∫ (1 / (SiNx + cosx)) DX, the integral interval is 0 to Pai / 2, it is better to use the universal formula and sin (x + Pai / 4) two methods

∫ (1 / (SiNx + cosx)) DX, the integral interval is 0 to Pai / 2, it is better to use the universal formula and sin (x + Pai / 4) two methods

∫dx/(sinx+cosx)
=∫dx/[√2sin(x+π/4)]
=(-1/√2)∫dcos(x+π/4)/[(1-cos(x+π/4))(1+cos(x+π/4))]
=(-1/√2)ln|1+cos(x+π/4)|/|sin(x+π/4| +C

Tana + COTA = Tan & sup2; a = cot & sup2; a, find Tan & sup3; a + cot & sup3; a

tan³a+cot³a
=(tana+cota)tan^2a
=(tana+cota)^2
=tan^2a+cot^2a+2
=2tana+2cota+2

Given 1 / (tana-1) = 1, find the value of 1 / (1 + sinacosa)

Tana = 2, the original formula = (Sina ^ 2 + cosa ^ 2) / (Sina ^ 2 + cosa ^ 2 + sinacosa) = (Tana ^ 2 + 1) / (Tana ^ 2 + 1 + Tana) = 5 / 7

Given Tana = - 3, find the value of sinacosa

Find the hidden 1 = Sin & # 178; a + cos & # 178; a, and then divide the numerator and denominator by cos & # 178; a
sinacosa/sin²a+cos²a=tana/（tan²a+1）=-3/10

Sin (π / 4 + α) sin (π / 4 - α) = 1 / 4 find 2Sin ^ 2 (a) - 1 + tana-1 / Tan

Sin (π / 4 + a) cos [π / 2 - (π / 4-A)] = 1 / 42sin (π / 4 + a) cos (π / 4 + a) = 1 / 2Sin [2 (π / 4 + a)] = 1 / 2Sin (π / 2 + 2a) = 1 / 2cos2a = 1 / 2Sin & sup2; 2A + cos & sup2; 2A = 1, so sin2a = ± √ 3 / 2tana-1 / Tana = Sina / cosa cosa / Sina = (Sin & sup2; a-cos & sup2; a)

If a is an acute angle and Tana + COTA = 2, what is Tana COTA?

The square of both sides leads to tan ^ 2A + cot ^ 2A + 2 = 4, so (Tana COTA) ^ 2 = Tan ^ 2A + cot ^ 2a-2 = 0
So Tana COTA = 0

Given that ∠ A is an acute angle, COTA = 2 / 4 root, then Tana=___

Tana = 1 / COTA = 2 times root 2

Tana + COTA = 9 / 4 (a is an acute angle) for Sina + cosa =?

Tana + COTA = 9 / 4tana + 1 / Tana = 9 / 4, Tana = 4 / 5sin2a = 2tana / (1 + Tana * Tana) = 40 / 41 (Sina + COSA) ^ 2 = 1 + 2sinacosa = 81 / 41, Sina + cosa = √ 81 / 41 or - √ 81 / 41, and a is an acute angle

It is known that in the triangle ABC, Tana and tanb are the two roots of the equation x square + 8x + 3 = 0. Find the value of Tan (a + b) and the value of COSC

Weida theorem: Tana + tanb = - 8, Tana * tanb = 3
Tan (a + b) = (Tana + tanb) / (1-tana * tanb) = - 8 / (1-3) = 4 > 0, a + B is the acute angle
tan^(A+B)+1=1/cos^2(A+B)=17
cos(A+B)=sqrt(17)/17

Tana = - 1 / 3, cos (a + b) = - 12 / 13, a, B belong to (π / 2, π)
Tan α = - 1 / 3, cos (α + β) = - 12 / 13, α, β belong to (π / 2, π) to find cos β

α ∈ (π / 2, π), Tan α = - 1 / 3, then sin α = 1 / √ 10, cos α = - 3 / √ 10, α + β ∈ (π, 2 π), cos (α + β) = - 12 / 13, then sin (α + β) = - 5 / 13, cos β = cos (α + β - α) = cos (α + β) cos α + sin (α + β) sin α = - 12 / 13 × (- 3 / √ 10) ＋ (- 5 / 13) × 1 / √ 10 = 31 /