Question Number 51367 by rahul 19 last updated on 26/Dec/18
![Evaluate: cot^(−1) [(((√(1−sinx))+(√(1+sinx)))/( (√(1−sinx))−(√(1+sinx))))] = ?](https://www.tinkutara.com/question/Q51367.png)
$${Evaluate}: \\ $$$$\mathrm{cot}^{−\mathrm{1}} \left[\frac{\sqrt{\mathrm{1}−\mathrm{sin}{x}}+\sqrt{\mathrm{1}+\mathrm{sin}{x}}}{\:\sqrt{\mathrm{1}−\mathrm{sin}{x}}−\sqrt{\mathrm{1}+\mathrm{sin}{x}}}\right]\:=\:? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18
$${case}−\mathrm{1}\:\:{a}>{b} \\ $$$${a}={sin}\frac{{x}}{\mathrm{2}}\:\:{b}={cos}\frac{{x}}{\mathrm{2}}\:\: \\ $$$$\frac{{N}_{{r}} }{{D}_{{r}} }=\frac{{a}−{b}+{a}+{b}}{{a}−{b}−{a}−{b}}=\frac{−{a}}{{b}}=−{tan}\frac{{x}}{\mathrm{2}}={cot}\left(\frac{\pi}{\mathrm{2}}+\frac{{x}}{\mathrm{2}}\right) \\ $$$${so}\:{ans}\:{is}\:\frac{\pi}{\mathrm{2}}+\frac{{x}}{\mathrm{2}} \\ $$$${if}\:{b}>{a} \\ $$$$\frac{{N}_{{r}} }{{D}_{{r}} }=\frac{{b}−{a}+{b}+{a}}{{b}−{a}−{b}−{a}}=\frac{−{b}}{{a}}=−{cot}\frac{{x}}{\mathrm{2}}={cot}\left(\frac{−{x}}{\mathrm{2}}\right) \\ $$$${ans}\:{is}\:\frac{−{x}}{\mathrm{2}} \\ $$$${pls}\:{check}…{x}\:{lies}\:{in}\:{which}\:{qusdrant}???? \\ $$
Commented by peter frank last updated on 26/Dec/18
$${i}\:{got}\:{the}\:{same}\:{ans}\: \\ $$$$ \\ $$
Answered by ajfour last updated on 26/Dec/18
![let cot θ = (((√(1−cos 2y))+(√(1+cos 2y)))/( (√(1−cos 2y))−(√(1+cos 2y)))) where x = (π/2)−2y ⇒ y = (π/4)−(x/2) ⇒ cot θ = ((sin y+cos y)/(sin y−cos y)) = ((cos (y−(π/4)))/(sin (y−(π/4)))) or cot θ = cot (y−(π/4)) = cot (−(x/2)) ⇒ or tan θ = tan (−(x/2)) θ = cot^(−1) [(((√(1−sinx))+(√(1+sinx)))/( (√(1−sinx))−(√(1+sinx))))] = n𝛑−(x/2) .](https://www.tinkutara.com/question/Q51376.png)
$${let}\:\mathrm{cot}\:\theta\:=\:\frac{\sqrt{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{y}}+\sqrt{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{y}}}{\:\sqrt{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{y}}−\sqrt{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{y}}} \\ $$$$\:\:{where}\:\:{x}\:=\:\frac{\pi}{\mathrm{2}}−\mathrm{2}{y}\:\:\Rightarrow\:\:{y}\:=\:\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{cot}\:\theta\:=\:\frac{\mathrm{sin}\:{y}+\mathrm{cos}\:{y}}{\mathrm{sin}\:{y}−\mathrm{cos}\:{y}}\:=\:\frac{\mathrm{cos}\:\left({y}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{sin}\:\left({y}−\frac{\pi}{\mathrm{4}}\right)} \\ $$$${or}\:\:\:\mathrm{cot}\:\theta\:=\:\mathrm{cot}\:\left({y}−\frac{\pi}{\mathrm{4}}\right)\:=\:\mathrm{cot}\:\left(−\frac{{x}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:{or}\:\:\mathrm{tan}\:\theta\:=\:\mathrm{tan}\:\left(−\frac{{x}}{\mathrm{2}}\right) \\ $$$$\:\theta\:=\:\mathrm{cot}^{−\mathrm{1}} \left[\frac{\sqrt{\mathrm{1}−\mathrm{sin}{x}}+\sqrt{\mathrm{1}+\mathrm{sin}{x}}}{\:\sqrt{\mathrm{1}−\mathrm{sin}{x}}−\sqrt{\mathrm{1}+\mathrm{sin}{x}}}\right] \\ $$$$\:\:\:\:=\:\boldsymbol{{n}\pi}−\frac{\boldsymbol{{x}}}{\mathrm{2}}\:. \\ $$
Answered by peter frank last updated on 26/Dec/18
$$\mathrm{cot}^{−\mathrm{1}} \left(\frac{{sin}\frac{{x}}{\mathrm{2}}−\mathrm{cos}\frac{{x}}{\mathrm{2}}+\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\:+\mathrm{cos}\:\frac{{x}}{\mathrm{2}}}{\mathrm{sin}\:\frac{{x}}{\mathrm{2}}−\mathrm{cos}\:\frac{{x}}{\mathrm{2}}−\mathrm{sin}\:\frac{{x}}{\mathrm{2}}−\mathrm{cos}\:\frac{{x}}{\mathrm{2}}}\right) \\ $$$$\mathrm{cot}^{−\mathrm{1}} \left(\frac{\mathrm{2sin}\:\frac{{x}}{\mathrm{2}}}{−\mathrm{2cos}\:\frac{{x}}{\mathrm{2}}}\right) \\ $$$$\mathrm{cot}^{−\mathrm{1}} \left(−\mathrm{tan}\:\frac{{x}}{\mathrm{2}\:}\right) \\ $$$$\mathrm{tan}^{−\mathrm{1}} {x}+\mathrm{cot}^{−\mathrm{1}} {x}=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{cot}^{−\mathrm{1}} {x}=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} {x} \\ $$$$\mathrm{cot}^{−\mathrm{1}} \left(−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left(−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right) \\ $$$$\frac{\pi}{\mathrm{2}}+\frac{{x}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by peter frank last updated on 26/Dec/18
$${rahul}\:{please}\:{check}\: \\ $$