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Question-212477




Question Number 212477 by Durganand last updated on 14/Oct/24
Answered by mehdee7396 last updated on 14/Oct/24
((sin2acosa−sinacos2a)/(sin2asina))  =((sina)/(sin2asina))=(1/(sin2a))  ✓
$$\frac{{sin}\mathrm{2}{acosa}−{sinacos}\mathrm{2}{a}}{{sin}\mathrm{2}{asina}} \\ $$$$=\frac{{sina}}{{sin}\mathrm{2}{asina}}=\frac{\mathrm{1}}{{sin}\mathrm{2}{a}}\:\:\checkmark \\ $$
Answered by A5T last updated on 14/Oct/24
((cosA)/(sinA))−((cos2A)/(sin2A))=((2cos^2 A−(cos^2 A−sin^2 A))/(sin2A))  =((cos^2 A+sin^2 A)/(sin2A))=(1/(sin2A))
$$\frac{{cosA}}{{sinA}}−\frac{{cos}\mathrm{2}{A}}{{sin}\mathrm{2}{A}}=\frac{\mathrm{2}{cos}^{\mathrm{2}} {A}−\left({cos}^{\mathrm{2}} {A}−{sin}^{\mathrm{2}} {A}\right)}{{sin}\mathrm{2}{A}} \\ $$$$=\frac{{cos}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {A}}{{sin}\mathrm{2}{A}}=\frac{\mathrm{1}}{{sin}\mathrm{2}{A}} \\ $$
Answered by Nadirhashim last updated on 15/Oct/24
  left side    ((coxA)/(sinA)) − ((cos2A)/(sin2A))     ((sin2AcosA−cos2AsinA)/(sin2AsinA))        ((sin(2A−A))/(sin2AsinA)) =((sinA)/(sin2AsinA))     =(1/(sin2A))
$$\:\:\boldsymbol{{left}}\:\boldsymbol{{side}} \\ $$$$\:\:\frac{\boldsymbol{{coxA}}}{\boldsymbol{{sinA}}}\:−\:\frac{\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{A}}}{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{A}}}\: \\ $$$$\:\:\frac{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{AcosA}}−\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{AsinA}}}{\boldsymbol{{sin}}\mathrm{2}{A}\boldsymbol{{sinA}}} \\ $$$$ \\ $$$$\:\:\:\:\frac{\boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{A}}−{A}\right)}{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{AsinA}}}\:=\frac{\boldsymbol{{sinA}}}{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{AsinA}}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{A}}}\: \\ $$$$ \\ $$

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