chứng minh

  • Thread starter hoangtubongdem831@yahoo.com.vn
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xuanquynh97

Ta có $sin^2xtanx+cos^2xcotx+2sinxcosx$

$=\dfrac{sin^3x}{cosx}+\dfrac{cos^3x}{sinx}+2sinxcosx$

$=\dfrac{sin^4x+cos^4x+2sin^2xcos^2x}{sinxcosx}$

$=\cfrac{(sin^2x+cos^2x)^2}{sinxcosx}$

$=\dfrac{1}{sinxcosx}=tanx+cotx$
 
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