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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 cosec 2x+cosec 4x+cosec 8x+cosec 16x+cosec 32x

A
cotxcot64x
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B
cot2xcot64x
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C
cotxcot32x
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D
cot2xcot32x
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Solution

The correct option is C cotxcot32x
cosec 2x+cot2x=cotxcosec 2x=cotxcot2x
Now,
cosec 4x=cot2xcot4xcosec 8x=cot4xcot8xcosec 16x=cot8xcot16xcosec 32x=cot16xcot32x
So,
cosec 2x+cosec 4x+cosec 8x+cosec 16x+cosec 32x=cotxcot32x

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