Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 116486 by ZiYangLee last updated on 04/Oct/20

Answered by Dwaipayan Shikari last updated on 04/Oct/20

((sin(A+B))/(cos(A+B)))=((sinAcosB+sinBcosA)/(cosAcosB−sinAsinB))=((((sinAcosB)/(cosAcosB))+((sinBcosA)/(cosAcosB)))/(1−tanAtanB))  =((tanA+tanB)/(1−tanAtanB))

$$\frac{\mathrm{sin}\left(\mathrm{A}+\mathrm{B}\right)}{\mathrm{cos}\left(\mathrm{A}+\mathrm{B}\right)}=\frac{\mathrm{sinAcosB}+\mathrm{sinBcosA}}{\mathrm{cosAcosB}−\mathrm{sinAsinB}}=\frac{\frac{\mathrm{sinAcosB}}{\mathrm{cosAcosB}}+\frac{\mathrm{sinBcosA}}{\mathrm{cosAcosB}}}{\mathrm{1}−\mathrm{tanAtanB}} \\ $$$$=\frac{\mathrm{tanA}+\mathrm{tanB}}{\mathrm{1}−\mathrm{tanAtanB}} \\ $$

Answered by Olaf last updated on 04/Oct/20

x = tan(a+b)(1−tanatanb) (1)  x = ((sin(a+b))/(cos(a+b)))(1−((sinasinb)/(cosacosb)))    x = (((e^(i(a+b)) −e^(−i(a+b)) )/(2i))/((e^(i(a+b)) +e^(−i(a+b)) )/2))(1−((((e^(ia) −e^(−ia) )/(2i)).((e^(ib) −e^(−ib) )/(2i)))/(((e^(ia) +e^(−ia) )/2).((e^(ib) +e^(ib) )/2))))  x = −i((e^(i(a+b)) −e^(−i(a+b)) )/(e^(i(a+b)) +e^(i(a+b)) ))(1+(((e^(ia) −e^(−ia) )(e^(ib) −e^(−ib) ))/((e^(ia) +e^(ia) )(e^(ib) +e^(−ib) ))))  After simplification :  x = −i((e^(ia) −e^(−ia) )/(e^(ia) +e^(ia) ))−i((e^(ib) −e^(−ib) )/(e^(ib) +e^(ib) ))  x = (((e^(ia) −e^(−ia) )/(2i))/((e^(ia) +e^(−ia) )/2))+(((e^(ib) −e^(−ib) )/(2i))/((e^(ib) +e^(−ib) )/2))  x = ((sina)/(cosa))+((sinb)/(cosb))  x = tana+tanb (2)    (1) and (2) :  ⇒ x = tan(a+b)(1−tanatanb) = tana+tanb

$${x}\:=\:\mathrm{tan}\left({a}+{b}\right)\left(\mathrm{1}−\mathrm{tan}{a}\mathrm{tan}{b}\right)\:\left(\mathrm{1}\right) \\ $$$${x}\:=\:\frac{\mathrm{sin}\left({a}+{b}\right)}{\mathrm{cos}\left({a}+{b}\right)}\left(\mathrm{1}−\frac{\mathrm{sin}{a}\mathrm{sin}{b}}{\mathrm{cos}{a}\mathrm{cos}{b}}\right) \\ $$$$ \\ $$$${x}\:=\:\frac{\frac{{e}^{{i}\left({a}+{b}\right)} −{e}^{−{i}\left({a}+{b}\right)} }{\mathrm{2}{i}}}{\frac{{e}^{{i}\left({a}+{b}\right)} +{e}^{−{i}\left({a}+{b}\right)} }{\mathrm{2}}}\left(\mathrm{1}−\frac{\frac{{e}^{{ia}} −{e}^{−{ia}} }{\mathrm{2}{i}}.\frac{{e}^{{ib}} −{e}^{−{ib}} }{\mathrm{2}{i}}}{\frac{{e}^{{ia}} +{e}^{−{ia}} }{\mathrm{2}}.\frac{{e}^{{ib}} +{e}^{{ib}} }{\mathrm{2}}}\right) \\ $$$${x}\:=\:−{i}\frac{{e}^{{i}\left({a}+{b}\right)} −{e}^{−{i}\left({a}+{b}\right)} }{{e}^{{i}\left({a}+{b}\right)} +{e}^{{i}\left({a}+{b}\right)} }\left(\mathrm{1}+\frac{\left({e}^{{ia}} −{e}^{−{ia}} \right)\left({e}^{{ib}} −{e}^{−{ib}} \right)}{\left({e}^{{ia}} +{e}^{{ia}} \right)\left({e}^{{ib}} +{e}^{−{ib}} \right)}\right) \\ $$$$\mathrm{After}\:\mathrm{simplification}\:: \\ $$$${x}\:=\:−{i}\frac{{e}^{{ia}} −{e}^{−{ia}} }{{e}^{{ia}} +{e}^{{ia}} }−{i}\frac{{e}^{{ib}} −{e}^{−{ib}} }{{e}^{{ib}} +{e}^{{ib}} } \\ $$$${x}\:=\:\frac{\frac{{e}^{{ia}} −{e}^{−{ia}} }{\mathrm{2}{i}}}{\frac{{e}^{{ia}} +{e}^{−{ia}} }{\mathrm{2}}}+\frac{\frac{{e}^{{ib}} −{e}^{−{ib}} }{\mathrm{2}{i}}}{\frac{{e}^{{ib}} +{e}^{−{ib}} }{\mathrm{2}}} \\ $$$${x}\:=\:\frac{\mathrm{sin}{a}}{\mathrm{cos}{a}}+\frac{\mathrm{sin}{b}}{\mathrm{cos}{b}} \\ $$$${x}\:=\:\mathrm{tan}{a}+\mathrm{tan}{b}\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right)\:: \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{tan}\left({a}+{b}\right)\left(\mathrm{1}−\mathrm{tan}{a}\mathrm{tan}{b}\right)\:=\:\mathrm{tan}{a}+\mathrm{tan}{b} \\ $$

Commented by ZiYangLee last updated on 05/Oct/20

★

$$\bigstar \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com