Question Number 11844 by minakshidahaval0202@gmail.co. last updated on 02/Apr/17
$$\mathrm{cot}\alpha+\mathrm{cosec}\alpha={k}\:{then}\:{find}\:\mathrm{cosec}\alpha−{cot}\alpha\:{and}\:{also}\:{find}\:{cot}\alpha \\ $$
Answered by sma3l2996 last updated on 02/Apr/17
$${we}\:{have} \\ $$$$\left({i}\right):{cot}\alpha+{cosec}\alpha=\frac{{cos}\alpha}{{sin}\alpha}+\frac{\mathrm{1}}{{sin}\alpha}={k} \\ $$$$=\frac{{cos}\alpha+\mathrm{1}}{{sin}\alpha}=\frac{\left({cos}\alpha+\mathrm{1}\right)\left({cos}\alpha−\mathrm{1}\right)}{{sin}\alpha\left({cos}\alpha−\mathrm{1}\right)}=\frac{−{sin}^{\mathrm{2}} \alpha}{{sin}\alpha\left({cos}\alpha−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\frac{\mathrm{1}}{{sin}\alpha}−\frac{{cos}\alpha}{{sin}\alpha}}=\frac{\mathrm{1}}{{cosec}\alpha−{cot}\alpha}={k} \\ $$$$\left({ii}\right):{cosec}\alpha−{cot}\alpha=\frac{\mathrm{1}}{{k}} \\ $$$${let}\:{do} \\ $$$$\left({i}\right)−\left({ii}\right):\:\mathrm{2}{cot}\alpha={k}−\frac{\mathrm{1}}{{k}} \\ $$$${cot}\alpha=\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}} \\ $$
Commented by sma3l2996 last updated on 02/Apr/17
$${I}\:{mean}\:{cot}\alpha=\frac{{k}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{k}} \\ $$
Answered by ajfour last updated on 02/Apr/17
$$\mathrm{cosec}\:^{\mathrm{2}} \alpha\:−\:\mathrm{cot}\:^{\mathrm{2}} \alpha\:=\mathrm{1} \\ $$$$\left(\mathrm{cosec}\:\alpha\:−\mathrm{cot}\:\alpha\right)\left(\mathrm{cosec}\:\alpha+\mathrm{cot}\:\alpha\right)=\mathrm{1} \\ $$$$\left(\mathrm{cosec}\:\alpha\:−\:\mathrm{cot}\:\alpha\:\right)\left({k}\right)\:=\mathrm{1} \\ $$$$\mathrm{cosec}\:\alpha\:−\mathrm{cot}\:\alpha\:=\:\frac{\mathrm{1}}{{k}}\:\:\:…\left({ii}\right) \\ $$$${and}\:{as}\:\:\mathrm{cosec}\:\alpha+\mathrm{cot}\:\alpha\:=\:{k}\:\:\:\:…\left({i}\right) \\ $$$$\left({i}\right)−\left({ii}\right)\:{gives} \\ $$$$\mathrm{2cot}\:\alpha=\:{k}−\frac{\mathrm{1}}{{k}} \\ $$$$\mathrm{cot}\:\alpha=\:\frac{{k}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{k}}\:. \\ $$