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Question

The value of (3+tan1)(3+tan2)(3+tan3)(3+tan28)(3+tan29) is

A
228
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B
215+23
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C
229
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D
214+23
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Solution

The correct option is C 229
Given : (3+tan1)(3+tan2)(3+tan3)(3+tan28)(3+tan29)

We know that
3+tan(30x)=3+tan30tanx1+tan30tanx=3+13tanx3+tanx=43+tanx

Now,
(3+tan1)(3+tan29)=43+tan29×(3+tan29)=4
Similarly,
(3+tan2)(3+tan28)=4

So,
(3+tan1)(3+tan2)(3+tan3)(3+tan28)(3+tan29)=414(3+tan15)=414(3+313+1)=414(2+233+1)=229

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