if tanθ=a/b then find (asinθ+bcosθ)/(asinθ-acosθ)
thinkdifferent:
I think the question would be bcos theta in the last
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tanθ=a/b, sinθ=a/√(a2+b2),cosθ=b/√(a2+b2),
(asinθ+bcosθ)/(asinθ-acosθ)=(a*
a/√(a2+b2)+b* b/√(a2+b2))/(a* a/√(a2+b2)-a*
b/√(a2+b2))
=(a2+b2)/(a(a-b))
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(asin
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