Trigonometry question?

Question

If tan theta + cot theta = 2, show tan^10 theta + cot^10 theta =2. thanks in advance.

Answer 1

Let tan theta = x

cot theta= 1/x

tan theta + cot theta= x+ 1/x = 2

x^2 + 1 = 2 x

or (x^2-2x+1) = 0

(x-1) ^2 = 0

x = 1

tan theta = cot theta = 1

tan ^10 theta + cot ^10 theta

=( 1)^10+(1)^10

= 2

tan ^10 theta + cot ^10 theta = 2

Answer 2

cosθ

Let tan θ = x

cot θ= 1/x

tan θ + cot θ= x+ 1/x = 2

x² + 1 = 2 x

or (x²-2x+1) = 0

(x-1)² = 0

(x-1)(x-1) = 0

x-1 = 0 ,x-1 = 0

x = 1

tan θ = cot θ = 1

tan¹⁰ θ + cot ¹⁰ θ

=( 1)¹⁰+(1)¹⁰

= 2

tan ¹⁰ θ + cot ¹⁰ θ = 2

There fore          tan ¹⁰ θ + cot ¹⁰ θ = 2

Answer 3

tanθ + cotθ = 2

tanθ + 1/tanθ = 2

(tanθ×tanθ + 1) / tanθ = 2

tan² θ + 1 = 2 tanθ

tan² θ - 2 tanθ + 1 = 0               [ This is in the form (a - b)² which is equal to a²- 2ab+b² ]

(tanθ - 1)² =0

(tanθ - 1)(tanθ - 1) = 0

tanθ - 1 = 0

tanθ = 1

=> cotθ = 1/tanθ = 1/1 = 1

Therefore tan¹⁰ θ + cot¹⁰ θ = (tanθ)² +(cotθ)² = 1¹⁰ + 1¹⁰ = 1+1 = 2.