Question

a=1/sin60°sin61°+1/sin62°sin63°+......1/sin118°sin119°. The value of (cosec1°/a)²

Answer 1

3

Answer 2

The given sum is $a = \frac{1}{\sin60^\circ\sin61^\circ} + \frac{1}{\sin62^\circ\sin63^\circ} + \dots + \frac{1}{\sin118^\circ\sin119^\circ}$.

Each term in the sum is of the form $\frac{1}{\sin x^\circ \sin y^\circ}$. We can use the identity: $\frac{\sin(y-x)}{\sin x \sin y} = \cot x - \cot y$. In our case, the difference between the angles in the denominator is $1^\circ$ (e.g., $61^\circ - 60^\circ = 1^\circ$). So, we can write $\frac{1}{\sin x^\circ \sin (x+1)^\circ} = \frac{1}{\sin 1^\circ} \frac{\sin((x+1)-x)}{\sin x^\circ \sin (x+1)^\circ} = \frac{1}{\sin 1^\circ} (\cot x^\circ - \cot (x+1)^\circ)$.

Let's apply this identity to each term in the sum for $a$: The general term is $T_k = \frac{1}{\sin (60+2(k-1))^\circ \sin (60+2(k-1)+1)^\circ}$, where $k$ is the term number. $T_k = \frac{1}{\sin 1^\circ} (\cot (60+2(k-1))^\circ - \cot (60+2(k-1)+1)^\circ)$.

To find the number of terms, observe the starting angles: $60^\circ, 62^\circ, \dots, 118^\circ$. This is an arithmetic progression. Let the starting angle be $A_k = 60+2(k-1)$. For the last term, $A_k = 118^\circ$. $118 = 60 + 2(k-1)$ $58 = 2(k-1)$ $29 = k-1$ $k = 30$. There are 30 terms in the sum.

Now let's write out the sum for $a$: $a = \frac{1}{\sin 1^\circ} [(\cot 60^\circ - \cot 61^\circ) + (\cot 62^\circ - \cot 63^\circ) + \dots + (\cot 118^\circ - \cot 119^\circ)]$.

We can pair terms from the beginning and the end of the sum. Let's consider the $k$-th term from the beginning, $T_k$, and the $k$-th term from the end, which is $T_{30-(k-1)} = T_{31-k}$.

$T_k = \frac{1}{\sin 1^\circ} (\cot (60+2k-2)^\circ - \cot (60+2k-1)^\circ)$. $T_{31-k}$ is the term corresponding to the starting angle $60+2((31-k)-1)^\circ = 60+2(30-k)^\circ = (120-2k)^\circ$. So, $T_{31-k} = \frac{1}{\sin 1^\circ} (\cot (120-2k)^\circ - \cot (121-2k)^\circ)$.

We use the identity $\cot(180^\circ - \theta) = -\cot\theta$. $\cot(120-2k)^\circ = \cot(180-(60+2k))^\circ = -\cot(60+2k)^\circ$. $\cot(121-2k)^\circ = \cot(180-(59+2k))^\circ = -\cot(59+2k)^\circ$. Substituting these into $T_{31-k}$: $T_{31-k} = \frac{1}{\sin 1^\circ} (-\cot(60+2k)^\circ - (-\cot(59+2k)^\circ)) = \frac{1}{\sin 1^\circ} (\cot(59+2k)^\circ - \cot(60+2k)^\circ)$.

Now, let's sum $T_k + T_{31-k}$: $T_k + T_{31-k} = \frac{1}{\sin 1^\circ} [(\cot (60+2k-2)^\circ - \cot (60+2k-1)^\circ) + (\cot (59+2k)^\circ - \cot (60+2k)^\circ)]$. Let's re-index $k$ for clarity. Let the first angle in $T_k$ be $x_k = 60+2(k-1)$. Then $T_k = \frac{1}{\sin 1^\circ} (\cot x_k^\circ - \cot (x_k+1)^\circ)$. The corresponding term from the end is $T_{31-k}$. Its angles are $180^\circ - (x_k+1)^\circ$ and $180^\circ - x_k^\circ$. Specifically, if $x_k = 60+2(k-1)$, the last term's angles are $118, 119$. $118 = 180 - 62$. $119 = 180 - 61$. So, $T_{30} = \frac{1}{\sin 1^\circ} (\cot 118^\circ - \cot 119^\circ) = \frac{1}{\sin 1^\circ} (-\cot 62^\circ - (-\cot 61^\circ)) = \frac{1}{\sin 1^\circ} (\cot 61^\circ - \cot 62^\circ)$.

