Solve tan x =cot x for 0 < x < pi

tan(x) = cot(x)

tan(x) = 1 / tan(x)

tan²(x) = 1

tan²(x) - 1 = 0

(tan(x) - 1) (tan(x) + 1) = 0

tan(x) - 1 = 0   or   tan(x) + 1 = 0

tan(x) = 1   or   tan(x) = -1

x = π/4 + nπ   or   x = -π/4 + nπ

where in both cases, n is any integer.

In the interval 0 < x < π, we get solutions:

• x = π/4 from the first family (when n = 0)

• x = 3π/4 from the second family (when n = 1)

So the equation has solutions x = π/4 or x = 3π/4.

answer

point o.

step-by-step explanation:

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.