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Question Number 88867 by liki last updated on 13/Apr/20

Commented by liki last updated on 13/Apr/20

..help me please question roman i,ii & ii

$$..{help}\:{me}\:{please}\:{question}\:{roman}\:{i},{ii}\:\&\:{ii} \\ $$

Answered by TANMAY PANACEA. last updated on 13/Apr/20

acosθ+bsinθ=c t=tan(θ/2) a(((1−t^2 )/(1+t^2 )))+b(((2t)/(1+t^2 )))=c a−at^2 +2bt=c+ct^2 t^2 (a+c)+t(−2b)+(c−a)=0 t_1 +t_2 =((2b )/(a+c)) and t_1 t_2 =((c−a)/(c+a)) t_1 =tan((α/2)) t_2 =tan((β/2)) sin(α+β)=sin2(((α+β)/2))=((2tan(((α+β)/2)))/(1+tan^2 (((α+β)/2)))) now sin(α+β)=((2×((tan(α/2)+tan(β/2))/(1−tan(α/2).tan(β/2))))/(1+(((tan(α/2)+tan(β/2))/(1−tan(α/2).tan(β/2))))^2 )) =((2×((t_1 +t_2 )/(1−t_1 t_2 )))/(1+(((t_1 +t_2 )/(1−t_1 t_2 )))^2 )) =((2(t_1 +t_2 )(1−t_1 t_2 ))/((1−t_1 t_2 )^2 +(t_1 +t_2 )^2 )) =((2(((2b)/(a+c)))(1−((c−a)/(c+a))))/((1−((c−a)/(c+a)))^2 +(((2b)/(c+a)))^2 )) =(((4b×2a)/((c+a)^2 ))/(((4a^2 )/((c+a)^2 ))+((4b^2 )/((c+a)^2 ))))=((8ab)/(4(a^2 +b^2 )))=((2ab)/(a^2 +b^2 )) 2)tan(α+β)=tan2(((α+β)/2))=((2tan(((α+β)/2)))/(1−tan^2 (((α+β)/2)))) =((2(((t_1 +t_2 )/(1−t_1 t_2 ))))/(1−(((t_1 +t_2 )/(1−t_1 t_2 )))^2 ))=((2(t_1 +t_2 )(1−t_1 t_2 ))/((1−t_1 t_2 )^2 −(t_1 +t_2 )^2 )) =((2(((2b)/(a+c)))(1−((c−a)/(c+a))))/((1−((c−a)/(c+a)))^2 −(((2b)/(c+a)))^2 )) =((4b×2a)/(4a^2 −4b^2 ))=((2ab)/(a^2 −b^2 )) 3)cos(α−β)=((1−tan^2 (((α−β)/2)))/(1+tan^2 (((α−β)/2)))) =((1−(((t_1 −t_2 )/(1+t_1 t_2 )))^2 )/(1+(((t_1 −t_2 )/(1+t_1 t_2 )))^2 )) =(((1+t_1 t_2 )^2 −(t_1 −t_2 )^2 )/((1+t_1 t_2 )^2 +(t_1 −t_2 )^2 )) =(((1+t_1 t_2 )^2 −(t_1 +t_2 )^2 +4t_1 t_2 )/((1+t_1 t_2 )^2 +(t_1 +t_2 )^2 −4t_1 t_2 )) =(((1+((c−a)/(c+a)))^2 −(((2b)/(c+a)))^2 +4.((c−a)/(c+a)))/((1+((c−a)/(c+a)))^2 +(((2b)/(c+a)))^2 −4.(((c−a)/(c+a))))) =(((4c^2 −4b^2 +4(c^2 −a^2 ))/((c+a)^2 ))/((4c^2 +4b^2 −4(c^2 −a^2 ))/((c+a)^2 ))) =((c^2 −b^2 +c^2 −a^2 )/(c^2 +b^2 −c^2 +a^2 )) ((2c^2 −a^2 −b^2 )/(a^2 +b^2 ))

