Question

Зная, что tgα + ctgα = 2,5, найдите tg2 α + ctg2а.

Answer 1

           1-й способ:
tg2a + ctg2a = 5
потому что если первое уравнение домножить на два(и правую и левую сторону) получится:
2*tga + 2*ctga = 2.5 *2
tg2a + ctg2a = 5.
            2-й способ:
tg2a + ctg2a - ?, можем вынести за скобки 2
2(tga + ctga) - ?
мы знаем что tga + ctga = 2.5, значит
2tga + 2ctga = 5

Answer 2

To find the value of tg(2α) + ctg(2α), we can use trigonometric identities and algebraic manipulations.

Let's start by using the double-angle formulas for tangent and cotangent:

1. tg(2α) = 2 * tg(α) / (1 - tg²(α))
2. ctg(2α) = ctg²(α) - 1 / 2 * ctg(α)

Given tg(α) + ctg(α) = 2.5, we can rewrite it as:

tg(α) + (ctg²(α) - 1 / tg(α)) = 2.5

To simplify the expression, we need to find ctg²(α) in terms of tg(α). Let's solve for ctg²(α):

ctg²(α) = 2.5 * tg(α) + 1 - tg²(α)

Now, let's find tg(2α) + ctg(2α) using the double-angle formulas:

tg(2α) + ctg(2α) = 2 * tg(α) / (1 - tg²(α)) + (ctg²(α) - 1) / (2 * tg(α))

Now, substitute ctg²(α) = 2.5 * tg(α) + 1 - tg²(α):

tg(2α) + ctg(2α) = 2 * tg(α) / (1 - tg²(α)) + (2.5 * tg(α) + 1 - tg²(α) - 1) / (2 * tg(α))

Simplify the expression:

tg(2α) + ctg(2α) = 2 * tg(α) / (1 - tg²(α)) + (2.5 * tg(α) - tg²(α)) / (2 * tg(α))

Now, find a common denominator for the fractions:

tg(2α) + ctg(2α) = [2 * tg(α) * 2 * tg(α) + (2.5 * tg(α) - tg²(α)) * (1 - tg²(α))] / [2 * tg(α) * (1 - tg²(α))]

tg(2α) + ctg(2α) = [4 * tg²(α) + (2.5 * tg(α) - tg²(α) - tg²(α) + tg⁴(α))] / [2 * tg(α) * (1 - tg²(α))]

Now, combine like terms:

tg(2α) + ctg(2α) = [tg⁴(α) + 2 * tg²(α) + 2.5 * tg(α)] / [2 * tg(α) * (1 - tg²(α))]

Now, we can use the given information tg(α) + ctg(α) = 2.5 to further simplify:

tg(2α) + ctg(2α) = [tg⁴(α) + 2 * tg²(α) + 2 * (tg(α) + ctg(α))] / [2 * tg(α) * (1 - tg²(α))]

tg(2α) + ctg(2α) = [tg⁴(α) + 2 * tg²(α) + 2 * 2.5] / [2 * tg(α) * (1 - tg²(α))]

tg(2α) + ctg(2α) = [tg⁴(α) + 2 * tg²(α) + 5] / [2 * tg(α) * (1 - tg²(α))]

Unfortunately, without a specific value for tg(α), we cannot find the exact numerical value of tg(2α) + ctg(2α). However, you can use the above expression if you have the value of tg(α).