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

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