Math, asked by mdimran, 1 year ago

if tanθ=a/b then find (asinθ+bcosθ)/(asinθ-acosθ)


thinkdifferent: I think the question would be bcos theta in the last

Answers

Answered by anshubansal
2
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))

Answered by thinkdifferent
1
(asin
Similar questions