If 6cos
2
θ+cosθ=2, 0
∘
<θ<90
∘
, then the value of (cscθ+cotθ+tanθ) is:

Given equation:
6cos
2
θ+cosθ−2=0
Let cosθ=x. The equation becomes a quadratic equation:

6x
2
+x−2=0
6x
2
+4x−3x−2=0
2x(3x+2)−1(3x+2)=0
(2x−1)(3x+2)=0
So, x=
2
1
​
or x=−
3
2
​
.
Since 0
∘
<θ<90
∘
, cosθ must be positive.

cosθ=
2
1
​
⟹θ=60
∘

Now, substituting θ=60
∘
into the required expression:

csc60
∘
+cot60
∘
+tan60
∘
=
3

​

2
​
+
3

​

1
​
+
3

​

=
3

​

2+1+3
​
=
3

​

6
​