几道送分题化简(1-cotα+cscα)(1-tanα+secα)求证sin^2α+sin^2β-sin^2α*sin^2β+cos^2α*cos^2β=1求证sin^2x*tanx+cos^2x*cotx+2sinx*cosx=tanx+cotx

Question

几道送分题
化简(1-cotα+cscα)(1-tanα+secα)
求证sin^2α+sin^2β-sin^2α*sin^2β+cos^2α*cos^2β=1
求证sin^2x*tanx+cos^2x*cotx+2sinx*cosx=tanx+cotx

Answer 1

用a代替α
(1-cotα+cscα)(1-tanα+secα)
=(1-cosa/sina+1/sina)(1-sina/cosa+1/cosa)
=[(sina-cosa+1)/sina][(cosa-sina+1)/(cosa]
=[1+(sina-cosa)][1-(sina-cosa)]/(sinacosa)
=[1-(sina-cosa)^2]/(sinacosa)
=[1-(sina)^2-(cosa)^2+2sinacosa/(sinacosa)
=(1-1+2sinacosa)/(sinacosa)
=2
用a代替α,b代替β
左边=sin^2α+sin^2β-sin^2α*sin^2β+cos^2α*cos^2β
=(sin^2a-sin^2asin^2b)+sin^2b+cos^2acos^2b
=sin^2a(1-sin^2b)+sin^2b+cos^2acos^2b
=sin^2acos^2b+sin^2b+cos^2acos^2b
=cos^2b(sin^2a+cos^2a)+sin^2b
=cos^2b*1+sin^2b
=1=右边
左边=sin^2x*tanx+cos^2x*cotx+2sinx*cosx
=sin^3x/cosx+cos^3x/sinx+2sinxcosx
=(sin^4x+cos^4x)/sinxcosx+2sinxcosx
=[(sin^2x+cos^2x)^2-2sin^2xcos^2x]/sinxcosx+2sinxcosx
=(1-2sin^2xcos^2x)/sinxcosx+2sin^2xcos^2x/sinxcosx
=(1-2sin^2xcos^2x+2sin^2xcos^2x)/sinxcosx
=1/sinxcosx
右边=sinx/cosx+cosx/sinx
=(sin^2x+cos^2x)/sinxcosx
=1/sinxcosx
所以左边=右边