Evaluate-cot-1-1-sinx-1-sinx-1-sinx-1-sinx-

Question Number 51367 by rahul 19 last updated on 26/Dec/18

E v a l u a t e :

\cot^{- 1} [\frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}}] = ?

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18

c a s e - 1 a > b

a = s i n \frac{x}{2} b = c o s \frac{x}{2}

\frac{N_{r}}{D_{r}} = \frac{a - b + a + b}{a - b - a - b} = \frac{- a}{b} = - t a n \frac{x}{2} = c o t (\frac{π}{2} + \frac{x}{2})

s o a n s i s \frac{π}{2} + \frac{x}{2}

i f b > a

\frac{N_{r}}{D_{r}} = \frac{b - a + b + a}{b - a - b - a} = \frac{- b}{a} = - c o t \frac{x}{2} = c o t (\frac{- x}{2})

a n s i s \frac{- x}{2}

p l s c h e c k \dots x l i e s i n w h i c h q u s d r a n t ? ? ? ?

Commented by peter frank last updated on 26/Dec/18

i g o t t h e s a m e a n s

Answered by ajfour last updated on 26/Dec/18

l e t \cot θ = \frac{\sqrt{1 - \cos 2 y} + \sqrt{1 + \cos 2 y}}{\sqrt{1 - \cos 2 y} - \sqrt{1 + \cos 2 y}}

w h e r e x = \frac{π}{2} - 2 y \Rightarrow y = \frac{π}{4} - \frac{x}{2}

\Rightarrow \cot θ = \frac{\sin y + \cos y}{\sin y - \cos y} = \frac{\cos (y - \frac{π}{4})}{\sin (y - \frac{π}{4})}

o r \cot θ = \cot (y - \frac{π}{4}) = \cot (- \frac{x}{2})

\Rightarrow o r \tan θ = \tan (- \frac{x}{2})

θ = \cot^{- 1} [\frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}}]

= n π - \frac{x}{2} .

Answered by peter frank last updated on 26/Dec/18

\cot^{- 1} (\frac{s i n \frac{x}{2} - \cos \frac{x}{2} + \sin \frac{x}{2} + \cos \frac{x}{2}}{\sin \frac{x}{2} - \cos \frac{x}{2} - \sin \frac{x}{2} - \cos \frac{x}{2}})

\cot^{- 1} (\frac{2 \sin \frac{x}{2}}{- 2 \cos \frac{x}{2}})

\cot^{- 1} (- \tan \frac{x}{2})

\tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}

\cot^{- 1} x = \frac{π}{2} - \tan^{- 1} x

\cot^{- 1} (- \tan \frac{x}{2}) = \frac{π}{2} - \tan^{- 1} (- \tan \frac{x}{2})

\frac{π}{2} + \frac{x}{2}

Commented by peter frank last updated on 26/Dec/18

r a h u l p l e a s e c h e c k