Question

Let $x=\sin 1^{\circ }$, then the value of the expression, $\frac{1}{\cos 0^{\circ }\cdot \cos 1^{\circ }}+\frac{1}{\cos 1^{\circ }\cdot \cos 2^{\circ }}+$ $\frac{1}{\cos 2^{\circ }\cdot \cos 3^{\circ }}+\dots +\frac{1}{\cos 44^{\circ }\cdot \cos 45^{\circ }}$ is equal to :

• A. $x$
• B. $1/x$
• C. $\sqrt{2}/x$
• D. $x/\sqrt{2}$

Answer 1

Answer: (B)

$\frac{1}{\sin 1^{\circ }}[\frac{\sin \left(1^{\circ }-0^{\circ }\right)}{\cos 2^{\circ }\cos 3^{\circ }}+\frac{\sin \left(2^{\circ }-1^{\circ }\right)}{\cos 1^{\circ }\cos 2^{\circ }}+$
$\frac{\sin \left(1^{\circ }-0^{\circ }\right)}{\cos 2^{\circ }\cos 3^{\circ }}+\frac{\sin \left(2^{\circ }-1^{\circ }\right)}{\cos 1^{\circ }\cos 2^{\circ }}+$
$\frac{\sin \left(3^{\circ }-2^{\circ }\right)}{\cos 2^{\circ }\cos 3^{\circ }}+\dots ..+\frac{\sin \left(45^{\circ }-44^{\circ }\right)}{\cos 44^{\circ }\cos 45^{\circ }}]$
$=\frac{1}{\sin 1^{\circ }}[\tan 1^{\circ }+\left(\tan 2^{\circ }-\tan 1^{\circ }\right)+\left(\tan 3^{\circ }-\tan 2^{\circ }\right)$
$+\left(\tan 4^{\circ }-\tan 3^{\circ }\right)+\dots .+\left(\tan 45^{\circ }-\tan 44^{\circ }\right)]=$
$\frac{1}{\sin 1^{\circ }}=\frac{1}{x}$