# How do you solve csc(x)+cot(x)=1 ?

Aug 12, 2018

$x = n \pi + {\left(- 1\right)}^{n} \cdot \left(\frac{\pi}{2}\right) , n \in \mathbb{Z}$.

#### Explanation:

Given that, $\csc x + \cot x = 1. \ldots \ldots \ldots \ldots . \left({\star}_{1}\right)$.

We know that, ${\csc}^{2} x - {\cot}^{2} x = 1$.

$\text{Factoring, } \left(\csc x + \cot x\right) \left(\csc x - \cot x\right) = 1$.

:. (1)(cscx-cotx)=1.........[because, (star_1).

$\therefore \csc x - \cot x = 1. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left({\star}_{2}\right)$.

$\left({\star}_{1}\right) + \left({\star}_{2}\right) \Rightarrow 2 \csc x = 2 , \mathmr{and} , \csc x = 1$.

$\therefore \sin x = 1 = \sin \left(\frac{\pi}{2}\right)$.

Since, $\sin \theta = \sin \alpha \Rightarrow \theta = n \pi + {\left(- 1\right)}^{n} \alpha , n \in \mathbb{Z}$,

$\therefore \sin x = \sin \left(\frac{\pi}{2}\right) \Rightarrow x = n \pi + {\left(- 1\right)}^{n} \cdot \left(\frac{\pi}{2}\right) , n \in \mathbb{Z}$.