Question

If tanA + cotA = 2. Find tan^n A + cot^n A = ???

Answer 1

Given Equation is tan A + cot A = 2 can be written as

We know that cot theta = 1/tan theta.

                               tan A + 1/tan A = 2

                               tan^2 A + 1 = 2 tan A

                               tan^2 A + 1 - 2 tan A = 0

We know that (a-b)^2 = a^2 + b^2 - 2 * a * b.

                               (tan A - 1)^2 = 0

                                tan A = 1.

                                cot A = 1/tan A = 1/1 = 1


Given tan^n A + cot^n A

         = (tan A)^n + (cot A)^n

We know that 1^a = 1.

         = 1^n + 1^n

         = 1 + 1

         = 2.


Therefore tan^n A + cot^n A = 2.


Hope this helps!

Answer 2

$tanA + cotA= 2 \\ tanA + \frac{1}{tanA} = 2 \\ {tan}^{2} A + 1- 2tanA = 0 \\ ( {tanA- 1})^{2} = 0 \\ tanA= 1 \\ \\ now \: \: {tan}^{n} A + {cot}^{n} A \\ = {1}^{n} + {1}^{n} \\ = 1 + 1 = 2$

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