tan−1[20∑k=0sec(5π12+kπ2)sec(5π12+(k+1)π2)]
=tan−1⎡⎢
⎢
⎢
⎢
⎢
⎢⎣20∑k=01cos(5π12+kπ2)cos(5π12+(k+1)π2)⎤⎥
⎥
⎥
⎥
⎥
⎥⎦
=tan−1⎡⎢
⎢
⎢
⎢
⎢
⎢⎣20∑k=0sin((5π12+(k+1)π2)−(5π12+kπ2))cos(5π12+kπ2)cos(5π12+(k+1)π2)⎤⎥
⎥
⎥
⎥
⎥
⎥⎦
=tan−1[20∑k=0tan(5π12+(k+1)π2)−tan(5π12+kπ2)]
=tan−1[(tan(5π12+π2)−tan(5π12)) +(tan(5π12+π)−tan(5π12+π2)) +...+(tan(5π12+21π2)−tan(5π12+20π2))]
=tan−1[tan(5π12+21π2)−tan(5π12)]
=tan−1[−cot(5π12)−tan(5π12)]
=tan−1⎡⎢
⎢
⎢⎣−⎛⎜
⎜
⎜⎝2sin5π6⎞⎟
⎟
⎟⎠⎤⎥
⎥
⎥⎦
=tan−1⎡⎢
⎢
⎢
⎢⎣−2sin(π−π6)⎤⎥
⎥
⎥
⎥⎦
=tan−1(−4)
=−tan−14