Given sin α + cos α = √ 2 / 2 F (x) = x + 1 / x, what is f (Tan α) equal to The process and thinking of solving problems

Given sin α + cos α = √ 2 / 2 F (x) = x + 1 / x, what is f (Tan α) equal to The process and thinking of solving problems

sinα+cosα=√2/2
(sinα+cosα)²=1/2
1+2sinacosa = 1/2
sinacosa = -1/4
f(tanα)
= tana+1/tana
=(sin²a+cos²a) / sinacosa
=1/sinacosa
= -4

If M = two and one seventh, n = four and two thirds, then M: one third=

45:98

We know that Tan ^ 2 α = 2tan ^ β + 1, and prove that sin ^ 2 β = asin ^ 2 α - 1
sin^2β= tan^2α-1 / 1+tan^2α
How did this come from,

Because sin and 178; a / COS and 178; a = 2Sin and 178; B / COS and 178; B + 1
So Sin & # 178; a / cos & # 178; a = (2Sin & # 178; B + cos & # 178; b) / cos & # 178; b
So Sin & # 178; a / cos & # 178; a = (Sin & # 178; B + 1) / cos & # 178; b
So Sin & # 178; a / cos & # 178; a = (Sin & # 178; B + 1) / (1-sin & # 178; b)
So sin and 178; a - sin and 178; asin and 178; b = COS and 178; asin and 178; B + COS and 178; a
So (cos-178; a + sin-178; a) sin-178; b = sin-178; a-cos-178; a
So Sin & # 178; b = Sin & # 178; a-cos & # 178; a
So Sin & # 178; b = (2Sin & # 178; a-SiN & # 178; a) - cos & # 178; a
So sin and#178; b = 2Sin and#178; a-SiN and#178; a-cos and#178; a
So Sin & # 178; b = 2Sin & # 178; a - (Sin & # 178; a + cos & # 178; a)
So Sin & # 178; b = 2Sin & # 178; A-1

A simplest true fraction is obtained by expanding its denominator by four times and reducing its numerator by two times. 11 out of 112 are obtained. Find the original fraction? Why

It's easy
If the denominator is expanded four times, the score value is reduced four times, the numerator is reduced two times, and the score value is reduced two times
Here, the score value is reduced by 2 * 4 = 8 times, that is to say, 8 times of 11 / 112 is it
11/112 * 8 =11/14
It's 11 out of 14

A fraction is a true fraction whose numerator and denominator are less than 10
A fraction is a true fraction whose numerator and denominator are less than 10. If you enlarge its numerator by five times and reduce its denominator by four times, you will get 13 and 1 / 13. What is the original true fraction?
(speedup telegram, urgent, due tomorrow)

The question is wrong. According to this question, the answer is 17 / 26, which does not meet the requirements

A simplest fraction, the sum of numerator and denominator is 10, has () fractions
A. 2B. 3C. 4

According to the stem analysis: the sum of numerator and denominator is 10, the simplest score is 19 and 37, a total of 2

The numerator of a true fraction is four times larger and the denominator is three times smaller than the original fraction ()

Let the original fraction be B / A,
Four times the molecular size is 4B,
The denominator is three times smaller than a / 3,
Then the new number is 12b / A,
The new fraction divided by the original fraction is 12 times the original fraction

If the numerator of a true fraction is expanded four times and the denominator is reduced three times, is the fraction smaller or larger than the original fraction?

Larger than the original is equivalent to multiplying by 4 * 3 = 12 > 1
Generally, true scores are discussed in the scope of certificates, so they are much larger