Question Number 176462 by blackmamba last updated on 19/Sep/22
$${In}\:\Delta{ABC}\:{given}\:\frac{\mathrm{2}{a}}{\mathrm{tan}\:{A}}\:=\:\frac{{b}}{\mathrm{tan}\:{B}}\: \\ $$$$\:{then}\:\frac{\mathrm{sin}\:^{\mathrm{2}} {A}−\mathrm{cos}\:^{\mathrm{2}} {B}}{\mathrm{cos}\:^{\mathrm{2}} {A}+\mathrm{cos}\:^{\mathrm{2}} {B}}=? \\ $$
Commented by cortano1 last updated on 19/Sep/22
$$\frac{\mathrm{3}}{\mathrm{5}} \\ $$
Answered by mnjuly1970 last updated on 20/Sep/22
$$\:\:\:\:\:\frac{\mathrm{4}{sin}\left({A}\right).{cos}\left({A}\right)}{{sin}\left({A}\right)}=\:\frac{\mathrm{2}{sin}\left({B}\right).{cos}\left({B}\right)}{{sin}\left({B}\right)} \\ $$$$\:\:\:\:\mathrm{2}{cos}\left({A}\right)\:=\:{cos}\left({B}\right) \\ $$$$\:\:\:\:\:−−−−− \\ $$$$\:\:\:\:\:\:\:\frac{{sin}^{\:\mathrm{2}} \left({A}\right)−\mathrm{4}{cos}^{\:\mathrm{2}} \left({A}\right)}{\mathrm{5}{cos}^{\:\mathrm{2}} \left({A}\right)} \\ $$$$\:\:\:\:{if}\::\:\:\:\:\:\:{sin}\left({A}\right)\:\overset{{numerator}} {\rightarrow}{cos}\left({A}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:{answer}\::=\:\:\:\frac{−\mathrm{3}}{\mathrm{5}}\:\: \\ $$$$\:\:{if}\::\:\:{sin}^{\:\mathrm{2}} \left({A}\right)−{cos}^{\:\mathrm{2}} \left({B}\right)\overset{{numerator}} {\rightarrow}{cos}^{\:\mathrm{2}} \left({B}\right)−{cos}^{\:\mathrm{2}} \left({A}\right) \\ $$$$\:\:\:\:\:\:{answer}\::\:\:\:\frac{\:\mathrm{3}}{\mathrm{5}}\:\:\:\:\:\:\:\:\:\:\blacksquare{m}.{n} \\ $$$$\:\:\:\:\:\:{result}\::\:{to}\:{me}\:{this}\:{question}\:{has}\:{a}\:{typo} \\ $$$$\:\:\:\:\:\:\:\:−−−−−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by cortano1 last updated on 20/Sep/22
$$\mathrm{only}\:\frac{\mathrm{3}}{\mathrm{5}}\:\mathrm{sir} \\ $$