Question

Пж решите. Номер 22.3 4);5);6)

Answer 1

Ответ:

$\boxed{\ sin^2a+cos^2a=1\ \ ,\ \ \ 1+tg^2a=\dfrac{1}{cos^2a}\ \ ,\ \ \ 1+ctg^2a=\dfrac{1}{sin^2a}\ }\\\\\\4)\ \ tga=\dfrac{3}{2}\ \ ,\ \ \dfrac{sina\cdot cosa}{sin^2a-cos^2a}=\dfrac{\dfrac{sina\cdot cosa}{cos^2a}}{\dfrac{sin^2a-cos^2a}{cos^2a}}=\dfrac{tga}{tg^2a-1}=\dfrac{\dfrac{3}{2}}{\dfrac{9}{4}-1}=\\\\\\=\dfrac{3\cdot 4}{2\cdot 5}=\dfrac{6}{5}=1,2$

$5)\ \ sina=\dfrac{2}{3}\ \ ,\ \ \dfrac{1-tg^2a}{1-ctg^2a}=\dfrac{1-\dfrac{1}{ctg^2a}}{1-(\dfrac{1}{sin^2a}-1)}=\dfrac{1-\dfrac{1}{\frac{1}{sin^2a}-1}}{2-\dfrac{1}{sin^2a}}=\\\\\\=\dfrac{1-\dfrac{sin^2a}{1-sin^2a}}{\dfrac{2sin^2a-1}{sin^2a}}=\dfrac{(1-2sin^2a)\cdot sin^2a}{(1-sin^2a)(2sin^2a-1)}=\dfrac{-sin^2a}{1-sin^2a}=-\dfrac{\dfrac{4}{9}}{1-\dfrac{4}{9}}=\\\\\\=-\dfrac{4\cdot 9}{9\cdot 5}=-\dfrac{4}{5}=-0,8$

$6)\ \ cosa=-\dfrac{1}{3}\ \ ,\ \ \dfrac{1-tg^2a}{1-ctg^2a}=\dfrac{1-\dfrac{sin^2a}{cos^2a}}{1-\dfrac{cos^2a}{sin^2a}}=\dfrac{(cos^2a-sin^2a)\cdot sin^2a}{cos^2a\cdot (sin^2a-cos^2a)}=\\\\\\=\dfrac{-sin^2a}{cos^2a}=-\dfrac{(1-cos^2a)}{cos^2a}=-\dfrac{1-\dfrac{1}{9}}{\dfrac{1}{9}}=-\dfrac{8}{1}=-8$