Question

Упростить выражение: а) tg²α*ctg²α-cos²α;
б) sin²α+cos²α+tg²α;
в) (1-sinα)(1+sinα);
г) cos²α-1/cos α sin α

Answer 1

а

${tg}^{2} \alpha \times {ctg}^{2} \alpha - { \cos }^{2} \alpha = \\ = \frac{1}{ {tg}^{2} \alpha } \times {tg}^{2} \alpha - { \cos }^{2} \alpha = \\ = 1 - { \cos }^{2} \alpha = { \sin}^{2} \alpha$

б

${ \sin }^{2} \alpha + { \cos }^{2} \alpha + {tg}^{2} \alpha = \\ = 1 + {tg}^{2} \alpha = 1 + \frac{ { \sin }^{2} \alpha }{ { \cos}^{2} \alpha } = \\ = \frac{ { \cos }^{2} \alpha + { \sin}^{2} \alpha }{ { \cos }^{2} \alpha } = \frac{1} { { \cos }^{2} \alpha }$

в

$(1 - \sin( \alpha ) )(1 + \sin( \alpha ) ) = \\ = {1}^{2} - { \sin }^{2} \alpha = { \cos }^{2} \alpha$

г

$\frac{{ \cos }^{2} \alpha - 1}{ \cos( \alpha ) \sin( \alpha ) } = \frac{ - \sin^{2}( \alpha ) }{ \cos( \alpha ) \sin( \alpha ) } =\\=-\frac{\sin(\alpha)}{\cos(\alpha)}=-tg(\alpha)$