Question

Задание видно......................

Answer 1

Ответ:

${ctg}^{2} \alpha - {tg}^{2} \alpha = \frac{ { \cos }^{2} \alpha }{ { \sin }^{2} \alpha } - \frac{ { \sin }^{2} \alpha }{ { \cos}^{2} \alpha } = \\ = \frac{ { \cos }^{4} \alpha - { \sin}^{4} \alpha }{ { \sin}^{2} \alpha \times { \cos }^{2} \alpha } \times \frac{4}{4} = \\ = \frac{4( { \cos }^{2} \alpha - { \sin}^{2} \alpha )( { \sin }^{2} \alpha + { \cos}^{2} \alpha ) }{4 { \sin }^{2} \alpha \times { \cos}^{2} \alpha } = \\ = \frac{4 \times \cos( 2\alpha ) \times 1 }{ { \sin}^{2}(2 \alpha ) } = \frac{4 \cos( 2\alpha ) }{ { \sin }^{2} (2 \alpha) }$