Question

22.7. Упростите ввражение(2;3;4)
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Answer 1

$2)\ \ \dfrac{2sin^2\alpha -1}{sin\alpha +cos\alpha }=\dfrac{2sin^2a-(sin^2a+cos^2a)}{sina+cosa}=\dfrac{sin^2a-cos^2a}{sina+cosa}=\\\\\\=\dfrac{(sina-cosa)(sina+cosa)}{sina+cosa}=sina-cosa=sina-sin(\frac{\pi}{2}-a)=\\\\\\=2\cdot sin\Big(\alpha -\dfrac{\pi}{4}\Big)\cdot cos\dfrac{\pi}{4}=\sqrt2\cdot sin\Big(\alpha -\dfrac{\pi}{4}\Big)$

$3)\ \ \dfrac{cos^2a-ctg^2a}{sin^2a-tg^3a}=\dfrac{cos^2a-\dfrac{cos^2a}{sin^2a}}{sin^2a-\dfrac{sin^3a}{cos^3a}}=\dfrac{(cos^2a\cdot sin^2a-cos^2a)\cdot cos^3a}{sin^2a\cdot (sin^2a\cdot cos^3a-sin^3a)}=\\\\\\=\dfrac{-cos^5a(1-sin^2a)}{sin^4a(cos^3a-sina)}=-\dfrac{cos^6a}{sin^4a(cos^3a-sina)}$

$4)\ \ ctg\beta +\dfrac{sin\beta }{1+cos\beta }=\dfrac{cos\beta }{sin\beta }+\dfrac{sin\beta }{1+cos\beta }=\dfrac{cos\beta +cos^2\beta +sin^2\beta }{sin\beta \, (1+cos\beta \beta )}=\\\\\\=\dfrac{cos\beta +1}{sin\beta \, (1+cos\beta )}=\dfrac{1}{sin\beta }$

$\star \ 3a)\ \ \dfrac{cos^2a-ctg^2a}{sin^2a-tg^2a}=\dfrac{cos^2a-\dfrac{cos^2a}{sin^2a}}{sin^2a-\dfrac{sin^2a}{cos^2a}}=\dfrac{(cos^2a\cdot sin^2a-cos^2a)\cdot cos^2a}{sin^2a\cdot (sin^2a\cdot cos^2a-sin^2a)}=\\\\\\=\dfrac{-cos^4a(1-sin^2a)}{-sin^4a(1-cos^2a)}=\dfrac{cos^5a}{sin^5a}=ctg^5a$