Question

Довести тождество
$1. \frac{\sin^{2}(\pi-\alpha)}{tg^{2}(\pi+\alpha)}*\frac{ctg(\frac{\pi}{2}-\alpha)}{tg(\frac{\pi}{2}+\alpha)} = -\sin^{2}\alpha\\ 2. \frac{tg(\pi-\alpha)}{\cos(\pi+\alpha)}*\frac{\sin(\frac{3\pi}{2}+\alpha)}{tg(\frac{3\pi}{2}+\alpha)}=tg^{2}\alpha$

Answer 1

Объяснение:

1.

$\frac{sin^2(\pi -\alpha )}{tg^2(\pi +\alpha )}*\frac{ctg(\frac{\pi }{2} -\alpha )}{tg(\frac{\pi }{2}+\alpha ) } =\frac{sin^2\alpha }{tg^2\alpha } *\frac{tg\alpha }{-ctg\alpha } =-\frac{sin^2\alpha }{\frac{sin^2\alpha }{cos^2\alpha } } *\frac{\frac{sin\alpha }{cos\alpha } }{ \frac{cos\alpha }{sin\alpha } } = -cos^2\alpha *\frac{sin^2\alpha }{cos^2\alpha }=-sin^2\alpha .$

2.

$\frac{tg(\pi -\alpha )}{cos(\pi +a)} *\frac{sin(\frac{3\pi }{2} +\alpha )}{tg(\frac{3\pi }{2} +a)}=\frac{-tg\alpha }{-cos\alpha }*\frac{-cos\alpha }{-ctg\alpha }=\frac{\frac{sin\alpha }{cos\alpha } }{cos\alpha }*\frac{cos\alpha }{\frac{cos\alpha }{sin\alpha } }=\frac{sin\alpha }{cos^2\alpha }*sin\alpha =\frac{sin^2\alpha }{cos^2\alpha }=tg^2\alpha .$

Answer 2

Ответ:

1.

$\frac{ { \sin}^{2} (\pi - \alpha) }{ {tg}^{2}( \pi + \alpha )} \times \frac{ctg( \frac{\pi}{2} - \alpha )}{tg( \frac{\pi}{2} + \alpha ) } = \\= \frac{ { \sin }^{2}( \alpha ) }{ {tg}^{2} \alpha } \times \frac{tg \alpha }{( - ctg \alpha )} = \\ = \frac{ { \sin}^{2} \alpha }{tg \alpha } \times ( - \frac{1}{ctg \alpha } ) = - \frac{ { \sin}^{2} \alpha }{tg \alpha } \times tg \alpha = - { \sin }^{2} \alpha$

2.

$\frac{tg(\pi - \alpha )}{ \cos(\pi + \alpha ) } \times \frac{ \sin( \frac{3\pi}{2} + \alpha ) }{tg( \frac{3\pi}{2} + \alpha ) } = \\= \frac{ - tg \alpha }{ - \cos( \alpha ) } \times \frac{( - \cos( \alpha )) }{( - ctg \alpha )} = \\= \frac{tg \alpha }{ctg \alpha } = tg \alpha \times tg \alpha = {tg}^{2} \alpha$