Find: ln (((2 tg 22,30°)/(1 − tg^2 22,30°))) = ?


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Question Number 207876 by hardmath last updated on 29/May/24

$Find : \ln (\frac{2 tg 22, 30 °}{1 - {tg}^{2} 22, 30 °}) = ?$
Commented by mr W last updated on 29/May/24

$u s e y o u r c a l c u l a t o r!$
	Answered by som(math1967) last updated on 29/May/24

	$l n (t a n 2 \times 22 ° 30^{'})$ $= l n t a n 45 = l n 1 = 0$
	Commented by hardmath last updated on 29/May/24

	$thank you professors$
	Commented by mr W last updated on 29/May/24

	$c a n y o u m a k e y o u r t h i n g s c l e a r$ $a n d y o u s h o u l d n o t l e t o t h e r p e o p l e$ $g u e s s . i f y o u m e a n 22 ° 30^{'} t h e n y o u$ $s h o u l d n o t w r i t e 22, 30 ° . i f y o u m a d e$ $t y p o, y o u c a n c o r r e c t i t .$
	Commented by hardmath last updated on 29/May/24

	$dear professor, yes, sorry . . .$ $22, 30 °$
	Commented by mr W last updated on 29/May/24

	$t h e n t h e r e s u l t i s n o t 0, b u t$ $\ln (\tan 44.6 °) f o r w h i c h y o u n e e d$ $a c a l c u l a t o r .$
	Commented by hardmath last updated on 29/May/24

	$thank you dear professor$
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