sin(540°+α)cos(360°-α)/sin(450°+α)tan(900°-α) Hurry!

sin(540°+α)cos(360°-α)/sin(450°+α)tan(900°-α) Hurry!

sin(540°+α)cos(360°-α)/[sin(450°+α)tan(900°-α)]
=sin(180°+α)cosα/[sin(90°+α)tan(180°-α)]
=-sinαcosα/[-cosα.tanα]
=-sinαcosα/[-sinα]
=cosα

Simplification of COS (α - π) Tan (α - 2 π) Tan (2 π - α) / sin (π + α)

cos(α -π )tan(α-2π)tan(2π-α)/sin(π+α)
=[-cos(α)]tan(α)[-tan(α)]/[-sin(α)]
=tan(α)[-tan(α)]/[tan(α)]
= -tan(α)

Simplify sin (- α) cos (2 PAI - α) by sin squared (PI - α) + Tan (- α)

Simplify sin (- α) cos (2 PAI - α) by sin squared (PI - α) + Tan (- α)
=sin²α+sinα*cosα/tanα
=sin²α+cos²α=1

Cos ^ 2 α - Sin ^ 2 α / cos ^ 2 α + sin ^ 2 α = 1-tan ^ 2 α / 1 + Tan ^ 2 α, which means square, how to simplify it

Divide the numerator and denominator by cosa ^ 2 at the same time
cos^2α-sin^2α/cos^2α+sin^2α
=1-tana^2/1+tana^2

If Tan α = - 2, then sin squared α + sin α cos α =?

(sina)²+sinacosa
=sin²a+sinacosa
=(sin²a+sinacosa)/(sin²a+cos²a)
=(tan²a+tana)/(tan²a+1)
=(2²+2)/(2²+1)
=6/5
=Six out of five

Tan (π - α) * sin ^ 2 (α + π / 2) * cos (2 π - α) / cos ^ 3 (- α - π) * Tan (2 π - α) sin ^ 2 is the square of sin

tan(π-α)=-tanα
sin(α+π/2)=-cosα
cos(2π-α)=cosα
cos(-α-π)=-cosα
tan(2π-α)=--tanα
Substitution = - 1

Tan α = 2, the square of (sin α + cos α), the process

0

It was proved that Tan A / 2 = (1-cos) / sin a = sin a / (COS a + 1)

Cos (A / 2) is not equal to 0
tan(a/2) = sin(a/2)/cos(a/2) = 2sin(a/2)cos(a/2)/{2[cos(a/2)]^2}
= sin(a)/[1 + cos(a)]
When sin (A / 2) is not equal to 0,
tan(a/2) = sin(a/2)/cos(a/2) = 2[sin(a/2)]^2/[2sin(a/2)cos(a/2)]
= [1 - cos(a)]/sin(a)

The results showed that: (1-sin α + cos α) / (1 + sin α + cos α) = Tan (π / 4 - α / 2)

It is proved that: (1-sin α + cos α) / (1 + sin α + cos α) = [sin ^ 2 (A / 2) + cos ^ 2 (A / 2) - 2 Sina / 2cosa / 2 + cos ^ 2 (A / 2) - Sin ^ 2 (A / 2) - Sin ^ 2 (A / 2) - Sin ^ 2 (A / 2) - Sin ^ 2 (A / 2) - Sin ^ 2 (A / 2) + 2sina / 2cosa / / 2 + cos ^ 2 (A / 2) - Sin ^ 2 (A / 2) - sin ^ 2 (A / 2) - Sin ^ 2 (A / 2) - Sin ^ 2 (A / 2) - 2 (A / 2) - 2 (A / 2) - 2 (A / 2) - 2 (A / 2) - 2sina / 2cosa / 2 / 2 / [2C...It's a good idea

It was proved that (1 + sin α / 1-sin α) = (1 / cos α + Tan α) ^ 2

It is proved that the left side = 1 + sin α / 1-sin α
Right = (1 / cos α + Tan α) ^ 2
=(1/cosα+sinα/cosα)^2
=(1+sinα/cosα) ^2
=(1+sinα) ^2/(cosα)^2
=(1+sinα) ^2/1-(sinα)^2
=(1+sinα) ^2/(1+sinα)(1-sinα)
=1+sinα/1-sinα
=Left
That's right