1) sin2a/(1+cos2a) 2) (sina+2sin(pi/3-a))/(2sin(pi/6-a)-cosa) 3)(sina+cosa)^2+(sina-cosa)^2 4) (1-(sina+cosa)^2)/(sina*cosa-ctga)

Предыдущий
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вопрос

1) We can simplify sin2a/(1+cos2a) by using the double angle identity sin2θ = 2sinθcosθ and the Pythagorean identity sin^2θ + cos^2θ = 1.

sin2a = 2sinacos2a
1 + cos2a = 1 + cos^2a - sin^2a = 1 + cos^2a - (1 - cos^2a) = 2cos^2a

Therefore, sin2a/(1+cos2a) = 2sinacos2a / 2cos^2a = sin2acos2a / cos^2a = sin2a/cos2a = tan2a

2) To simplify (sina+2sin(pi/3-a))/(2sin(pi/6-a)-cosa), we can convert the sine and cosine of π/3 and π/6 using trigonometric identities sin(π/3) = √3/2, sin(π/6) = 1/2, and cos(π/6) = √3/2.

(sina+2sin(pi/3-a))/(2sin(pi/6-a)-cosa)
= (sina+2sin(π/3-a))/(2sin(π/6-a)-cosa)
= (sina + 2√3/2cosa)/(2(1/2)cosa - cosa)
= (sina + √3cosa)/(cosa - cosa)
= (sina + √3cosa)/0
= Undefined

3) (sina+cosa)^2 + (sina-cosa)^2
= sin^2a + 2sina cosa + cos^2a + sin^2a - 2sina cosa + cos^2a
= 2(sin^2a + cos^2a)
= 2(1)
= 2

4) (1 - (sina+cosa)^2)/(sinacosa - ctga)
= (1 - (sin^2a + 2sina cosa + cos^2a))/(sina cosa - ctga)
= (1 - 1)/(sina cosa - ctga)
= 0/(sina * cosa - ctga)
= 0

Therefore, the simplified forms of the given expressions are:
1) tan2a
2) Undefined
3) 2
4) 0