Question

化简:(3/sin^20)-(1/cos^20)+64sin^20

Answer 1

(3/sin^20)-(1/cos^20)+64sin^20

=[3(cos20)^2-(sin20)^2]/(sin20)^2(cos20)^2+64(sin20)^2

=[(√3cos20+sin20)(√3cos20-sin20)]/(sin20cos20)^2+64(sin20)^2

=[2sin(60+20)*2sin(60-20)]/[(sin40)/2]^2+64(sin20)^2

=16sin80sin40/sin40*sin40+64(sin20)^2

=16sin80/sin40+64(sin20)^2

=32sin40cos40/sin40+64(sin20)^2

=32cos40+64(sin20)^2

=32[1-2(sin20)^2]+64(sin20)^2

=32