change-in-simplest-form-tan-1-1-x-2-1-x-2-1-x-2-1-x-2-

Question Number 58500 by Kunal12588 last updated on 24/Apr/19

c h a n g e i n s i m p l e s t f o r m :

\tan^{- 1} \frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}

Answered by MJS last updated on 24/Apr/19

\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{{(\sqrt{a} + \sqrt{b})}^{2}}{(\sqrt{a} - \sqrt{b}) (\sqrt{a} + \sqrt{b})} = \frac{a + 2 \sqrt{a b} + b}{a - b}

\frac{1 + x^{2} + 2 \sqrt{(1 + x^{2}) (1 - x^{2})} + 1 - x^{2}}{1 + x^{2} - (1 - x^{2})} =

= \frac{1 + \sqrt{1 - x^{4}}}{x^{2}}

\arctan \frac{1 + \sqrt{1 - x^{4}}}{x^{2}} = t

\Rightarrow x = \pm \sqrt{\sin 2 t} \Rightarrow t = \frac{2 n π + \arcsin x^{2}}{2} \lor t = \frac{(2 n + 1) π - \arcsin x^{2}}{2}

testing values we get

\arctan \frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}} = \frac{π - \arcsin x^{2}}{2}

Commented by Kunal12588 last updated on 24/Apr/19

t h a n k y o u s i r, I h a v e a l s o w r i t t e n a n a n s

p l s c h e c k .

Answered by Kunal12588 last updated on 24/Apr/19

{t a n}^{- 1} \frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}

l e t x^{2} = c o s y

\Rightarrow {c o s}^{- 1} x^{2} = y

{t a n}^{- 1} \frac{\sqrt{1 + c o s y} + \sqrt{1 - c o s y}}{\sqrt{1 + c o s y} - \sqrt{1 - c o s y}}

= {t a n}^{- 1} \frac{\sqrt{2} c o s \frac{y}{2} + \sqrt{2} s i n \frac{y}{2}}{\sqrt{2} c o s \frac{y}{2} - \sqrt{2} s i n \frac{y}{2}}

= {t a n}^{- 1} \frac{1 + t a n \frac{y}{2}}{1 - t a n \frac{y}{2}}

= {t a n}^{- 1} \frac{t a n \frac{π}{4} + t a n \frac{y}{2}}{1 - t a n \frac{π}{4} t a n \frac{y}{2}}

= {t a n}^{- 1} (t a n (\frac{π}{4} + \frac{y}{2}))

= \frac{π}{4} + \frac{y}{2}

= \frac{π}{4} + \frac{1}{2} {c o s}^{- 1} x^{2}

Commented by MJS last updated on 24/Apr/19

good!

\frac{π}{4} + \frac{1}{2} \arccos x^{2} = \frac{π}{2} - \frac{1}{2} \arcsin x^{2}

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