If $6 tanA (tanA+1)=5- tanA$, Given that $0 <A<\frac{\pi}{2}$ what is the value of $(sinA+cosA)$?

$\frac{3}{\sqrt{5}}$

6tanA(tanA + 1 ) = 5 - tanA

6tan²A + 6tanA = 5 - taanA

6tan²A + 7tanA - 5 = 0

on solving tanA = \(\frac{1}{2}\) 

P = 1 & B = 2

P² + B² = H²

1² + 2² = H²

H = √5

Now , ( sinA + cosA )

= ( \(\frac{1}{√5}\)  + \(\frac{2}{√5}\) )

= \(\frac{3}{√5}\)