If tan⁡θ = 4, then the value of $$\frac{(4cos⁡θ+2sin⁡θ)}{(2sin⁡θ-cos⁡θ)}$$​ is:

Correct option is B

Given:

$$\tan\theta = 4$$​​

Formula Used:

​$$\tan\theta \text{ =} \frac{\sin\theta}{\cos\theta}$$​​

Solution:

​$$\\\frac{\sin\theta}{\cos\theta} = 4 \implies \sin\theta = 4\cos\theta \\\ \\ \ \text{Now putting the values}\\ \ \\\frac{4\cos\theta + 2\sin\theta}{2\sin\theta - \cos\theta} = \frac{4\cos\theta + 2(4\cos\theta)}{2(4\cos\theta) - \cos\theta} \\\ \\= \frac{4\cos\theta + 8\cos\theta}{8\cos\theta - \cos\theta} \\\ \\= \frac{12\cos\theta}{7\cos\theta} \\\ \\= \frac{12}{7} \\$$​​