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Question Number 159466 by abdullah_ff last updated on 17/Nov/21
if tanA = (a/b)  then prove that sinA = ± (a/( (√(a^   + b^  ))))  please help..
$$\mathrm{if}\:{tanA}\:=\:\frac{{a}}{{b}} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:{sinA}\:=\:\pm\:\frac{{a}}{\:\sqrt{{a}^{ } \:+\:{b}^{ } }} \\ $$$$\mathrm{please}\:\mathrm{help}.. \\ $$
Answered by Ar Brandon last updated on 17/Nov/21
tanA=(a/b)⇒cotA=(b/a)  cot^2 A+1=cosec^2 A  (b^2 /a^2 )+1=cosec^2 A=((a^2 +b^2 )/a^2 )  ⇒sinA=±(a/( (√(a^2 +b^2 ))))
$$\mathrm{tan}{A}=\frac{{a}}{{b}}\Rightarrow\mathrm{cot}{A}=\frac{{b}}{{a}} \\ $$$$\mathrm{cot}^{\mathrm{2}} {A}+\mathrm{1}=\mathrm{cosec}^{\mathrm{2}} {A} \\ $$$$\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\mathrm{1}=\mathrm{cosec}^{\mathrm{2}} {A}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{sin}{A}=\pm\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$
Answered by MJS_new last updated on 17/Nov/21
tan α =((sin α)/(cos α))=((sin α)/( (√(1−sin^2  α))))  ⇒  tan^2  α =((sin^2  α)/(1−sin^2  α))  ⇒  sin α =±((tan α)/( (√(1+tan^2  α))))  tan α =(a/b) ⇒ sin α =±((a/b)/( (√(1+(a^2 /b^2 )))))=±(a/( (√(a^2 +b^2 ))))
$$\mathrm{tan}\:\alpha\:=\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\alpha}=\frac{\mathrm{sin}\:\alpha}{\:\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\alpha}} \\ $$$$\Rightarrow \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\alpha\:=\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\alpha} \\ $$$$\Rightarrow \\ $$$$\mathrm{sin}\:\alpha\:=\pm\frac{\mathrm{tan}\:\alpha}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\alpha}} \\ $$$$\mathrm{tan}\:\alpha\:=\frac{{a}}{{b}}\:\Rightarrow\:\mathrm{sin}\:\alpha\:=\pm\frac{\frac{{a}}{{b}}}{\:\sqrt{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}}=\pm\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$

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