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If `tantheta=a/b` , show that `(asintheta-bcostheta)/(asintheta+bcostheta)=(a^2-b^2)/(a^2+b^2)` |
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Answer» `((asin theta- b cos theta ))/(( a sin theta+b cos theta ))=((atan theta-b))/((a tan theta +b))" " ["dividing num , and denom , by cos "theta]` `=((axx(a)/(b)-b))/((axx(a)/(b)+b))=((a^(2)-b^(2)))/((a^(2)+b^(2))).` |
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