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[homerinio]

Jak to zrobić? Nic mi nie wychodzi.

a) \(\displaystyle{ \frac{sin \alpha }{1+cos \alpha } + \frac{1+cos \alpha }{sin \alpha }= \frac{2}{sin \alpha }}\)
b) \(\displaystyle{ \frac{tg \alpha }{1+tg ^{2} \alpha }=sin \alpha *cos \alpha}\)

[plocman]

a) \(\displaystyle{ cos* \frac{cos}{sin} +sin= \frac{cos^{2}}{sin} + \frac{sin^{2}}{sin} = \frac{1}{sin}}\)

[homerinio]

no dobra, na punkt b lub c ma ktoś pomysł ?

[Charles90]

b)
\(\displaystyle{ \frac{cos \alpha }{sin \alpha -1}+ \frac{sin \alpha }{cos \alpha } = \frac{cos^2 \alpha }{cos \alpha (sin \alpha -1)}+ \frac{sin \alpha (sin \alpha -1)}{cos \alpha (sin \alpha -1)}= \frac{1-sin \alpha }{cos \alpha (sin \alpha -1)}= \frac{-(sin \alpha -1)}{cos \alpha (sin \alpha -1)}=- \frac{1}{cos \alpha }}\) \(\displaystyle{ \Rightarrow L=P}\)

-- 26 października 2009, 17:26 --

c)
\(\displaystyle{ sin^{4}\alpha +cos^{4}\alpha = (sin^{2}\alpha+cos^{2}\alpha)^{2}-2sin^{2}\alpha cos^{2}\alpha = sin^4 \alpha +2sin^2 \alpha cos^2 \alpha +cos^4 \alpha -2sin^2 \alpha cos^2 \alpha =1^2-2sin^2 \alpha cos^2 \alpha =1-2sin^2 \alpha cos^2 \alpha}\) \(\displaystyle{ \Rightarrow L=P}\)

[rodzyn7773]

\(\displaystyle{ \frac{sin x }{1+cos x } + \frac{1+cos x }{sin x }= \frac{sin^2 x+1+2cos x +cos^2 x}{sin x (1+cos x)}= \frac{2(1+cos x )}{(1+cosx)sin x}= \frac{2}{sin x}}\)

-- 28 paź 2009, o 16:30 --

\(\displaystyle{ \frac{tg x }{1+tg ^{2} x }= \frac{tg x}{ \frac{cos^2 x+sin^2 x}{cos^2x} } = \frac{sin x}{cos x} * cos^2 x=sinx * cos x}\)-- 29 paź 2009, o 15:58 --\(\displaystyle{ ctg x- \frac{cos^3x}{sin x}= \frac{cos x-cos^3 x}{sin x}= \frac{cos x}{sin x}*(1-cos^2x)= \frac{cos x}{sin x}*sin^2x=cosx*sinx}\)