Question

Let $K=1^{\circ }$ , then prove that $\sum _{n=0}^{88}\frac{1}{\cos nK.\cos \left(n+1\right)K}=\frac{\cos K}{{\sin }^{2}K}$

Answer 1

$\sum _{n=0}^{88}\frac{1}{\cos k.\cos (n+1)k}$

$[\frac{1}{\cos k.\cos k}+\frac{1}{\cos k.\cos 2k}+...\dots .+\frac{1}{\cos 88k.\cos 89k}]$

Now $\frac{1}{\cos nk.\cos (n+1)k}=\frac{1}{\sin k}\left[\frac{\sin k}{\cos nk.\cos (n+1)k}\right]$

$=\frac{1}{\sin k}\left[\frac{\sin \left[\right(n+1)k-nk]}{\cos nk\cos (n+1)k}\right]$

$=\frac{1}{\sin k}\left[\frac{\sin (n+1)k\cos nk-\cos (n+1)k\sin nk}{\cos nk\cos (n+1)k}\right]$

$=\frac{1}{\sin x}\left[\tan \right(n+1)k-\tan nk]$

$\therefore \frac{1}{\cos ok\cdot cosk}=(\tan k-\tan ok)\times \frac{1}{\sin k}$

$=\frac{1}{\cos k\cos 2k}=(\tan 2k-\tan k)\times \frac{1}{\sin k}$

$\frac{1}{\cos 88k\cos 89k}=(\tan 89k-\tan 88k)\times \frac{1}{\sin k}$

$\sum _{n=0}^{88}\frac{1}{\cos k\cdot \cos (n+1)k}=\tan 89k\times \frac{1}{\sin k}$

$\cos k=1^o$

$\therefore$ sum $=\tan {89}^o\times \frac{1}{\sin 1^o}$

$=\cot 1^o\times \frac{1}{\sin 1^o}$

$=\frac{\cos 1^o}{(\sin 1^o{)}^2}$

$=\frac{\cos k}{(\sin k{)}^2}=$RHS

Hence proved.