Question

The integer part of the number $\sum _{k=0}^{44}\frac{1}{\cos k^{\circ }\cos (k+1{)}^{\circ }}$ is

• A. $50$
• B. $52$
• C. $57$
• D. $59$

Answer 1

Answer: (C)

$57$

$\sum _0^{44}\frac{1}{cos\left(k\right)cos(k+1))}$

$=\frac{1}{sin\left(1^{\circ }\right)}\sum _0^{44}\frac{sin(k+1-k)}{cos\left(k\right)(k+1)}$

$=cosec\left(1^{\circ }\right)\sum _0^{44}\frac{sin(k+1)cos\left(k\right)}{cos\left(k\right)cos(k+1)}-\frac{sin\left(k\right)cos(k+1)}{cos\left(k\right)cos(k+1)}$

$=cosec\left(1^{\circ }\right){\sum }_0^{44}Tan(k+1)-Tan\left(k\right)$

$=cosec\left(1^{\circ }\right)\left[Tan\right(1^{\circ })-Tan(0)+Tan(2^{\circ })-Tan(1^{\circ })+.........+Tan({45}^{\circ })-Tan({44}^{\circ }\left)\right]$

$=cosec\left(1^{\circ }\right)\left[Tan\right({45}^{\circ })-Tan(0\left)\right]$

$=cosec\left(1^{\circ }\right)$