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First Principle of Differentiation

Solve: cos20...

Question

Solve:
$c o s (2001) π + c o t (2001) \frac{π}{2} + s e c (2001) \frac{π}{3} + t a n (2001) \frac{π}{4} + c o s e c (2001) \frac{π}{6}$ equals to

A

0

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B

1

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C

-2

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D

Not defined

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Solution

The correct option is A 0
Given

$cos (2001) π + cot (2001) \frac{π}{2} + sec (2001) \frac{π}{3} + tan (2001) \frac{π}{4} + csc (2001) \frac{π}{6}$

$cos (2001) π$

$cos (2000 π + π)$

$cos π$ $(cos (2 n π + θ) = cos θ)$

$- 1$

$cot (2001) \frac{π}{2}$

$cot (1000 π + \frac{π}{2})$

$cot \frac{π}{2}$ $(cot (2 n π + θ))$

$0$

$sec (2001) \frac{π}{3}$

$sec (667 π)$

$sec (666 π + π)$

$sec π$ $(sec (2 n π + θ) = sec θ)$

$- 1$

$tan (2001) \frac{π}{4}$

$tan (500 π + \frac{π}{4})$

$tan (\frac{π}{4})$ $(tan (2 n π + θ) = tan θ)$

$1$

$csc (2001) \frac{π}{6}$

$csc (667 \frac{π}{2})$

$csc (333 π + \frac{π}{2})$

$csc (\frac{π}{2})$ $(csc (2 n π + θ) = csc θ)$

$1$

From Above conclusions,

$⟹ - 1 + 0 - 1 + 1 + 1 = 0$

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