Question Number 155810 by zainaltanjung last updated on 05/Oct/21
$$\mathrm{Verify}\:\mathrm{the}\:\mathrm{identity}\:\mathrm{in}\:\mathrm{Excercise}\:\mathrm{below} \\ $$$$\left.\mathrm{1}\right).\:\mathrm{cos}\:\theta\mathrm{sec}\:\theta=\mathrm{1} \\ $$$$\left.\mathrm{2}\right).\:\left(\mathrm{1}+\mathrm{cos}\:\beta\right)\left(\mathrm{1}−\mathrm{cos}\:\beta\right)=\mathrm{sin}\:^{\mathrm{2}} \beta \\ $$$$\left.\mathrm{3}\right).\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\left(\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)=\mathrm{sin}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\left.\mathrm{4}\right).\:\frac{\mathrm{sin}\:\mathrm{t}}{\mathrm{cosec}\:\mathrm{t}}+\frac{\mathrm{cos}\:\mathrm{t}}{\mathrm{sec}\:\mathrm{t}}=\mathrm{1} \\ $$$$\left.\mathrm{5}\right).\:\frac{\mathrm{cosec}\:^{\mathrm{2}} \theta}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta}\:=\:\mathrm{cot}\:^{\mathrm{2}} \theta \\ $$
Answered by puissant last updated on 05/Oct/21
$$\left.\mathrm{1}\right) \\ $$$${cos}\theta{sec}\theta={cos}\theta×\frac{\mathrm{1}}{{cos}\theta}=\mathrm{1}.. \\ $$$$\left.\mathrm{2}\right) \\ $$$$\left(\mathrm{1}−{cos}\beta\right)\left(\mathrm{1}+{cos}\beta\right)=\mathrm{1}−{cos}^{\mathrm{2}} \beta={sin}^{\mathrm{2}} \beta.. \\ $$$$\left.\mathrm{3}\right) \\ $$$${cos}^{\mathrm{2}} {x}\left({sec}^{\mathrm{2}} {x}−\mathrm{1}\right)={cos}^{\mathrm{2}} {x}\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}−\mathrm{1}\right) \\ $$$$=\mathrm{1}−{cos}^{\mathrm{2}} {x}\:=\:{sin}^{\mathrm{2}} {x}.. \\ $$$$\left.\mathrm{4}\right) \\ $$$$\frac{{sint}}{{cosect}}+\frac{{cost}}{{sect}}=\frac{{sint}}{\frac{\mathrm{1}}{{sint}}}+\frac{{cost}}{\frac{\mathrm{1}}{{cost}}}={sin}^{\mathrm{2}} {t}+{cos}^{\mathrm{2}} {t}=\mathrm{1}.. \\ $$$$\left.\mathrm{5}\right) \\ $$$$\frac{{cosec}^{\mathrm{2}} \theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}=\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \theta}×\frac{\mathrm{1}}{\mathrm{1}+\frac{{sin}^{\mathrm{2}} \theta}{{cos}^{\mathrm{2}} \theta}}=\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \theta}×\frac{{cos}^{\mathrm{2}} \theta}{{cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta} \\ $$$$=\:\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \theta}×\frac{{cos}^{\mathrm{2}} \theta}{\mathrm{1}}=\left(\frac{{cos}\theta}{{sin}\theta}\right)^{\mathrm{2}} =\:{cot}^{\mathrm{2}} \theta.. \\ $$
Commented by zainaltanjung last updated on 05/Oct/21
$$\mathrm{All}\:\mathrm{of}\:\mathrm{your}\:\mathrm{answere}\:\mathrm{is}\:\mathrm{true} \\ $$