If $6 tanA (tanA+1)=5- tanA$, Given that $0 <A<\frac{\pi}{2}$ what is the value of $(sinA+cosA)$?
Answer & explanation
Correct answer: option 4
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 )
= ( \(\frac{1}{√5}\) + \(\frac{2}{√5}\) )
= \(\frac{3}{√5}\)