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Question

If tan1[20k=0sec(5π12+kπ2)sec(5π12+(k+1)π2)]=tan1a, then the value of a is

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Solution

tan1[20k=0sec(5π12+kπ2)sec(5π12+(k+1)π2)]

=tan1⎢ ⎢ ⎢ ⎢ ⎢ ⎢20k=01cos(5π12+kπ2)cos(5π12+(k+1)π2)⎥ ⎥ ⎥ ⎥ ⎥ ⎥

=tan1⎢ ⎢ ⎢ ⎢ ⎢ ⎢20k=0sin((5π12+(k+1)π2)(5π12+kπ2))cos(5π12+kπ2)cos(5π12+(k+1)π2)⎥ ⎥ ⎥ ⎥ ⎥ ⎥

=tan1[20k=0tan(5π12+(k+1)π2)tan(5π12+kπ2)]

=tan1[(tan(5π12+π2)tan(5π12)) +(tan(5π12+π)tan(5π12+π2)) +...+(tan(5π12+21π2)tan(5π12+20π2))]

=tan1[tan(5π12+21π2)tan(5π12)]

=tan1[cot(5π12)tan(5π12)]

=tan1⎢ ⎢ ⎢⎜ ⎜ ⎜2sin5π6⎟ ⎟ ⎟⎥ ⎥ ⎥

=tan1⎢ ⎢ ⎢ ⎢2sin(ππ6)⎥ ⎥ ⎥ ⎥
=tan1(4)
=tan14

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