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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 tanx+12tanx2+122tanx22++1210tanx210 is

A
1211cotx2112cot2x
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B
1210cotx2102cot2x
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C
1210cotx2102cotx
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D
None of the above
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Solution

The correct option is B 1210cotx2102cot2x
cotAtanA=2cot2AtanA=cotA2cot2A
Now,
tanx=cotx2cot2x12tanx2=12(cotx22cotx)122tanx22=122(cotx222cotx2)...1210tanx210=1210(cotx2102cotx29)
So,
tanx+12tanx2+122tanx22++1210tanx210=1210cotx2102cot2x

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