The smallest and the largest values of tan^(−1) (((1−x)/(1+x))) , 0≤ x ≤ 1 are


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Question Number 43180 by gunawan last updated on 08/Sep/18

$The smallest and the largest values of$ $\tan^{- 1} (\frac{1 - x}{1 + x}), 0 ⩽ x ⩽ 1 are$
Commented by maxmathsup by imad last updated on 08/Sep/18

$l e t φ (x) = a r c t a n (\frac{1 - x}{1 + x}) w i t h x \in [0, 1] c h a n g e m e n t x = c o s θ g i v e$ $φ (x) = φ (c o s θ) = a r c t a n (\frac{1 - c o s θ)}{1 + c o s θ}) = a r c t a n ({t a n}^{2} (\frac{θ}{2}))$ $w e h a v e 0 ⩽ x ⩽ 1 \Rightarrow 0 ⩽ θ ⩽ \frac{π}{2} \Rightarrow 0 ⩽ t a n (\frac{π}{2}) ⩽ 1 \Rightarrow 0 ⩽ {t a n}^{2} (\frac{θ}{2}) ⩽ 1 \Rightarrow$ $0 ⩽ φ (c o s θ) ⩽ \frac{π}{4} \Rightarrow φ (x) \in [0, \frac{π}{4}] (b o t h f u n c t i o n t a n a n d a r c t a n a r e i n c r e a s i n g)$
	Answered by tanmay.chaudhury50@gmail.com last updated on 08/Sep/18

	$y = {t a n}^{- 1} 1 - {t a n}^{- 1} x$ $y = \frac{Π}{4} - {t a n}^{- 1} x$ $w h e n x = 0$ ${t a n}^{- 1} x = 0$ $y = \frac{Π}{4}$ $w h e n x = 1$ ${t a n}^{- 1} (1) = \frac{Π}{4}$ $s o y = \frac{Π}{4} - \frac{Π}{4} = 0$ $s o m a x o f y = \frac{Π}{4} m i n o f y = 0$
	Commented by gunawan last updated on 08/Sep/18

	$Nice Sir$
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