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Question

For all permissible values of A,2A, following holds true.
(i)cotA+tanA=1sinAcosA=2cosec 2A(ii)cotAtanA=cos2Asin2AsinAcosA=2cot2A(iii)2cotA=2(cosec 2A+cot2A) cosec 2A+cot2A=cotA

Also to evaluate a series of form f(x)+f(2x)+f(4x)++f(2nx) when f(x) can be expressed as g(x)g(2x), we can use the following technique,
f(x)+f(2x)+f(4x)++f(2nx)=(g(x)g(2x))+(g(2x)g(4x))+(g(2nx)g(2n+1x))=g(x)g(2n+1x)

Based on the above information, solve the following questions for all permissible values of x.

The value of cot3712 is

A
6+4+32
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B
6+432
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C
6+4+3+2
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D
6432
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Solution

The correct option is B 6+432
cotA=cosec 2A+cot2A
Putting A=3712,
cot3712=cosec 75+cot75cot3712=223+1+23cot3712=62+23cot3712=6+423

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