If 16 cot A = 12, then find the value of $$\tfrac{sinA+cosA}{sinA-cosA}$$​.

Correct option is A

Given:

16 cot A = 12

​$$\frac{\sin A + \cos A}{\sin A - \cos A}$$​​

Concept Used:

Using the identity:

​$$\cot A = \frac{\cos A}{\sin A}$$​​

Solution:  

$$\cot A = \frac{12}{16} = \frac{3}{4}\\\ \\\cos A = \frac{3}{5}, \quad \sin A = \frac{4}{5}\\\ \\\text{Since, } \sin^2 A + \cos^2 A = 1\\\ \\\left(4k\right)^2 + \left(3k\right)^2 = 1\\\ \\16k^2 + 9k^2 = 1\\\ \\25k^2 = 1\\\ \\k^2 = \frac{1}{25}, \quad k = \frac{1}{5}\\\ \\\sin A = \frac{4}{5}, \quad \cos A = \frac{3}{5}\\\ \\\sin A + \cos A = \frac{4}{5} + \frac{3}{5} = \frac{7}{5}\\\ \\\sin A - \cos A = \frac{4}{5} - \frac{3}{5} = \frac{1}{5}\\\ \\ \frac{sinA + cosA}{sinA - cosA} =\frac{\frac{7}{5}}{\frac{1}{5}} = 7 \\\ \\\boxed{\text{Option (A) is right.}}\\\ \\$$​

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​Alternate Method 2:​

$$\cot A = \frac{12}{16} = \frac{3}{4}$$

Then,   $$\sin A = \frac{4}{5}, \quad \cos A = \frac{3}{5}$$​​

$$\frac{\sin A + \cos A}{\sin A - \cos A} = \frac{\frac{4}{5} + \frac{3}{5}}{\frac{4}{5} - \frac{3}{5}}$$=$$\frac{\frac{7}{5}}{\frac{1}{5}}$$=7

Option (A) is right.