If a belongs to (0,90), B belongs to (90180) and (1 + Tana) (1 + tanb) = 2, then a + B? 2: On X inequality ix-2i + ix-3i

If a belongs to (0,90), B belongs to (90180) and (1 + Tana) (1 + tanb) = 2, then a + B? 2: On X inequality ix-2i + ix-3i

1. A + B = 225 degrees, which can be obtained by using the known equation and Tan (a + b) = (Tana + tanb) / (1-tanatanb)
2. A > 1 is enough (calculated with the coordinate axis)

Mathematics Tana = tanb (90 ° - a)

This is a function! When you go to senior one, you will be given it! His meaning is the same as Sina = SINB (90 degrees minus a)! That tanb (90 degrees minus a) is actually Tana Tana is also equal to Sina than cosa These are all from high school!

Tanb = Tana, sin2a = sin2b, how to prove a + B = 90?

sin2A-sin2B=0
Sum difference product
2sin(A-B)cos(A+B)=0
If sin (a-b) = 0, then a = B is an isosceles triangle
If cos (a + b) = 0, then a + B = 90 ° is a right triangle
So the original triangle is isosceles triangle or right triangle

5＋5²＋5³＋… ＋5²ºº²

Using the summation formula of equal ratio sequence:
Sn = A1 (Q ^ n-1) / (Q-1) (A1 is the first term, q is the common ratio)
Sn=5^0+5^1+5^2+…… +5^2002
=1×（5^2003-1）/（5-1）
=（5^2003-1）/4

2+2²+2³.+2²ºº=?

The sum of the first 100 items of the arithmetic sequence is 2, and the common ratio is 2, so S100 = 2 (1-2 ^ 100) / (1-2) = 2 ^ 101-2

If the algebraic formula (X-2) & ordm; + (4 + 2x) & sup2; is meaningful, then what is the condition that x should satisfy?

X ≠ 2, and X ≠ - half

1. (X & sup2; - Y & sup2;) - 4 (2x & sup2; - 3Y & sup2;) = () 2. If M and N are reciprocal, then the value of Mn & sup2; - (n-1) is（
3. If the value of formula 2A & sup2; + 3A + 1 is 6, find the value of formula 6A & sup2; + 9A + 5
4. Calculate x + Y / 2-x-y / 2
5. If 3A & sup2; - A-2 = 0, then 5 + 2a-6a & sup2; = ()
PS: the third question is the process,

1. -7x²+11y²
2. mn²-（n-1）=n-（n-1）=1
3. Because 2A & sup2; + 3A + 1 = 6, then 2A & sup2; + 3A = 5 is replaced as a whole
6a²+9a+5=3（2a²+3a）+5
=3*5+5
=20
4. =0
If 5.3a & sup2; - A-2 = 0, 3A & sup2; - a = 2
5+2a-6a²=5-2（3a²-a）=5-2*2
=1

If a and B are opposite to each other and m and N are reciprocal to each other, then (a + b) & sup2; + Mn =?

Because a and B are opposite numbers, a + B = 0
And because m and N are reciprocal to each other, Mn = 1
So, (a + b) & sup2; + Mn = 1

It is known that AB is opposite to each other and Mn is reciprocal to each other. B = 1, find a & sup2; - (x + y + Mn) a + (x + y) & sup2; =?

∵ B = 1, AB is opposite to each other
∴ A=-B=-1
Mn is reciprocal to each other
∴ MN=1
∴ A^2-（X+Y+MN）A+（X+Y)^2=(-1)^2-(X+Y+1)*(-1)+(X+Y)^2
=1+X+Y+1+X^2+2XY+Y^2
=X^2+2XY+Y^2+X+Y+2

If M and N are reciprocal, the value of Mn2 - (n-1) is___ ．

Because m and N are reciprocal, we can get Mn = 1, so Mn2 - (n-1) = n - (n-1) = 1