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Question Number 193467 by aba last updated on 14/Jun/23

$$\mathrm{Proof}:$$$${\cot }^{-1}\left(\frac{\sqrt{1+\mathrm{sint}}+\sqrt{1-\mathrm{sint}}}{\sqrt{1+\mathrm{sint}}-\sqrt{1-\mathrm{sint}}}\right)=\frac{\mathrm{t}}{2}$$

Commented by MM42 last updated on 14/Jun/23

$$attention$$$$inallthreeargumentsweusedequalities$$$$thattherelationshipsuchas$$$$‘‘\sqrt{1-{sin}^{2}x}=cosx{}^{″}or‘‘{(sinx+cosx)}^2=1+sin2x^{″}or‘‘{(sinx-cosx)}^2=1-sin2x^{″}$$$$whichmaynotalwayshold.$$$$$$

Commented by Frix last updated on 15/Jun/23

$$\mathrm{Yes}.\mathrm{We}\mathrm{get}$$$${\cot }^{-1}\frac{1+\mid \cos t\mid }{\sin t}=\frac{t}{2}$$$$\mathrm{Which}\mathrm{is}\mathrm{only}\mathrm{true}\mathrm{for}$$$$-\frac{\pi }{2}⩽t⩽\frac{\pi }{2}$$

Commented by MM42 last updated on 15/Jun/23

$$attention$$$$iff\left(x\right)=tanx\Rightarrow for\mathrm{\forall }x\in (\frac{(2k-1)\pi }{2},\frac{(2k+1)\pi }{2})\to R_f=\mathbb{R}$$$$\Rightarrow \mathrm{\exists }{f}^{-1}\left(x\right)={tan}^{-1}\left(x\right)\mathrm{\&}{D}_{f^{-1}}=\mathbb{R}\mathrm{\&}{R}_{f^{-1}}=(-\frac{\pi }{2},\frac{\pi }{2})$$$$tan:(\frac{(2k-1)\pi }{2},\frac{(2k+1)\pi }{2})\to \mathbb{R}\iff {tan}^{-1}:\mathbb{R}\to (-\frac{\pi }{2},\frac{\pi }{2})$$$$$$$$iff\left(x\right)=cotx\Rightarrow for\mathrm{\forall }x\in (k\pi ,(k+1)\pi )\to R_f=\mathbb{R}$$$$\Rightarrow \mathrm{\exists }{f}^{-1}\left(x\right)={cot}^{-1}\left(x\right)\mathrm{\&}{D}_{f^{-1}}=\mathbb{R}\mathrm{\&}{R}_{f^{-1}}=(0,\pi )$$$$cot:(k\pi ,(k-1)\pi )\to \mathbb{R}\iff {cot}^{-1}:\mathbb{R}\to (0,\pi )$$$$$$$$thereforeif{cot}^{-1}\left(\frac{}{}\right)=\frac{t}{2}\Rightarrow$$$$0<\frac{t}{2}<\pi \Rightarrow 0<t<2\pi$$

Answered by Frix last updated on 14/Jun/23

$$\frac{\sqrt{1+s}+\sqrt{1-s}}{\sqrt{1+s}-\sqrt{1-s}}=$$$$=\frac{{(\sqrt{1+s}+\sqrt{1-s})}^2}{(\sqrt{1+s}-\sqrt{1-s})(\sqrt{1+s}+\sqrt{1-s})}=$$$$=\frac{1+\cos t}{\sin t}=\cot \frac{t}{2}$$

Answered by Subhi last updated on 14/Jun/23

$$\left(\frac{\sqrt{1+sin\left(t\right)}+\sqrt{1-sin\left(t\right)}}{\sqrt{1+sin\left(t\right)}-\sqrt{1-sin\left(t\right)}}\right).\left(\frac{\sqrt{1+sin\left(t\right)}}{\sqrt{1+sin\left(t\right)}}\right)=cot\left(\frac{t}{2}\right)$$$$\frac{1+sin\left(t\right)+\sqrt{1-{sin}^2\left(t\right)}}{1+sin\left(t\right)-\sqrt{1-{sin}^2\left(t\right)}}=\frac{1+sin\left(t\right)+cos\left(t\right)}{1+sin\left(t\right)-cos\left(t\right)}$$$$\frac{1+2sin\left(\frac{t}{2}\right).cos\left(\frac{t}{2}\right)+2{cos}^2\left(\frac{t}{2}\right)-1}{1+2sin\left(\frac{t}{2}\right)cos\left(\frac{t}{2}\right)-1+2{sin}^2\left(\frac{t}{2}\right)}$$$$\frac{2cos\left(\frac{t}{2}\right)(sin\left(\frac{t}{2}\right)+cos\left(\frac{t}{2}\right))}{2sin\left(\frac{t}{2}\right)(sin\left(\frac{t}{2}\right)+cos\left(\frac{t}{2}\right))}=cot\left(\frac{t}{2}\right)$$$$$$

Answered by MM42 last updated on 14/Jun/23

$$S1$$$$A={cot}^{-1}\left(\frac{\sqrt{1+sint}+\sqrt{1-sint}}{\sqrt{1+sint}-\sqrt{1-sint}}\right)=$$$${cot}^{-1}\left(\frac{\sqrt{{(cos\frac{t}{2}+sin\frac{t}{2})}^2}+\sqrt{{(cos\frac{t}{2}-sin\frac{t}{2})}^2}}{\sqrt{{(cos\frac{t}{2}+sin\frac{t}{2})}^2}-\sqrt{{(cos\frac{t}{2}-sin\frac{t}{2})}^2}}\right)$$$$\Rightarrow A={cot}^{-1}\left(\frac{cos\frac{t}{2}}{sin\frac{t}{2}}\right)=\frac{t}{2}$$$$S2lettan\frac{t}{2}=u$$$$A={cot}^{-1}\left(\frac{\sqrt{1+\frac{2u}{1+u^2}}+\sqrt{1-\frac{2u}{1+u^2}}}{\sqrt{1+\frac{2u}{1+u^2}}-\sqrt{1-\frac{2u}{1+u^2}}}\right)=$$$${cot}^{-1}\left(\frac{\sqrt{{(1+u)}^2}+\sqrt{{(1-u)}^2}}{\sqrt{{(1+u)}^2}-\sqrt{{(1-u)}^2}}\right)=$$$${cot}^{-1}\left(\frac{2}{2u}\right)={cot}^{-1}\left(u\right)=\frac{t}{2}$$$$$$

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