Let's sum $T_k$ with $T_{31-k}$ for $k=1, \dots, 15$. $k=1$: $T_1 + T_{30} = \frac{1}{\sin 1^\circ} [(\cot 60^\circ - \cot 61^\circ) + (\cot 61^\circ - \cot 62^\circ)] = \frac{1}{\sin 1^\circ} (\cot 60^\circ - \cot 62^\circ)$. $k=2$: $T_2 + T_{29} = \frac{1}{\sin 1^\circ} [(\cot 62^\circ - \cot 63^\circ) + (\cot 63^\circ - \cot 64^\circ)] = \frac{1}{\sin 1^\circ} (\cot 62^\circ - \cot 64^\circ)$. This pattern continues. The sum $a$ is the sum of these 15 pairs: $a = \sum_{k=1}^{15} (T_k + T_{31-k})$ $a = \frac{1}{\sin 1^\circ} [(\cot 60^\circ - \cot 62^\circ) + (\cot 62^\circ - \cot 64^\circ) + \dots + (\cot (60+2(15-1))^\circ - \cot (60+2(15-1)+2)^\circ)]$. The last pair is for $k=15$. $T_{15} + T_{16} = \frac{1}{\sin 1^\circ} [(\cot (60+28)^\circ - \cot (60+29)^\circ) + (\cot (60+30)^\circ - \cot (60+31)^\circ)]$. $T_{15} = \frac{1}{\sin 1^\circ} (\cot 88^\circ - \cot 89^\circ)$. $T_{16} = \frac{1}{\sin 1^\circ} (\cot 90^\circ - \cot 91^\circ)$. Using $\cot(180-\theta) = -\cot\theta$: $T_{16} = \frac{1}{\sin 1^\circ} (\cot 90^\circ - (-\cot 89^\circ)) = \frac{1}{\sin 1^\circ} (0 + \cot 89^\circ) = \frac{1}{\sin 1^\circ} \cot 89^\circ$. (Since $\cot 90^\circ = 0$) So, $T_{15} + T_{16} = \frac{1}{\sin 1^\circ} [(\cot 88^\circ - \cot 89^\circ) + \cot 89^\circ] = \frac{1}{\sin 1^\circ} \cot 88^\circ$.

Let's re-examine the sum of pairs. The sum is a telescoping series: $a = \frac{1}{\sin 1^\circ} [(\cot 60^\circ - \cot 62^\circ) + (\cot 62^\circ - \cot 64^\circ) + \dots + (\cot 88^\circ - \cot 90^\circ) + (\cot 90^\circ - \cot 92^\circ) + \dots + (\cot 116^\circ - \cot 118^\circ)]$. This is not the sum of pairs as I wrote it. The sum of pairs is: $a = \sum_{k=1}^{15} \frac{1}{\sin 1^\circ} (\cot (60+2(k-1))^\circ - \cot (60+2(k-1)+2)^\circ)$. Let $X_k = 60+2(k-1)$. Then the sum for each pair is $\frac{1}{\sin 1^\circ} (\cot X_k^\circ - \cot (X_k+2)^\circ)$. The sum $a$ is: $a = \frac{1}{\sin 1^\circ} [(\cot 60^\circ - \cot 62^\circ) + (\cot 62^\circ - \cot 64^\circ) + \dots + (\cot (60+2(15-1))^\circ - \cot (60+2(15-1)+2)^\circ)]$ $a = \frac{1}{\sin 1^\circ} [(\cot 60^\circ - \cot 62^\circ) + (\cot 62^\circ - \cot 64^\circ) + \dots + (\cot 88^\circ - \cot 90^\circ)]$. This is a telescoping sum. The intermediate terms cancel out. $a = \frac{1}{\sin 1^\circ} (\cot 60^\circ - \cot 90^\circ)$. Since $\cot 90^\circ = 0$: $a = \frac{1}{\sin 1^\circ} (\cot 60^\circ - 0)$. $a = \frac{\cot 60^\circ}{\sin 1^\circ}$. We know $\cot 60^\circ = \frac{1}{\sqrt{3}}$. So, $a = \frac{1}{\sqrt{3}\sin 1^\circ}$.

We need to find the value of $(\text{cosec}1^\circ/a)^2$. $\text{cosec}1^\circ = \frac{1}{\sin 1^\circ}$. Substitute the value of $a$: $\frac{\text{cosec}1^\circ}{a} = \frac{1/\sin 1^\circ}{1/(\sqrt{3}\sin 1^\circ)} = \frac{1}{\sin 1^\circ} \cdot \sqrt{3}\sin 1^\circ = \sqrt{3}$.

Finally, we need to calculate $(\text{cosec}1^\circ/a)^2$: $(\text{cosec}1^\circ/a)^2 = (\sqrt{3})^2 = 3$.

The final answer is $\boxed{3}$.