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Question Number 73706 by Faradtimmy last updated on 15/Nov/19
If ((tan 3A)/(tan A)) = k, then ((sin 3A)/(sin A)) is equal to
$$\mathrm{If}\:\:\frac{\mathrm{tan}\:\mathrm{3}{A}}{\mathrm{tan}\:{A}}\:=\:{k},\:\mathrm{then}\:\frac{\mathrm{sin}\:\mathrm{3}{A}}{\mathrm{sin}\:{A}}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$
Answered by Tanmay chaudhury last updated on 15/Nov/19
tanA=t ((3t−t^3 )/(t(1−3t^2 )))=k→k−3kt^2 =3−t^2 t^2 (1−3k)=3−k tan^2 A=((3−k)/(1−3k)) cot^2 A=((1−3k)/(3−k))→cosec^2 A=((1−3k+3−k)/(3−k))=((4−4k)/(3−k)) sin^2 A=((3−k)/(4−4k)) ((sin3A)/(sinA)) =3−4sin^2 A =3−4(((3−k)/(4−4k)))=((12−12k−12+4k)/(4−4k))=((−8k)/(4−4k))=((2k)/(k−1))
$${tanA}={t} \\ $$$$\frac{\mathrm{3}{t}−{t}^{\mathrm{3}} }{{t}\left(\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}} \right)}={k}\rightarrow{k}−\mathrm{3}{kt}^{\mathrm{2}} =\mathrm{3}−{t}^{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{3}{k}\right)=\mathrm{3}−{k}\:\:\:\:\:\:\:\:\:\:\:{tan}^{\mathrm{2}} {A}=\frac{\mathrm{3}−{k}}{\mathrm{1}−\mathrm{3}{k}} \\ $$$${cot}^{\mathrm{2}} {A}=\frac{\mathrm{1}−\mathrm{3}{k}}{\mathrm{3}−{k}}\rightarrow{cosec}^{\mathrm{2}} {A}=\frac{\mathrm{1}−\mathrm{3}{k}+\mathrm{3}−{k}}{\mathrm{3}−{k}}=\frac{\mathrm{4}−\mathrm{4}{k}}{\mathrm{3}−{k}} \\ $$$${sin}^{\mathrm{2}} {A}=\frac{\mathrm{3}−{k}}{\mathrm{4}−\mathrm{4}{k}} \\ $$$$\frac{{sin}\mathrm{3}{A}}{{sinA}} \\ $$$$=\mathrm{3}−\mathrm{4}{sin}^{\mathrm{2}} {A} \\ $$$$=\mathrm{3}−\mathrm{4}\left(\frac{\mathrm{3}−{k}}{\mathrm{4}−\mathrm{4}{k}}\right)=\frac{\mathrm{12}−\mathrm{12}{k}−\mathrm{12}+\mathrm{4}{k}}{\mathrm{4}−\mathrm{4}{k}}=\frac{−\mathrm{8}{k}}{\mathrm{4}−\mathrm{4}{k}}=\frac{\mathrm{2}{k}}{{k}−\mathrm{1}} \\ $$$$ \\ $$$$ \\ $$

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