∝ ∡ ostry , korzystam z jedynki trygonometrycznej: sin²∝ + cos²∝ = 1 oraz ze wzorów skróconego mnożenia.
[tex]c)~~L=\dfrac{1-2cos^{2} \alpha }{sin\alpha +cos\alpha } =\dfrac{sin^{2} \alpha +cos^{2} \alpha -2cos^{2} \alpha }{sin\alpha +cos\alpha } =\dfrac{sin^{2} \alpha -cos^{2} \alpha }{sin\alpha +cos\alpha } =[/tex]
[tex]=\dfrac{(sin\alpha +cos\alpha )\cdot (sin\alpha -cos\alpha ) }{(sin\alpha +cos\alpha ) } =sin\alpha +cos\alpha =P ~~~~cbdu\\\\\\1=sin^{2} \alpha +cos^{2} \alpha \\\\sin^{2} \alpha -cos^{2} \alpha =(sin\alpha -cos\alpha )\cdot (sin\alpha +cos\alpha )[/tex]
[tex]d)~~L=cos^{4} \alpha -sin^{4} \alpha =(cos^{2} \alpha )^{2} -(sin^{2} \alpha )^{2}=(cos^{2} \alpha -sin^{2} \alpha )\cdot (cos^{2} \alpha +sin^{2} \alpha )=cos^{2} \alpha -sin^{2} \alpha =P~~~~cbdu[/tex]
[tex]e)~~L=cos^{4} \alpha +sin^{4} \alpha = (cos^{2} \alpha +sin^{2} \alpha )^{2} -2sin^{2} \alpha cos^{2} \alpha =1-2sin^{2} \alpha cos^{2} \alpha =P~~~~cbdu\\\\(cos^{2} \alpha +sin^{2} \alpha )^{2}=cos^{4} \alpha + 2sin^{2} \alpha cos^{2} \alpha +sin^{4} \alpha~~\Rightarrow ~~cos^{4} \alpha +sin^{4} \alpha= (cos^{2} \alpha +sin^{2} \alpha )^{2} -2sin^{2} \alpha cos^{2} \alpha[/tex]
[tex]f)~~L=\dfrac{sin^{4} \alpha -sin^{2} \alpha }{cos^{4} \alpha -cos^{2} \alpha } =\dfrac{sin^{2} \alpha \cdot (sin^{2} \alpha -1)}{cos^{2} \alpha\cdot (cos^{2} \alpha -1)} =\dfrac{sin^{2} \alpha \cdot (-cos^{2} \alpha )}{cos^{2} \alpha\cdot (-sin^{2} \alpha )} =1=P~~~~cbdu\\\\sin^{2} \alpha +cos^{2} \alpha =1~~\Rightarrow ~~-sin^{2} \alpha =cos^{2} \alpha -1\\\\sin^{2} \alpha +cos^{2} \alpha =1~~\Rightarrow ~~-cos^{2} \alpha =sin^{2} \alpha -1[/tex]
