one side of the equation:
$$1=\sqrt{5}\tan Y-6\tan Y$$

Factor out $\tan Y$ on the right-hand side:
$$1=(\sqrt{5}-6)\tan Y$$

Solve for $\tan Y$:
$$\tan Y=\frac{1}{\sqrt{5}-6}$$

Step 3: Rationalize the denominator
To express the answer in standard form, multiply both the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{5}+6$:
$$\tan Y=\frac{1}{\sqrt{5}-6}·\frac{\sqrt{5}+6}{\sqrt{5}+6}$$

Using the difference of squares identity, $(a-b)(a+b)=a^2-b^2$, the denominator becomes:
$${\left(\sqrt{5}\right)}^2-6^2=5-36=-31$$

Thus, we have:
$$\tan Y=\frac{\sqrt{5}+6}{-31}$$

Move the negative sign to the numerator to match the options:
$$\tan Y=\frac{-(6+\sqrt{5})}{31}=\frac{-6-\sqrt{5}}{31}$$