Cot 5^o - tan5^o - 2 tan10^o - 4 tan 20^o - 8 cot40^o is equal to

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3.Trigonometrical Ratios, Functions and Identities

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$cot 5^o$ -$tan5^o$ -$2$ $tan10^o$ -$4$ $tan 20^o$ -$8$ $cot40^o$ is equal to

A

$0$

B

$4\tan {40^o}$

C

$8\tan {40^o}$

D

$8\cot {40^o}$

Solution

Use the relation $\cot \theta – \tan \theta = 2\cot 2\theta$

Standard 11

Mathematics

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