1-89-log-2-tan-r-

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} + \dots + {l n}_{2} t a n 89^{o}

= {l n}_{2} (t a n 1^{o} t a n 2^{2} \dots t a n 45^{o} \dots 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

\dots

\dots

s o

{l n}_{2} (t a n 1^{o} t a n 2^{o} \dots t a n 89^{o})

= {l n}_{2} (1) = 0

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