# Prove that cosec(x/4)+cosec (x/2)+cosecx=cot(x/8)-cotx ?

Apr 28, 2018

$L H S = \cos e c \left(\frac{x}{4}\right) + \cos e c \left(\frac{x}{2}\right) + \cos e c x$

$= \cos e c \left(\frac{x}{4}\right) + \cos e c \left(\frac{x}{2}\right) + \cos e c x + \cot x - \cot x$

$= \cos e c \left(\frac{x}{4}\right) + \cos e c \left(\frac{x}{2}\right) + \textcolor{b l u e}{\frac{1}{\sin} x + \cos \frac{x}{\sin} x} - \cot x$

$= \cos e c \left(\frac{x}{4}\right) + \cos e c \left(\frac{x}{2}\right) + \textcolor{b l u e}{\frac{1 + \cos x}{\sin} x} - \cot x$

$= \cos e c \left(\frac{x}{4}\right) + \cos e c \left(\frac{x}{2}\right) + \textcolor{b l u e}{\frac{2 {\cos}^{2} \left(\frac{x}{2}\right)}{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}} - \cot x$

$= \cos e c \left(\frac{x}{4}\right) + \cos e c \left(\frac{x}{2}\right) + \textcolor{b l u e}{\cos \frac{\frac{x}{2}}{\sin} \left(\frac{x}{2}\right)} - \cot x$
$= \cos e c \left(\frac{x}{4}\right) + \textcolor{g r e e n}{\cos e c \left(\frac{x}{2}\right) + \cot \left(\frac{x}{2}\right)} - \cot x$

$\textcolor{m a \ge n t a}{\text{Proceeding in similar manner as before}}$

$= \cos e c \left(\frac{x}{4}\right) + \textcolor{g r e e n}{\cot} \left(\frac{x}{4}\right) - \cot x$

$= \cot \left(\frac{x}{8}\right) - \cot x = R H S$

Apr 29, 2018

Kindly go through a Proof given in the Explanation.

#### Explanation:

Setting $x = 8 y$, we have to prove that,

$\cos e c 2 y + \cos e c 4 y + \cos e c 8 y = \cot y - \cot 8 y$.

Observe that, $\cos e c 8 y + \cot 8 y = \frac{1}{\sin 8 y} + \frac{\cos 8 y}{\sin 8 y}$,

$= \frac{1 + \cos 8 y}{\sin 8 y}$,

$= \frac{2 {\cos}^{2} 4 y}{2 \sin 4 y \cos 4 y}$,

$= \frac{\cos 4 y}{\sin 4 y}$.

$\text{Thus, } \cos e c 8 y + c o 8 y = \cot 4 y \left[= \cot \left(\frac{1}{2} \cdot 8 y\right)\right] \ldots \ldots . . \left(\star\right)$.

Adding, $\cos e c 4 y$,

$\cos e c 4 y + \left(\cos e c 8 y + c o 8 y\right) = \cos e c 4 y + \cot 4 y$,

$= \cot \left(\frac{1}{2} \cdot 4 y\right) \ldots \ldots \ldots \left[\because , \left(\star\right)\right]$.

$\therefore \cos e c 4 y + \cos e c 8 y + c o 8 y = \cot 2 y$.

Re-adding $\cos e c 2 y$ and re-using $\left(\star\right)$,

$\cos e c 2 y + \left(\cos e c 4 y + \cos e c 8 y + c o 8 y\right) = \cos e c 2 y + \cot 2 y$,

$= \cot \left(\frac{1}{2} \cdot 2 y\right)$.

$\therefore \cos e c 2 y + \cos e c 4 y + \cos e c 8 y + c o 8 y = \cot y , i . e . ,$

$\cos e c 2 y + \cos e c 4 y + \cos e c 8 y = \cot y - \cot 8 y$, as desired!

May 1, 2018

Another approach I seem to have learned previously from respected sir dk_ch.

#### Explanation:

$R H S = \cot \left(\frac{x}{8}\right) - \cot x$

$= \cos \frac{\frac{x}{8}}{\sin} \left(\frac{x}{8}\right) - \cos \frac{x}{\sin} x$

$= \frac{\sin x \cdot \cos \left(\frac{x}{8}\right) - \cos x \cdot \sin \left(\frac{x}{8}\right)}{\sin x \cdot \sin \left(\frac{x}{8}\right)}$

$= \sin \frac{x - \frac{x}{8}}{\sin x \cdot \sin \left(\frac{x}{8}\right)} = \sin \frac{\frac{7 x}{8}}{\sin x \cdot \sin \left(\frac{x}{8}\right)}$

$= \frac{2 \sin \left(\frac{7 x}{8}\right) \cdot \cos \left(\frac{x}{8}\right)}{2 \cdot \sin \left(\frac{x}{8}\right) \cdot \cos \left(\frac{x}{8}\right) \cdot \sin x}$

$= \frac{\sin x + \sin \left(\frac{3 x}{4}\right)}{\sin x \cdot \sin \left(\frac{x}{4}\right)} = \frac{\cancel{\sin x}}{\cancel{\sin x} \cdot \sin \left(\frac{x}{4}\right)} + \frac{2 \sin \left(\frac{3 x}{4}\right) \cdot \cos \left(\frac{x}{4}\right)}{\sin x \cdot 2 \cdot \sin \left(\frac{x}{4}\right) \cdot \cos \left(\frac{x}{4}\right)}$

$= \cos e c \left(\frac{x}{4}\right) + \frac{\sin x + \sin \left(\frac{x}{2}\right)}{\sin x \cdot \sin \left(\frac{x}{2}\right)} = \cos e c x + \cos e c \left(\frac{x}{2}\right) + c o e s c \left(\frac{x}{4}\right) = L H S$