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