Question

$\frac{ { \tan }^{2} \alpha -1 }{ \cos(2 \alpha ) } \\ \\ \frac{4 \tan( \alpha ) (1 - { \tan }^{2}) }{1 + { \tan }^{2} \alpha }$
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Answer 1

б) $\frac{tg^{2}a - 1}{cos2a } = \frac{\frac{sin^{2}a - cos^{2}a}{cos^{2}a} }{cos2a } = - \frac{\frac{cos2a}{cos^{2}a} }{cos2a} = - \frac{1}{cos^{2}a} = - 1 - tg^{2}a$