Simplify. 1+i/1-I + 1/ 1 + i. if i^2 = -1

I guess you mean

(1 + i ) / (1 - i ) + 1 / (1 + i )

in which case, notice that the denominator of both fractions are conjugates of one another. Combine the fractions by multiplying either fraction by the appropriate conjugate:

[(1 + i ) / (1 - i ) • (1 + i ) / (1 + i )]+ [1 / (1 + i ) • (1 - i ) / (1 - i )]

In either denominator, we have a difference of squares:

(1 + i ) (1 - i ) = 1² + i - i - i ² = 1² - i ² = 1 + 1 = 2

→   (1 + i )² / 2 + (1 - i ) / 2

→   ((1 + i )² + (1 - i )) / 2

Expand the numerator:

(1 + i )² + (1 - i ) = (1² + 2i + i ²) + 1 - i = 1 + 2 - 1 + 1 - i = 3 - i

and so

(1 + i ) / (1 - i ) + 1 / (1 + i ) = (3 - i ) / 2

∠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.