If tanalpha + cotalpha = 2, then tan20 alpha + cot20 alpha . . .

Question : If tanα + cotα = 2, then tan²⁰ α + cot²⁰α =

a) 0
b) 2
c) 20
d) 2²⁰

Doubt by Nimit

Solution :

tanα + cotα = 2
tanα + (1/tanα)= 2  
[tan²α +1]/tanα =2
tan²α +1=2tanα
tan²α - 2tanα+1=0

Let tanα = x
x²-2x+1=0
x2-2×x×1+1=0
(x-1)²=0
x-1=√0
x-1=0
x=1

tanα = 1
tanα = tan45°
α=45°

Now,
= tan²⁰ α + cot²⁰α
= tan²⁰ 45°  + cot²⁰45° 
= (tan 45°)²⁰ + (cot 45°)²⁰
= (1)²⁰ + (1)²⁰
= 1 + 1
= 2 

Hence, b) is the correct option.