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Numerical Ability
Geometry
Tan (A+B) = 1/2, tan (A-B) = 1/3, then find the value of SIN 2A
Read Solution (Total 2)
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- since tan(a+b)= tan a +tan b/(1-tan a*tan b)
Assume
tan[(A+B)+(A-B)]=tan(A+B) + tan(A-B)/ 1- (tan(A+B)* tan(A_B))
tan(2A)= (1/2 + 1/3) /(1- (1/2*1/3))
tan(2A) = 1
tan= perpendicular/base
thus sin = perpendicular/hypotaneous
so sin(2A)= 1/square root of 2 - 8 years agoHelpfull: Yes(0) No(0)
- tan(2A)=tan{(A+B)+(A-B)}=1/sqr root2
- 8 years agoHelpfull: Yes(0) No(1)
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