已知tan(π/12+A)=根号2,tan(B-π/3)=2根号2,求
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已知tan(π/12+A)=根号2,tan(B-π/3)=2根号2,求
(1)tan(A+B-π/4)
(2)tan(A+B)的值
(1)tan(A+B-π/4)
(2)tan(A+B)的值
1、
tan(A+B-π/4)
=tan[(A+π/12)+(B-π/3)]
=[tan(A+π/12)+tan(B-π/3)]/[1-tan(A+π/12)tan(B-π/3)]
=-√2
2、
tan(A+B-π/4)
=[tan(A+B)-tanπ/4]/[1+tan(A+B)tanπ/4]
=[tan(A+B)-1]/[1+tan(A+B)]=-√2
tan(A+B)=-3+2√2
tan(A+B-π/4)
=tan[(A+π/12)+(B-π/3)]
=[tan(A+π/12)+tan(B-π/3)]/[1-tan(A+π/12)tan(B-π/3)]
=-√2
2、
tan(A+B-π/4)
=[tan(A+B)-tanπ/4]/[1+tan(A+B)tanπ/4]
=[tan(A+B)-1]/[1+tan(A+B)]=-√2
tan(A+B)=-3+2√2
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