Wzor na tangens [tex]\boxed{\begin{array}{lclcl}\multicolumn{5}{c}{tg\alpha=\dfrac{sin\alpha}{cos\alpha}}\\\\\cline{1-5}\\tg^2\alpha=\dfrac{sin^2\alpha}{cos^2\alpha}&\wedge&tg^2\alpha=\dfrac{1-cos^2\alpha}{cos^2\alpha}&\wedge&tg^2\alpha=\dfrac{sin^2\alpha}{1-sin^2\alpha}\end{array}}[/tex]
Odpowiedź:
Równość nie jest tożsamością trygonometryczną
Wzory trygonometryczne - przeksztacenia
[tex]\boxed{\begin{array}{ccc}\multicolumn{=3}{c}{sin^2\alpha+cos^2\alpha=1}\\\cline{1-3}\\sin^2\alpha=1-cos^2\alpha&\wedge&cos^2\alpha=1-sin^2\alpha\end{array}}[/tex]
[tex]\boxed{\begin{array}{lclcl}\multicolumn{5}{c}{tg\alpha=\dfrac{sin\alpha}{cos\alpha}}\\\\\cline{1-5}\\tg^2\alpha=\dfrac{sin^2\alpha}{cos^2\alpha}&\wedge&tg^2\alpha=\dfrac{1-cos^2\alpha}{cos^2\alpha}&\wedge&tg^2\alpha=\dfrac{sin^2\alpha}{1-sin^2\alpha}\end{array}}[/tex]
Wzory skróconego mnożenia
[tex]\huge\boxed{\begin{array}{l}(a-b)^2=(a-b)(a-b)=a^2-2ab+b^2\\(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2\\(a-b)(a+b)=a^2-b^2\end{array}}[/tex]
Rozwiązanie:
[tex]\dfrac{cos\alpha}{1-sin\alpha}-tg\alpha=\dfrac1{cos\alpha} |^2\\\\\dfrac{cos^2\alpha}{(1-sin\alpha)^2}-\left(\dfrac{sin\alpha}{cos\alpha}\right)^2=\dfrac{1}{cos^2\alpha}\\\\\dfrac{1-sin^2\alpha}{(1-sin\alpha)^2}-\dfrac{sin^2\alpha}{cos^2\alpha}=\dfrac{1}{cos^2\alpha} |+\dfrac{sin^2\alpha}{cos^2\alpha}\\\\\dfrac{(1-sin\alpha)(1+sin\alpha)}{(1-sin\alpha)(1-sin\alpha)}=\dfrac{1+sin^2\alpha}{cos^2\alpha}\\\\\dfrac{1+sin\alpha}{1-sin\alpha} \neq \dfrac{1+sin^2\alpha}{1-sin^2\alpha}[/tex]
Verified answer
Odpowiedź:
[tex]\displaystyle \frac{cos\alpha }{1-sin\alpha } -tg\alpha =\frac{1}{cos\alpha } \quad \alpha \ne\frac{\pi }{2} +k\pi \quad \ k\in C\\L=\frac{cos\alpha }{1-sin\alpha } -tg\alpha =\frac{cos\alpha }{1-sin\alpha }-\frac{sin\alpha }{cos\alpha } =\frac{cos^{2} \alpha -sin\alpha +sin^2\alpha }{cos\alpha (1-sin\alpha )} =\frac{1-sin\alpha }{cos\alpha (1-sin\alpha )} =\\\frac{1}{cos\alpha } =P[/tex]