Question

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

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

Answer 1

Answer: (B)

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