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1) ¨Ò¡ÃÙ» ÁØÁ ADC = 90° ,ÁØÁ DAC = 30° , ÁØÁ CAB = 45° , ÁØÁ ABC = 30°´éÒ¹ DC ÂÒÇ 10 ˹èÇ ´éÒ¹ BC ÂÒÇ¡Õè˹èÇ ![]() 2) ¤èҢͧ tan 15° + tan 22.5°+ cot 15° + cot 22.5° à·èҡѺà·èÒã´ ¢Í¢Íº¾ÃФسÅèǧ˹éÒ¤ÃѺ:please: |
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$\rm{tan}(15)+\rm{tan}(22.5)+\rm{cot}(22.5)+\rm{cot}(15)$
=$\dfrac{\rm{tan^2}(15)+1}{\rm{tan}(15)}+\dfrac{\rm{tan^2}(22.5)+1}{\rm{tan}(22.5)}$ =$\dfrac{\rm{sec^2}(15)}{\rm{tan}(15)}+\dfrac{\rm{sec^2}(22.5)}{\rm{tan}(22.5)}$ =$2[\dfrac{1}{\rm{sin}(30)}+\dfrac{1}{\rm{sin}(45)}]$ =$4+2\sqrt{2}$ |
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