Σ_(1°) ^(89°) log


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Question Number 59147 by Pranay last updated on 05/May/19

$\underset{1 °}{\sum^{89 °}} {l o g}_{2} t a n r °$
	Answered by ajfour last updated on 05/May/19

	$= \log_{2} 1 = 0 .$
	Answered by tanmay last updated on 05/May/19

	$= {l n}_{2} t a n 1^{o} + {l n}_{2} t a n 2^{o} + . . . + {l n}_{2} t a n 89^{o}$ $= {l n}_{2} (t a n 1^{o} t a n 2^{2} . . . t a n 45^{o} . . . t a n 89^{o})$ $n o w l o o k$ $t a n 89^{0} = t a n (90^{o} - 1^{o}) = c o t 1^{o}$ $s o t a n 1^{o} \times t a n 89^{o}$ $= t a n 1^{o} \times c o t 1^{o}$ $= 1$ $s i m i l a r l y t a n 2^{o} \times t a n 87^{o} = 1$ $. . .$ $. . .$ $s o$ ${l n}_{2} (t a n 1^{o} t a n 2^{o} . . . t a n 89^{o})$ $= {l n}_{2} (1) = 0$
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