If sin A=$$\frac{8 }{17}$$​ then what is the value of cot A + sec A?

Correct option is B

Given:
sin A =$$\frac{8}{17}$$​​
Formula Used:
​$$cotA = \frac{\cos A}{\sin A}\\\ \\secA = \frac{1}{\cos A}$$​​
​$$\sin^2 A + \cos^2 A = 1$$​​
Solution:
​$$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{8}{17}\right)^2}\\\cos A = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}.$$​​

​$$\cot A = \frac{\cos A}{\sin A} = \frac{\frac{15}{17}}{\frac{8}{17}} = \frac{15}{8}.$$​​
​$$\sec A = \frac{1}{\cos A} = \frac{17}{15}.$$​​
​$$\cot A + \sec A = \frac{15}{8} + \frac{17}{15} = \frac{225 + 136}{120} = \frac{361}{120}= 3 \frac1{120}$$​​
Alternate Method:
​Sin A = $$\frac{ 8}{17} = \frac PH$$​​

As we know, 8, 15, 17 are Pythagorean triplet, then Base = 15,

Cot A + Sec A

=> $$\frac BP + \frac HB$$​​

​$$\rightarrow \frac{15}{8} + \frac{17}{15} = \frac{225 + 136}{120} = \frac{361}{120}= 3 \frac1{120}$$​​