$${acos}\theta+{bsin}\theta={c} \\ $$$${t}={tan}\frac{\theta}{\mathrm{2}} \\ $$$${a}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)+{b}\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)={c} \\ $$$${a}−{at}^{\mathrm{2}} +\mathrm{2}{bt}={c}+{ct}^{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} \left({a}+{c}\right)+{t}\left(−\mathrm{2}{b}\right)+\left({c}−{a}\right)=\mathrm{0} \\ $$$${t}_{\mathrm{1}} +{t}_{\mathrm{2}} =\frac{\mathrm{2}{b}\:\:}{{a}+{c}}\:\:\:{and}\:\:{t}_{\mathrm{1}} {t}_{\mathrm{2}} =\frac{{c}−{a}}{{c}+{a}} \\ $$$${t}_{\mathrm{1}} ={tan}\left(\frac{\alpha}{\mathrm{2}}\right)\:\:\:{t}_{\mathrm{2}} ={tan}\left(\frac{\beta}{\mathrm{2}}\right) \\ $$$${sin}\left(\alpha+\beta\right)={sin}\mathrm{2}\left(\frac{\alpha+\beta}{\mathrm{2}}\right)=\frac{\mathrm{2}{tan}\left(\frac{\alpha+\beta}{\mathrm{2}}\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{\alpha+\beta}{\mathrm{2}}\right)} \\ $$$${now}\:{sin}\left(\alpha+\beta\right)=\frac{\mathrm{2}×\frac{{tan}\frac{\alpha}{\mathrm{2}}+{tan}\frac{\beta}{\mathrm{2}}}{\mathrm{1}−{tan}\frac{\alpha}{\mathrm{2}}.{tan}\frac{\beta}{\mathrm{2}}}}{\mathrm{1}+\left(\frac{{tan}\frac{\alpha}{\mathrm{2}}+{tan}\frac{\beta}{\mathrm{2}}}{\mathrm{1}−{tan}\frac{\alpha}{\mathrm{2}}.{tan}\frac{\beta}{\mathrm{2}}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}×\frac{{t}_{\mathrm{1}} +{t}_{\mathrm{2}} }{\mathrm{1}−{t}_{\mathrm{1}} {t}_{\mathrm{2}} }}{\mathrm{1}+\left(\frac{{t}_{\mathrm{1}} +{t}_{\mathrm{2}} }{\mathrm{1}−{t}_{\mathrm{1}} {t}_{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}\left({t}_{\mathrm{1}} +{t}_{\mathrm{2}} \right)\left(\mathrm{1}−{t}_{\mathrm{1}} {t}_{\mathrm{2}} \right)}{\left(\mathrm{1}−{t}_{\mathrm{1}} {t}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({t}_{\mathrm{1}} +{t}_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}\left(\frac{\mathrm{2}{b}}{{a}+{c}}\right)\left(\mathrm{1}−\frac{{c}−{a}}{{c}+{a}}\right)}{\left(\mathrm{1}−\frac{{c}−{a}}{{c}+{a}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}{b}}{{c}+{a}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\frac{\mathrm{4}{b}×\mathrm{2}{a}}{\left({c}+{a}\right)^{\mathrm{2}} }}{\frac{\mathrm{4}{a}^{\mathrm{2}} }{\left({c}+{a}\right)^{\mathrm{2}} }+\frac{\mathrm{4}{b}^{\mathrm{2}} }{\left({c}+{a}\right)^{\mathrm{2}} }}=\frac{\mathrm{8}{ab}}{\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}=\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right){tan}\left(\alpha+\beta\right)={tan}\mathrm{2}\left(\frac{\alpha+\beta}{\mathrm{2}}\right)=\frac{\mathrm{2}{tan}\left(\frac{\alpha+\beta}{\mathrm{2}}\right)}{\mathrm{1}−{tan}^{\mathrm{2}} \left(\frac{\alpha+\beta}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{2}\left(\frac{{t}_{\mathrm{1}} +{t}_{\mathrm{2}} }{\mathrm{1}−{t}_{\mathrm{1}} {t}_{\mathrm{2}} }\right)}{\mathrm{1}−\left(\frac{{t}_{\mathrm{1}} +{t}_{\mathrm{2}} }{\mathrm{1}−{t}_{\mathrm{1}} {t}_{\mathrm{2}} }\right)^{\mathrm{2}} }=\frac{\mathrm{2}\left({t}_{\mathrm{1}} +{t}_{\mathrm{2}} \right)\left(\mathrm{1}−{t}_{\mathrm{1}} {t}_{\mathrm{2}} \right)}{\left(\mathrm{1}−{t}_{\mathrm{1}} {t}_{\mathrm{2}} \right)^{\mathrm{2}} −\left({t}_{\mathrm{1}} +{t}_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}\left(\frac{\mathrm{2}{b}}{{a}+{c}}\right)\left(\mathrm{1}−\frac{{c}−{a}}{{c}+{a}}\right)}{\left(\mathrm{1}−\frac{{c}−{a}}{{c}+{a}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{2}{b}}{{c}+{a}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}{b}×\mathrm{2}{a}}{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} }=\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right){cos}\left(\alpha−\beta\right)=\frac{\mathrm{1}−{tan}^{\mathrm{2}} \left(\frac{\alpha−\beta}{\mathrm{2}}\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{\alpha−\beta}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}−\left(\frac{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }{\mathrm{1}+{t}_{\mathrm{1}} {t}_{\mathrm{2}} }\right)^{\mathrm{2}} }{\mathrm{1}+\left(\frac{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }{\mathrm{1}+{t}_{\mathrm{1}} {t}_{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$$=\frac{\left(\mathrm{1}+{t}_{\mathrm{1}} {t}_{\mathrm{2}} \right)^{\mathrm{2}} −\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}+{t}_{\mathrm{1}} {t}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{\left(\mathrm{1}+{t}_{\mathrm{1}} {t}_{\mathrm{2}} \right)^{\mathrm{2}} −\left({t}_{\mathrm{1}} +{t}_{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{t}_{\mathrm{1}} {t}_{\mathrm{2}} }{\left(\mathrm{1}+{t}_{\mathrm{1}} {t}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({t}_{\mathrm{1}} +{t}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{t}_{\mathrm{1}} {t}_{\mathrm{2}} } \\ $$$$=\frac{\left(\mathrm{1}+\frac{{c}−{a}}{{c}+{a}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{2}{b}}{{c}+{a}}\right)^{\mathrm{2}} +\mathrm{4}.\frac{{c}−{a}}{{c}+{a}}}{\left(\mathrm{1}+\frac{{c}−{a}}{{c}+{a}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}{b}}{{c}+{a}}\right)^{\mathrm{2}} −\mathrm{4}.\left(\frac{{c}−{a}}{{c}+{a}}\right)} \\ $$$$=\frac{\frac{\mathrm{4}{c}^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} +\mathrm{4}\left({c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}{\left({c}+{a}\right)^{\mathrm{2}} }}{\frac{\mathrm{4}{c}^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{2}} −\mathrm{4}\left({c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}{\left({c}+{a}\right)^{\mathrm{2}} }} \\ $$$$=\frac{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} +{a}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}{c}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$

Commented by liki last updated on 13/Apr/20

...thank you so much sir blessed

$$...{thank}\:{you}\:{so}\:{much}\:{sir}\:{blessed} \\ $$

Commented by TANMAY PANACEA. last updated on 13/Apr/20

most welcome

$${most}\:{welcome} \\ $$

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