If tan2θ+cot2θ = 2, where θ is an acute angle, the value of tan6θ+cot6θ is equal to . . . .

Question : If tan²θ+cot²θ = 2, where θ is an acute angle, the value of tan⁶θ+cot⁶θ is equal to . . . . 

a)4
b)8
c)3
d)2

Doubt by Muskan

Solution :
tan²θ+1/tan²θ = 2
(tan⁴θ+1)/tan²θ = 2
tan⁴θ+1=2tan²θ
tan⁴θ+1-2tan²θ = 0
(tan²θ)2+(1)²-2tan²θ =0
(tan²θ-1)²=0
tan²θ-1=0
tan2θ=1
tanθ=√1
tanθ=1
tanθ=tan45°
θ=45°

Now,
tan⁶θ+cot⁶θ
= (tanθ)⁶+(cotθ)⁶
=(tan45°)⁶+(cot45°)⁶
=1⁶+1⁶
=1+1
=2

∴ d) is the correct option.