# Formula:KLS:14.02:22

$\displaystyle {\displaystyle (1-\alpha q^x)(1-\beta\delta q^x)(1-\gamma q^x)(1-\gamma\delta q^x)\qRacah{n}@{\mu(x)}{\alpha}{\beta}{\gamma}{\delta}{q} {}-(1-q^x)(1-\delta q^x)(\alpha-\gamma\delta q^x)(\beta-\gamma q^x)\qRacah{n}@{\mu(x-1)}{\alpha}{\beta}{\gamma}{\delta}{q} {}=q^x(1-\alpha)(1-\beta\delta)(1-\gamma)(1-\gamma\delta q^{2x}) {} \qRacah{n+1}@{\mu(x)}{\alpha q^{-1}}{\beta q^{-1}}{\gamma q^{-1}}{\delta}{q} }$

## Substitution(s)

$\displaystyle {\displaystyle \mu(x)=q^{-x}+\gamma\delta q^{x+1} =\lambda(x)=q^{-x}+cq^{x-N} =q^{-x}+q^{x+\gamma+\delta+1} =2a\cos@@{\theta}}$ &

$\displaystyle {\displaystyle \lambda(x)=x(x+\gamma+\delta+1)}$ &
$\displaystyle {\displaystyle \mu(x):=q^{-x}+\gamma\delta q^{x+1}}$ &
$\displaystyle {\displaystyle \mu(x)=q^{-x}+\gamma\delta q^{x+1} =\lambda(x)=q^{-x}+cq^{x-N} =q^{-x}+q^{x+\gamma+\delta+1} =2a\cos@@{\theta}}$ &

$\displaystyle {\displaystyle \lambda(x)=x(x+\gamma+\delta+1)}$

## Proof

We ask users to provide proof(s), reference(s) to proof(s), or further clarification on the proof(s) in this space.

## Symbols List

& : logical and
$\displaystyle {\displaystyle R_{n}}$  : $\displaystyle {\displaystyle q}$ -Racah polynomial : http://dlmf.nist.gov/18.28#E19
$\displaystyle {\displaystyle \mathrm{cos}}$  : cosine function : http://dlmf.nist.gov/4.14#E2