1.

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

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


Discussion

No Comment Found