If a ∈ (0, π 2) and sin2 α + Cos2 α = 14, then the value of Tan α is equal to______ .

If a ∈ (0, π 2) and sin2 α + Cos2 α = 14, then the value of Tan α is equal to______ .

∵ sin2 α + Cos2 α = 14, ∵ 1 − Cos2 α 2 + Cos2 α = 14, the solution is Cos2 α = - 12, and α ∈ (0, π 2), ∵ 2 α ∈ (0, π), ∵ 2 α = 2 π 3, α = π 3, ∵ Tan α = 3
-How to reduce SiNx / cosx to 1 / cos ^ 2T
You should make the title clear
1 ^ A + 2 ^ A + 3 ^ A +. + n ^ a = sum
Especially when a is greater than 4
When a = 0, n ones are added, and the result is n
When a = 1, 1 + 2 +... + n = (1 + n) n / 2
When a = 2, 1 ^ 2 + 2 ^ 2 +... + n ^ 2 = n (n + 1) (2n + 1) / 6
I don't know the rest
Four decimal operations in Grade 5: 71.5 + 11 × 71.51-51.11
If we first calculate 11 × 71.51, the result is 786.61
If we add 11 × 71.51 to 71.5, that is 71.5 + 786.61, we get 858.11
Finally, we calculate 71.5 + 11 × 71.51-51.11, that is 858.11-51.11, and get 807
Multiplication, addition and subtraction, 807.01
This is a simple algorithm that's a little hidden. Only by careful observation can we find the decimal rule behind. Correct must adopt!!
71.5+11×71.51-51.11
=71.5+11×（71.5+0.01）-51.11
=71.5+11×71.5+0.11-51.11
=71.5×（1+11）-51
=71.5×12-51
=858-51
=807
This is a simple algorithm that's a little hidden. ... unfold
This is a simple algorithm that's a little hidden. Only by careful observation can we find the decimal rule behind. Correct must adopt!!
71.5+11×71.51-51.11
=71.5+11×（71.5+0.01）-51.11
=71.5+11×71.5+0.11-51.11
=71.5×（1+11）-51
=71.5×12-51
=858-51
=807
This is a simple algorithm that's a little hidden. Only by careful observation can we find the decimal rule behind. Put it away
Four mixed operations of decimal and integer
60 decimal and 40 integer
(0.75+0.2)/0.25*25% 12/0.75+7.2/2.4 605*8+3.5-44 10.9-(6.6+0.125/12.5%) 980-9.8)*0.6-2.12 (0.125*8-0.5)*4 1+0.45/0.9-0.75 168.1/(4.3*2-0.4) 1.21*42-(4.46+0.14) 1375+450/18*25 18/1.5-0.5*0.3 1.9-1.9*(1...
Yes, I can help you, but no problem?
What about the title?
How to check the division
To judge whether quotient multiplied by divisor plus remainder is equal to divisor
How to sum 1 / N (n + 1) (n + 2)
1/n(n+1)(n+2)
=1/2[2/n(n+1)(n+2)]
=1/2[(n+2)-n]/n(n+1)(n+2)]
=(1/2)[(n+2)/n(n+1)(n+2)-n/n(n+1)(n+2)]
=(1/2)[1/n(n+1)-1/(n+1)(n+2)]
So sum = (1 / 2) [1 / 1 * 2-1 / 2 * 3 + 1 / 2 * 3-1 / 3 * 4 + +1/n(n+1)-1/(n+1)(n+2)]
=(1/2)[1/1*2-1/(n+1)(n+2)]
=n(n+3)/[4(n+1)(n+2)]
1/n(n+1)(n+2)=1/2n-1/(n+1)+1/2(n+2)
Add up:
a1=1/2*1-1/2+1/2*3
a2=1/2*2-1/3+1/2*4
a3=1/2*3-1/4+1/2*5
…………………………
an=1/2n-1/(n+1)+1/2(n+2)
Sn=1/4+1/2(n+1)-1/(n+1)+1/2*(n+2)
=1/4-1/2(n+1)(n+2)
As shown in the picture
Grade 5 decimal four operations, three calculations, to be fast!
Just 10
（2×50）÷4＋25=100
25.86+74.14-100
There should be decimal addition and subtraction, simple calculation (Law of combination of addition, nature of subtraction, law of combination of multiplication, nature of division, law of distribution of multiplication, law of splitting, missing items), rate of progression (example: 4.4cm = (44mm)). It should be 7.5cm
1.25＋300÷52.（25＋300）÷53.25＋75÷5＋754.200－（35＋48÷24）5.（72＋32×4）÷406.45＋55－45＋557.2－2×0＋28.10－10÷10×109.（75＋25÷5）－4010.2800÷25×4＋8011.2800÷25×4＋8012.（380－80×4）...
50.4－19.64＝
I'll see if I can get the result of minus 19.76
--! You mean you want answers
thirty point seven six