已知tan(a+6π/5)=m(m≠1),求[sin(11π/5+a)+3cos(a-9π/5)]/[sin(14π/5
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已知tan(a+6π/5)=m(m≠1),求[sin(11π/5+a)+3cos(a-9π/5)]/[sin(14π/5-a)+cos(a+16π/5)]的值.
![已知tan(a+6π/5)=m(m≠1),求[sin(11π/5+a)+3cos(a-9π/5)]/[sin(14π/5](/uploads/image/z/7400098-10-8.jpg?t=%E5%B7%B2%E7%9F%A5tan%28a%2B6%CF%80%2F5%29%3Dm%28m%E2%89%A01%29%2C%E6%B1%82%5Bsin%2811%CF%80%2F5%2Ba%29%2B3cos%28a-9%CF%80%2F5%29%5D%2F%5Bsin%2814%CF%80%2F5)
tan(a+6π/5)=m
则tan(a+π/5)=m
[sin(π/5+a)+3cos(a+π/5)]/[sin(-π/5-a)-cos(a+π/5)]
=[sin(π/5+a)+3cos(a+π/5)]/[-sin(π/5+a)-cos(a+π/5)]
上下除以cos(a+π/5)
且sin(a+π/5)/cos(a+π/5)=tan(a+6π/5)
所以原式=[tan(a+6π/5)+3]/[-tan(a+6π/5)-1]
=-(m+3)/(m+1)
则tan(a+π/5)=m
[sin(π/5+a)+3cos(a+π/5)]/[sin(-π/5-a)-cos(a+π/5)]
=[sin(π/5+a)+3cos(a+π/5)]/[-sin(π/5+a)-cos(a+π/5)]
上下除以cos(a+π/5)
且sin(a+π/5)/cos(a+π/5)=tan(a+6π/5)
所以原式=[tan(a+6π/5)+3]/[-tan(a+6π/5)-1]
=-(m+3)/(m+1)
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