If 3sec
4
θ+8=10sec
2
θ, then the value of tanθ can be:

Let sec
2
θ=x. The given equation becomes:

3x
2
−10x+8=0
Let's solve this quadratic equation:

3x
2
−6x−4x+8=0
3x(x−2)−4(x−2)=0
(3x−4)(x−2)=0
This gives two possible values for x:

x=2⟹sec
2
θ=2

x=
3
4
​
⟹sec
2
θ=
3
4
​


We know the trigonometric identity: tan
2
θ=sec
2
θ−1.

Case 1: tan
2
θ=2−1=1⟹tanθ=1

Case 2: tan
2
θ=
3
4
​
−1=
3
1
​
⟹tanθ=
3

​

1
​


Thus, tanθ can be 1 or
3

​

1
​
.