What is the value of sinθ+cosθ, if θ satisfies the equation $$cot^2θ$$ -(√3+1)cotθ+√3=0 ; 0<θ<π/4?

$$cot^2θ-3cotθ-cotθ+3=0$$

⇒

$$cotθ(cotθ-3)-(cotθ-3)=0$$

⇒

$$(cotθ-3)(cotθ-1)=0$$

⇒

$$cotθ=3,1 but  θ  \epsilon  0<θ<π/4$$

$$∴ cotθ=3  at  θ  \epsilon  π/6$$

$$Now, sinθ+cosθ  at  θ =π/6,$$

$$= sin(π/6)+cos(π/6)$$

$$= 1/2 +\sqrt{3}/2$$  

Hence the answer (C) is

$$\frac{√3+1}{2}$$