Question

A) 0,5 + sin’a + cos²a; ə) 2 - sin’a - cosa;6) 2sin’a + cos-a - 1;B) (1 - sin a)(1 + sin a);1) tgʻa - sin’a – tgʻasin’a;F) sinta + cos-a-costa.​

Answer 1

1) 0,5 + Sin²α + Cos²α = 0,5 + (Sin²α + Cos²α) = 0,5 + 1 = 1,5

2) 2 - Sin²α - Cos²α = 2 - (Sin²α + Cos²α ) = 2 - 1 = 1

3) 2Sin²α + Cos²α - 1 = 2Sin²α - Sin²α = Sin²α

4) (1 - Sinα)(1 + Sinα) = 1 - Sin²α = Cos²α

$5)tg^{2} \alpha-Sin^{2} \alpha-tg\alpha ^{2} Sin^{2}\alpha=(tg^{2}\alpha-tg^{2}\alpha Sin^{2}\alpha)-Sin^{2}\alpha=\\\\=tg^{2}\alpha(1-Sin^{2} \alpha)-Sin^{2}\alpha= \frac{Sin^{2}\alpha}{Cos^{2}\alpha}*Cos^{2}\alpha-Sin^{2}\alpha=Sin^{2} \alpha-Sin^{2}\alpha=0$

Последнее задание неверно записано

Answer 2

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