Question Number 224606 by gregori last updated on 21/Sep/25
$$\:\:\: \alpha\: \beta\:=\:\mathrm{9}\:\: \\ $$$$\:\:\:\:\frac{ \left(\alpha+\beta\right)}{ \left(\alpha−\beta\right)}\:=?\: \\ $$
Answered by som(math1967) last updated on 21/Sep/25
![((10sin αsin β)/(cosαcosβ))=9 ⇒((cosαcosβ)/(sinαsinβ))=((10)/9) ⇒((cosαcosβ+sinαsinβ)/(cosαcosβ−sinαsinβ))=((10+9)/(10−9)) [using comp.and div] ⇒((cos(α−β))/(cos(α+β)))=((19)/1) ∴((cos(α+β))/(cos(α−β)))=(1/(19))](Q224607.png)
$$\:\frac{\mathrm{10sin}\:\alpha\mathrm{sin}\:\beta}{{cos}\alpha{cos}\beta}=\mathrm{9} \\ $$$$\Rightarrow\frac{{cos}\alpha{cos}\beta}{{sin}\alpha{sin}\beta}=\frac{\mathrm{10}}{\mathrm{9}} \\ $$$$\Rightarrow\frac{{cos}\alpha{cos}\beta+{sin}\alpha{sin}\beta}{{cos}\alpha{cos}\beta−{sin}\alpha{sin}\beta}=\frac{\mathrm{10}+\mathrm{9}}{\mathrm{10}−\mathrm{9}} \\ $$$$\left[\boldsymbol{{using}}\:\boldsymbol{{comp}}.\boldsymbol{{and}}\:\boldsymbol{{div}}\right] \\ $$$$\Rightarrow\frac{{cos}\left(\alpha−\beta\right)}{{cos}\left(\alpha+\beta\right)}=\frac{\mathrm{19}}{\mathrm{1}} \\ $$$$\therefore\frac{{cos}\left(\alpha+\beta\right)}{{cos}\left(\alpha−\beta\right)}=\frac{\mathrm{1}}{\mathrm{19}} \\ $$$$ \\ $$