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If `tan theta=a/b` then `(a sin theta-b cos theta)/(a sin theta+bcos theta)=`

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If `tan theta=a/b` then `(a sin theta-b cos theta)/(a sin theta+bcos theta)=`
A. `((a^(2)+b^(2)))/((a^(2)-b^(2)))`
B. `((a^(2)-b^(2)))/((a^(2)+b^(2)))`
C. `(a^(2))/((a^(2)+b^(2)))`
D. `(b^(2))/((a^(2)+b^(2)))`
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Correct Answer - B
Dividing num. and denom by `costheta`, we get
`((asintheta-bcostheta))/((asintheta+bcostheta))=((atantheta-b))/((atantheta+b))=((axx(a)/(b)-b))/((axx(a)/(b)+b))=((a^(2)-b^(2)))/((a^(2)+b^(2))).`
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