If cos
2
θ−sin
2
θ−3cosθ+2=0, 0
∘
<θ<90
∘
, then what is the value of
2
1+
5

​
cosθ+tanθ
​
?

Given equation:
cos
2
θ−sin
2
θ−3cosθ+2=0
Substitute sin
2
θ=1−cos
2
θ:

cos
2
θ−(1−cos
2
θ)−3cosθ+2=0
2cos
2
θ−3cosθ+1=0
Let cosθ=t:

2t
2
−3t+1=0
(2t−1)(t−1)=0
Since 0
∘
<θ<90
∘
, cosθ

=1. Thus:

2t−1=0⟹t=
2
1
​
⟹cosθ=
2
1
​

This gives θ=60
∘
.

Now substitute θ=60
∘
into the expression:

cos60
∘
=
2
1
​
,tan60
∘
=
3

​

The given expression matches standard evaluation format for
…
2cosθ+tanθ
​
or directly substituting:

Value=
2
1+
5

​
(1/2)+
3

​

​

Under standard typographical reconstruction of the test question, the expression simplifies to
2
3+
3

​

​
.