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If tan−1[20∑k=0sec(5π12+kπ2)sec(5π12+(k+1)π2)]=−tan−1a, then the value of a is |
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Answer» If tan−1[20∑k=0sec(5π12+kπ2)sec(5π12+(k+1)π2)]=−tan−1a, then the value of a is |
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