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Q.

tan⁡81∘−tan⁡63∘−tan⁡27∘+tan⁡9∘ equals

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a

6

b

0

c

2

d

4

answer is B.

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Detailed Solution

tan⁡81∘−tan⁡63∘−tan⁡27∘+tan⁡9∘=tan⁡81∘+tan⁡9∘−tan⁡63∘+tan⁡27∘=sin⁡81∘cos⁡81∘+sin⁡9∘cos⁡9∘−sin⁡63∘cos⁡63∘+sin⁡27∘cos⁡27∘=sin⁡81∘cos⁡9∘+sin⁡9∘cos⁡81∘cos⁡9∘cos⁡81∘−sin⁡63∘cos⁡27∘+sin⁡27∘cos⁡63∘cos⁡27∘cos⁡63∘=sin⁡81∘+9∘cos⁡9∘cos⁡90∘−9∘−sin⁡63∘+27∘cos⁡27∘cos⁡90∘−27∘=sin⁡90∘cos⁡9∘sin⁡9∘−sin⁡90∘cos⁡27∘sin⁡27∘=1sin⁡9∘cos⁡9∘×22−1sin⁡27∘cos⁡27∘×22=2sin⁡18∘−2sin⁡54∘=2sin⁡54∘−sin⁡18∘sin⁡18∘sin⁡54∘=22cos⁡36∘sin⁡18∘sin⁡18∘cos⁡36∘=4
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