Given that $tan\; B\; =\; cot(4B\; -\; 45°)$, find the value of $B$.

The correct answer is 27°.

We are given the trigonometric equation:

$\tan \left(B\right)=\cot (4B-\mathrm{45°})$

Recall the co-function identity which states that tangent and cotangent are complementary functions. Specifically:

$\tan \left(\theta \right)=\cot (\mathrm{90°}-\theta )$

Therefore, if $\tan \left(X\right)=\cot \left(Y\right)$, the angles $X$ and $Y$ must add up to $\mathrm{90°}$:

$X+Y=\mathrm{90°}$

Here, setting $X=B$ and $Y=4B-\mathrm{45°}$, we can write the equation as:

$B+(4B-\mathrm{45°})=\mathrm{90°}$

Combine the like terms on the left side of the equation:

$5B-\mathrm{45°}=\mathrm{90°}$

Add $\mathrm{45°}$ to both sides of the equation:

$5B=\mathrm{90°}+\mathrm{45°}$

$5B=\mathrm{135°}$

Finally, divide both sides by $5$ to solve for $B$:

$B=\frac{\mathrm{135°}}{5}$

$B=\mathrm{27°}$

Thus, the value of $B$ is 27°.