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8 votes
8 votes

$\log \tan 1^o + \log \tan 2^o + \dots + \log \tan 89^o$ is $\ldots$

  1. $1$
  2. $1/\sqrt{2}$
  3. $0$
  4. $−1$
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2 Comments

Before looking at answer try to guess the logic by just looking at answer options.
1
1
If $y=\tan 1^{\circ}\cdot \tan 2^{\circ}\cdot \tan 3^{\circ}\cdot \tan 4^{\circ}\dots \tan 89^{\circ},$ then $y=1$
0
0

1 Answer

13 votes
13 votes
Best answer
$(\tan 1 . \tan 89) . (\tan 2 . \tan 88) \ldots (\tan 44 . \tan 46) . \tan 45$
$= ( \tan 1 . \cot 1).(\tan 2 . \cot 2)\ldots (\tan 44. \cot 44). \tan 45$
$= 1 . 1 . 1 \ldots . 1 . \tan 45$
$= 1$
So, the given expression reduces to $\log 1 = 0.$

Correct Option: (C) 0
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1 comment

$=>tan89^{\circ}$

$=>tan(90^{\circ}-1^{\circ})$ $(1^{st}$ Quadrant, All are positive.)

$=>cot1^{\circ}$

1
1
Answer:

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