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Proof-cot-1-1-sint-1-sint-1-sint-1-sint-t-2-

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Question Number 193467 by aba last updated on 14/Jun/23
Proof : cot^(−1) ((((√(1+sint))+(√(1−sint)))/( (√(1+sint))−(√(1−sint)))))=(t/2)
$$\mathrm{Proof}\:: \\ $$$$\mathrm{cot}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+\mathrm{sint}}+\sqrt{\mathrm{1}−\mathrm{sint}}}{\:\sqrt{\mathrm{1}+\mathrm{sint}}−\sqrt{\mathrm{1}−\mathrm{sint}}}\right)=\frac{\mathrm{t}}{\mathrm{2}}\: \\ $$
Commented by MM42 last updated on 14/Jun/23
attention in all three arguments we used equalities that the relationship such as “(√(1−sin^2 x))=cosx ” or “ (sinx+cosx)^2 =1+sin2x” or “(sinx−cosx)^2 =1−sin2x” which may not always hold.
$${attention} \\ $$$${in}\:{all}\:\:{three}\:{arguments}\:\:{we}\:{used}\:{equalities} \\ $$$${that}\:{the}\:{relationship}\:{such}\:{as} \\ $$$$“\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {x}}={cosx}\:''\:{or}\:“\:\left({sinx}+{cosx}\right)^{\mathrm{2}} =\mathrm{1}+{sin}\mathrm{2}{x}''\:{or}\:“\left({sinx}−{cosx}\right)^{\mathrm{2}} =\mathrm{1}−{sin}\mathrm{2}{x}'' \\ $$$$\:{which}\:{may}\:{not}\:\:{always}\:{hold}. \\ $$$$ \\ $$
Commented by Frix last updated on 15/Jun/23
Yes. We get cot^(−1) ((1+∣cos t∣)/(sin t)) =(t/2) Which is only true for −(π/2)≤t≤(π/2)
$$\mathrm{Yes}.\:\mathrm{We}\:\mathrm{get} \\ $$$$\mathrm{cot}^{−\mathrm{1}} \:\frac{\mathrm{1}+\mid\mathrm{cos}\:{t}\mid}{\mathrm{sin}\:{t}}\:=\frac{{t}}{\mathrm{2}} \\ $$$$\mathrm{Which}\:\mathrm{is}\:\mathrm{only}\:\mathrm{true}\:\mathrm{for} \\ $$$$−\frac{\pi}{\mathrm{2}}\leqslant{t}\leqslant\frac{\pi}{\mathrm{2}} \\ $$
Commented by MM42 last updated on 15/Jun/23
attention if f(x)=tanx ⇒ for ∀x ∈( (((2k−1)π)/2) , (((2k+1)π)/2)) → R_f = R ⇒ ∃ f^(−1) (x)=tan^(−1) (x) & D_f^(−1) =R & R_f^(−1) =(− (π/2) , (π/2)) tan : ( (((2k−1)π)/2) , (((2k+1)π)/2)) → R ⇔ tan^(−1) : R → (−(π/2) , (π/2)) if f(x)=cotx ⇒ for ∀x ∈( kπ, (k+1)π) → R_f = R ⇒ ∃ f^(−1) (x)=cot^(−1) (x) & D_f^(−1) =R & R_f^(−1) =(0 , π) cot : ( kπ , (k−1)π) → R ⇔ cot^(−1) : R → (0, π) therefore if cot^(−1) ((/))=(t/2)⇒ 0<(t/2)<π⇒0<t<2π
$${attention} \\ $$$${if}\:\:{f}\left({x}\right)={tanx}\:\Rightarrow\:{for}\:\:\forall{x}\:\in\left(\:\:\frac{\left(\mathrm{2}{k}−\mathrm{1}\right)\pi}{\mathrm{2}}\:,\:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)\:\:\rightarrow\:\:{R}_{{f}} =\:\mathbb{R}\: \\ $$$$\Rightarrow\:\:\exists\:\:{f}^{−\mathrm{1}} \left({x}\right)={tan}^{−\mathrm{1}} \left({x}\right)\:\:\:\:\&\:\:{D}_{{f}^{−\mathrm{1}} } =\mathbb{R}\:\:\:\&\:\:{R}_{{f}^{−\mathrm{1}} } =\left(−\:\frac{\pi}{\mathrm{2}}\:,\:\frac{\pi}{\mathrm{2}}\right) \\ $$$$\:{tan}\::\:\left(\:\:\frac{\left(\mathrm{2}{k}−\mathrm{1}\right)\pi}{\mathrm{2}}\:,\:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)\:\:\rightarrow\:\mathbb{R}\:\:\:\Leftrightarrow\:{tan}^{−\mathrm{1}} \::\:\mathbb{R}\:\rightarrow\:\left(−\frac{\pi}{\mathrm{2}}\:,\:\frac{\pi}{\mathrm{2}}\right)\: \\ $$$$ \\ $$$${if}\:\:{f}\left({x}\right)={cotx}\:\Rightarrow\:{for}\:\:\forall{x}\:\in\left(\:\:{k}\pi,\:\left({k}+\mathrm{1}\right)\pi\right)\:\:\rightarrow\:\:{R}_{{f}} =\:\mathbb{R}\: \\ $$$$\Rightarrow\:\:\exists\:\:{f}^{−\mathrm{1}} \left({x}\right)={cot}^{−\mathrm{1}} \left({x}\right)\:\:\:\:\&\:\:{D}_{{f}^{−\mathrm{1}} } =\mathbb{R}\:\:\:\&\:\:{R}_{{f}^{−\mathrm{1}} } =\left(\mathrm{0}\:,\:\pi\right) \\ $$$$\:{cot}\::\:\left(\:\:{k}\pi\:,\:\left({k}−\mathrm{1}\right)\pi\right)\:\:\rightarrow\:\mathbb{R}\:\:\:\Leftrightarrow\:{cot}^{−\mathrm{1}} \::\:\mathbb{R}\:\rightarrow\:\left(\mathrm{0},\:\pi\right)\: \\ $$$$ \\ $$$${therefore}\:{if}\:\:\:{cot}^{−\mathrm{1}} \left(\frac{}{}\right)=\frac{{t}}{\mathrm{2}}\Rightarrow \\ $$$$\mathrm{0}<\frac{{t}}{\mathrm{2}}<\pi\Rightarrow\mathrm{0}<{t}<\mathrm{2}\pi \\ $$
Answered by Frix last updated on 14/Jun/23
(((√(1+s))+(√(1−s)))/( (√(1+s))−(√(1−s))))= =((((√(1+s))+(√(1−s)))^2 )/( ((√(1+s))−(√(1−s)))((√(1+s))+(√(1−s)))))= =((1+cos t)/(sin t))=cot (t/2)
$$\frac{\sqrt{\mathrm{1}+{s}}+\sqrt{\mathrm{1}−{s}}}{\:\sqrt{\mathrm{1}+{s}}−\sqrt{\mathrm{1}−{s}}}= \\ $$$$=\frac{\left(\sqrt{\mathrm{1}+{s}}+\sqrt{\mathrm{1}−{s}}\right)^{\mathrm{2}} }{\:\left(\sqrt{\mathrm{1}+{s}}−\sqrt{\mathrm{1}−{s}}\right)\left(\sqrt{\mathrm{1}+{s}}+\sqrt{\mathrm{1}−{s}}\right)}= \\ $$$$=\frac{\mathrm{1}+\mathrm{cos}\:{t}}{\mathrm{sin}\:{t}}=\mathrm{cot}\:\frac{{t}}{\mathrm{2}} \\ $$
Answered by Subhi last updated on 14/Jun/23
((((√(1+sin(t)))+(√(1−sin(t))))/( (√(1+sin(t)))−(√(1−sin(t)))))).(((√(1+sin(t)))/( (√(1+sin(t))))))=cot((t/2)) ((1+sin(t)+(√(1−sin^2 (t))))/(1+sin(t)−(√(1−sin^2 (t)))))=((1+sin(t)+cos(t))/(1+sin(t)−cos(t))) ((1+2sin((t/2)).cos((t/2))+2cos^2 ((t/2))−1)/(1+2sin((t/2))cos((t/2))−1+2sin^2 ((t/2)))) ((2cos((t/2))(sin((t/2))+cos((t/2))))/(2sin((t/2))(sin((t/2))+cos((t/2)))))=cot((t/2))
$$\left(\frac{\sqrt{\mathrm{1}+{sin}\left({t}\right)}+\sqrt{\mathrm{1}−{sin}\left({t}\right)}}{\:\sqrt{\mathrm{1}+{sin}\left({t}\right)}−\sqrt{\mathrm{1}−{sin}\left({t}\right)}}\right).\left(\frac{\sqrt{\mathrm{1}+{sin}\left({t}\right)}}{\:\sqrt{\mathrm{1}+{sin}\left({t}\right)}}\right)={cot}\left(\frac{{t}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{1}+{sin}\left({t}\right)+\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left({t}\right)}}{\mathrm{1}+{sin}\left({t}\right)−\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left({t}\right)}}=\frac{\mathrm{1}+{sin}\left({t}\right)+{cos}\left({t}\right)}{\mathrm{1}+{sin}\left({t}\right)−{cos}\left({t}\right)} \\ $$$$\frac{\mathrm{1}+\mathrm{2}{sin}\left(\frac{{t}}{\mathrm{2}}\right).