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 $$\frac{√3+1}{2}$$