Решите пожалуйста (б,г)

Question

Решите пожалуйста (б,г)

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Answer 1

[sin(π+α)cos(2π-α)]/[tg(π-α)cos(α-π)]=[-sin(α)cos(α)]/[-tg(α)(-cos(α))]=
=[sin(α))]/[-tg(α)]=-cos(α)

[sin(π+α)sin(α+2π)]/[tg((π+α)cos(1,5π+α)]=[-sin(α)sin(α)]/[tg(α)sin(α)]=-cos(α)