Question

Tan x =5/2 then find sec x and root over(sec x+1)÷(sec x-1)

Answer 1

${sec}^{2} x = 1 + {tan}^{2} x \\ {sec} \: x = \sqrt{29} \div2 \\ \\ \sqrt{ \frac{sec \: x + 1}{sec \: x - 1}} = \sqrt{ \frac{33 + 4 \sqrt{29} }{25} } \\ = \sqrt{33 + 4 \sqrt{29} } \div 5$