Question

8.50. Kąt a jest ostry. Zapisz dane wyrażenia w prostszej postaci. PROSZE NA SZYBKO DAJE NAJ​

Answer 1

[tex]\textrm{Wiemy:} \ \alpha < 90^{\circ} \\\\a)\ sin \alpha \cdot \frac{cos\alpha}{sin\alpha} =cos\alpha \\b)\ sin\alpha(cos^2\alpha + sin^2\alpha)=sin\alpha \\c)\ \frac{sin\alpha}{cos\alpha} \cdot \frac{1}{sin\alpha} =\frac{1}{cos\alpha} \\\\d) \ \frac{1}{sin\alpha} - cos\alpha \frac{cos\alpha}{sin\alpha} = \frac{1}{sin\alpha} (1-cos^2\alpha) = sin\alpha[/tex]

[tex]e) \ cos\alpha + cos\alpha \frac{sin^2\alpha}{cos^2\alpha} =cos\alpha+\frac{sin^2\alpha}{cos\alpha} =\frac{sin^2\alpha+cos^2\alpha}{cos\alpha}} =\frac{1}{cos\alpha} \\f) \ sin\alpha + cos\alpha\cdot\frac{cos\alpha}{sin\alpha} =sin\alpha+ \frac{cos^2\alpha}{sin\alpha} =\frac{sin^2\alpha+cos^2\alpha}{sin\alpha} =\frac{1}{sin\alpha}[/tex]

[tex]g) \ \frac{1}{sin^2\alpha} \cdot sin^2\alpha = 1\\h) \ (cos\alpha + \frac{sin\alpha}{cos\alpha} \cdot sin\alpha) \cdot\frac{cos\alpha}{sin\alpha} = (cos\alpha + \frac{sin^2\alpha}{cos\alpha} )\cdot\frac{cos\alpha}{sin\alpha}= \frac{1}{cos\alpha} \cdot\frac{cos\alpha}{sin\alpha} = \frac{1}{sin\alpha}[/tex]