tan 1°+tan 5°+tan 9°+…+tan 177°=...


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Question Number 57373 by gunawan last updated on 03/Apr/19

$\tan 1 ° + \tan 5 ° + \tan 9 ° + \dots + \tan 177 ° = . . .$
Commented by mr W last updated on 03/Apr/19

$= 45$
	Answered by tanmay.chaudhury50@gmail.com last updated on 03/Apr/19

	$l e t$ $y = c o s x c o s 5 x c o s 9 x . . . c o s (x + (n - 1) 4 x)$ $y = c o s x c o s 5 x c o s 9 x . . . c o s (4 n x - 3 x)$ $l n y = l n c o s x + l n c o s 5 x + l n c o s 9 x + . . + l n c o s (4 n x - 3 x 3$ $\frac{1}{y} \times \frac{d y}{d x} = - [(t a n x) + t s n 5 x + t a n 9 x + . . t a n (4 n x - 3 x)]$ $t h i s m e t h o d a p p l i c a b l e i f w e k n o w v a l u e o f y$
	Commented by mr W last updated on 03/Apr/19

	${i t}^{'} s w r o n g s i r . p l e a s e c h e c k :$ $\frac{d (\ln \cos n x)}{d x} = - n \tan n x \neq - \tan n x$
	Commented by tanmay.chaudhury50@gmail.com last updated on 03/Apr/19

	$y e s y e s . . . y o u a r e c o r r e c t . . i a m j u s t g i v i n g e x a m p l e . .$ $l e t m e g i v e r e a l p r o b l e m e x a m p l e . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 04/Apr/19

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