Question

Cell constituted by two electrodes a and b is .35v and .42 v find e 0 cell

Answer 1

Explanation:

Step-by-step explanation:

$\huge \underline \mathfrak {Solution:-}$

$\begin{lgathered}\frac{ 1}{ \sin(x) } + \frac{ \cos(x) }{ \sin(x) } = k \\ \\ \frac{(1 + \cos(x) )}{ \sin(x) } = k \\ \\ \frac{2 \cos {}^{2} ( \frac{x}{2} ) }{2 \sin(x ) \cos(x) } = k \\ \\ \frac{ \cos( \frac{x}{2} ) }{ \sin( \frac{x}{2} ) } = k \: \: \: \: \: \: \: \: ......(1)\end{lgathered}$

$\begin{lgathered}\frac{ {k}^{2} - 1}{ {k}^{2} + 1} \\ \\ = \frac{ \frac{ \cos {}^{2} ( \frac{x}{2} ) }{ \sin {}^{2} ( \frac{x}{2} ) } - 1 }{ \frac{ \cos {}^{2} ( \frac{x}{2} ) }{ \sin {}^{2} ( \frac{x}{2} ) } + 1 } \\ \\ = \frac{\cos {}^{2} ( \frac{x}{2} ) -\sin {}^{2} ( \frac{x}{2} ) }{\cos {}^{2} ( \frac{x}{2} ) + \sin {}^{2} ( \frac{x}{2} )} \\ \\ = \frac{ \cos(x) }{1} \\ \\ = \cos(x)\end{lgathered}$