1.计算:cot(-15π/4)注:括号内是四分之十五派2.证明:(tanα+secα-1)/(tanα-secα+1)=(1+sinα)/cosα3.已知:(sin^2A)/(sin^2B)+cos^2A·cos^2C=1,求证:tan^2A·cot^2B=sin^2C4.已知tan^2α=2tan^2β+1,求证:sin^2β=2sin^2α-15.证明:(1+si

Question

1.计算:cot(-15π/4) 注:括号内是四分之十五派
2.证明:(tanα+secα-1)/(tanα-secα+1)=(1+sinα)/cosα
3.已知:(sin^2A)/(sin^2B)+cos^2A·cos^2C=1,
求证:tan^2A·cot^2B=sin^2C
4.已知tan^2α=2tan^2β+1,求证:sin^2β=2sin^2α-1
5.证明:(1+sinα+cosα)/(1+sinα-cosα)+(1+sinα-cosα)/(1+sinα+cosα)=2cscα
题目中所有"/"都代表分数线!

Answer 1

1.cot(-15π/4)=cot(-15π/4+4π)=cot(π/4)=1
2.α 太难打,我用A代替了阿
左=(tanA+1/cosA-1)/(tanA-1/cosA+1)=(sinA+1-cosA)/(sinA-1+cosA)
=(2sin(A/2)cos(A/2)+1-1+2sin^2(A/2))/(2sin(A/2)cos(A/2)-1+1-2sin^2(A/2))
=(cos(A/2)+sin(A/2))/(cos(A/2)-sin(A/2))
=(cos(A/2)+sin(A/2))*(cos(A/2)+sin(A/2))/((cos(A/2)-sin(A/2))*(cos(A/2)+sin(A/2)))
=(1+sinA)/cosA
3.(sin^2A)/(sin^2B)+cos^2A·cos^2C=1
(tan^2A)/(sin^2B)+cos^2C=1/(cos^2A)=1+tan^2A
1-cos^2C=tan^2A*(1/sin^2B-1)=tan^2A*cos^2B/sin^2B
tan^2A·cot^2B=sin^2C
4.sin^2A/cos^2A=2sin^2B/cos^2B+1=(1+sin^2B)/cos^2B
sin^2A*cos^2B=cos^2A*(1+sin^2B)
sin^2B=2sin^2A-1
5.(1+sinα+cosα)/(1+sinα-cosα)=(1+2sin(A/2)cos(A/2)+2cos^2(A/2)-1)/(1+2sin(A/2)cos(A/2)-1+2sin^2(A/2))
=(tan(A/2)+1)/(tan^2(A/2)+tan(A/2))=cot(A/2)
同理,(1+sinα-cosα)/(1+sinα+cosα)=tan(A/2)
原式左=tan(A/2)+cot(A/2)=cos(A/2)/sin(A/2)+sin(A/2)/cos(A/2)=1/(sin(A/2)cos(A/2))=2cscA
下次这么多题分开问吧,答的人会多点