The correct option is C −4+√73
cosx+sinx=12
Squaring both sides,
⇒1+sin2x=14
⇒sin2x=−34
When x∈(0,π2)→2x∈(0,π)
Then sin2x→ positive
When x∈(π2,π)→2x∈(π,2π)
Then sin2x→ negative
So,
x∈(π2,π)
Now,
sin2x=−34⇒2tanx1+tan2x=−34
⇒3tan2x+8tanx+3=0
⇒tanx=−8±√64−366⇒tanx=−4±√73
As both values are negative, so both are acceptable.
Therefore the maximum possible value is,
tanx=−4+√73