{cos}\left(\frac{{t}}{\mathrm{2}}\right)+\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{t}}{\mathrm{2}}\right)−\mathrm{1}}{\mathrm{1}+\mathrm{2}{sin}\left(\frac{{t}}{\mathrm{2}}\right){cos}\left(\frac{{t}}{\mathrm{2}}\right)−\mathrm{1}+\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{t}}{\mathrm{2}}\right)} \\ $$$$\frac{\mathrm{2}{cos}\left(\frac{{t}}{\mathrm{2}}\right)\left({sin}\left(\frac{{t}}{\mathrm{2}}\right)+{cos}\left(\frac{{t}}{\mathrm{2}}\right)\right)}{\mathrm{2}{sin}\left(\frac{{t}}{\mathrm{2}}\right)\left({sin}\left(\frac{{t}}{\mathrm{2}}\right)+{cos}\left(\frac{{t}}{\mathrm{2}}\right)\right)}={cot}\left(\frac{{t}}{\mathrm{2}}\right) \\ $$$$ \\ $$
Answered by MM42 last updated on 14/Jun/23
S1 A=cot^(−1) ((((√(1+sint))+(√(1−sint)))/( (√(1+sint))−(√(1−sint)))))= cot^(−1) ((((√((cos(t/2)+sin(t/2))^2 ))+(√((cos(t/2)−sin(t/2))^2 )))/( (√((cos(t/2)+sin(t/2))^2 ))−(√((cos(t/2)−sin(t/2))^2 ))))) ⇒A=cot^(−1) (((cos(t/2))/(sin(t/2))))=(t/2) S2 let tan(t/2)=u A=cot^(−1) ((((√(1+((2u)/(1+u^2 ))))+(√(1−((2u)/(1+u^2 )))))/( (√(1+((2u)/(1+u^2 ))))−(√(1−((2u)/(1+u^2 )))))))= cot^(−1) ((((√((1+u)^2 ))+(√((1−u)^2 )))/( (√((1+u)^2 ))−(√((1−u)^2 )))))= cot^(−1) ((2/( 2u)))=cot^(−1) (u)=(t/2)
$${S}\mathrm{1} \\ $$$${A}={cot}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+{sint}}+\sqrt{\mathrm{1}−{sint}}}{\:\sqrt{\mathrm{1}+{sint}}−\sqrt{\mathrm{1}−{sint}}}\right)= \\ $$$${cot}^{−\mathrm{1}} \left(\frac{\sqrt{\left({cos}\frac{{t}}{\mathrm{2}}+{sin}\frac{{t}}{\mathrm{2}}\right)^{\mathrm{2}} }+\sqrt{\left({cos}\frac{{t}}{\mathrm{2}}−{sin}\frac{{t}}{\mathrm{2}}\right)^{\mathrm{2}} }}{\:\sqrt{\left({cos}\frac{{t}}{\mathrm{2}}+{sin}\frac{{t}}{\mathrm{2}}\right)^{\mathrm{2}} }−\sqrt{\left({cos}\frac{{t}}{\mathrm{2}}−{sin}\frac{{t}}{\mathrm{2}}\right)^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow{A}={cot}^{−\mathrm{1}} \left(\frac{{cos}\frac{{t}}{\mathrm{2}}}{{sin}\frac{{t}}{\mathrm{2}}}\right)=\frac{{t}}{\mathrm{2}} \\ $$$${S}\mathrm{2}\:\:\:{let}\:\:{tan}\frac{{t}}{\mathrm{2}}={u} \\ $$$${A}={cot}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}+\sqrt{\mathrm{1}−\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}−\sqrt{\mathrm{1}−\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}}\right)= \\ $$$${cot}^{−\mathrm{1}} \left(\frac{\sqrt{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{1}−{u}\right)^{\mathrm{2}} }}{\:\sqrt{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }−\sqrt{\left(\mathrm{1}−{u}\right)^{\mathrm{2}} }}\right)= \\ $$$${cot}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\mathrm{2}{u}}\right)={cot}^{−\mathrm{1}} \left({u}\right)=\frac{{t}}{\mathrm{2}} \\ $$$$ \\ $$

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