Question

If tan theta + cot theta = 2, then find the value of tan^100 theta + cot^100 theta.​

Answer 1

Answer:

tan Ø + cot Ø = 2

We know that: cot Ø = 1/tan Ø

=> tan Ø + 1/tan Ø = 2

=> tan^2 Ø + 1 = 2tan Ø

=> tan^2 Ø - 2tan Ø + 1 = 0

Now, treat tan Ø as a variable, such that tan Ø = x

=> x^2 - 2x + 1 = 0

=> x^2 - x - x + 1 = 0

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

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

=> x = 1

Thus, tan Ø = 1

This is only possible when Ø = 45°

tan^100 Ø + cot^100 Ø = (1)^100 + (1)^100 = 1 + 1 = 2

Your answer is 2.

Thank you!

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Answer 2

If tan theta + cot theta = 2, then find the value of tan^100 theta + cot^100 theta.​

Solution -

tanθ + cotθ = 2  

We know that  ,

 (   cot θ = 1 / tan θ )                  

tanθ + 1 / tanθ = 2                                

$tan^{2}$θ + 1 = 2tanθ

$tan^{2}$θ - 2tanθ + 1 = 0                     ∴$(a - b)^{2} = a^{2} + b^{2} - 2ab$

( tanθ - 1 )² = 0

 tanθ = 1

cot θ =  1/ tanθ

         =  1/1

        = 1

$(tan)^100$θ + $(cot)^100$θ = 1 + 1 ⇒ 2

so, the answer of this question is  2.

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