If $$\frac{(\sec⁡A+ \tan⁡A)}{(\sec⁡A-\tan⁡A)} = 2\frac{51}{79}$$​, then the value of sin⁡A is equal to:

Correct option is B

Given:
​$$\frac{\sec A + \tan A}{\sec A - \tan A} = 2 \frac{51}{79} = \frac{209}{79}$$​​

Formula Used:

​$$(\sec A + \tan A)(\sec A - \tan A) = \sec^2 A - \tan^2 A = 1$$​​
​​​​$$\sec A = \frac{1}{\cos A}$$

$$\tan A = \frac{\sin A}{\cos A}$$​​​

Solution:​​

Let’s directly compute $$\sin A$$​ using the identity:

​$$\ \frac{\sec A + \tan A}{\sec A - \tan A} = \frac{1 + \sin A}{1 - \sin A} = \frac{209}{79}$$​

​$$(1 + \sin A) \times 79 = (1 - \sin A) \times 209$$​

​$$79 + 79 \sin A = 209 - 209 \sin A$$​

​$$79 \sin A + 209 \sin A = 209 – 79$$​

​$$288 \sin A = 130$$​

​$$\sin A = \frac{130}{288} = \frac{65}{144}$$​​
Thus, the correct option is (b) $$\frac{65}{144}$$​​