Question

(cos 8x)/(sin8x-tg4x)

Answer 1

(cos 8x)/(sin8x-tg4x)=(cos8x)/(2sin4xcos4x-sin4x/cos4x)=(cos8x)/(sin4x(2cos4x-1/cos4x))=(cos8x)/(sin4x($2cos^{2}4x$-1)/cos4x)=(cos8xcos4x)/(sin4xcos8x)=cos4x/sin4x=ctg4x