tan(arcsin(1/3)+arccos(-1/5))求值
问题描述：
tan(arcsin(1/3)+arccos(-1/5))求值
答
设sina=1/3 cosb=-1/5
则cosa=√(1-sin²a)=2√2/3
sinb=√(1-cos²b)=2√6/5
所以tana=sina/cosa=√2/4
tanb=sinb/cosb=-2√6
故tan(arcsin(1/3)+arccos(-1/5))
=tan(a+b)
=(tana+tanb)/(1-tana*tanb)
=(√2/4-2√6)/(1+√2/4*2√6)
=(√2-8√6)/(4+4√3)
=(9√6-25√2)/8