Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity:
$\dfrac{\sin \left(2 \alpha + \beta\right)}{\sin \alpha} - 2 \cos \left(\alpha + \beta\right) = \dfrac{\sin \beta}{\sin \alpha}$

LHS $= \dfrac{\sin \left(2 \alpha + \beta\right)}{\sin \alpha} - 2 \cos \left(\alpha + \beta\right)$

$= \dfrac{\sin \left[\left(\alpha + \alpha\right) + \beta\right] - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha }$

$= \dfrac{\sin \left[\alpha + \left(\alpha + \beta\right)\right] - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha}$

$= \dfrac{\sin \alpha \cos \left(\alpha + \beta\right) + \cos \alpha \sin \left(\alpha + \beta\right) - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha}$

$= \dfrac{\sin \left(\alpha + \beta\right) \cos \alpha - \cos \left(\alpha + \beta\right) \sin \alpha}{\sin \alpha}$

$= \dfrac{\sin \left[\left(\alpha + \beta\right) - \alpha\right]}{\sin \alpha}$

$= \dfrac{\sin \beta}{\sin \alpha} = $ RHS

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity:
$\sec \left(\dfrac{\pi}{4} + \alpha\right) \sec \left(\dfrac{\pi}{4} - \alpha\right) = 2 \sec 2 \alpha$

LHS $=\sec \left(\dfrac{\pi}{4} + \alpha\right) \sec \left(\dfrac{\pi}{4} - \alpha\right)$

$= \dfrac{1}{\cos \left(\dfrac{\pi}{4} + \alpha\right) \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{1}{\cos \left[\left(\dfrac{\pi}{2} - \dfrac{\pi}{4}\right) + \alpha \right] \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{1}{\cos \left[\dfrac{\pi}{2} - \left(\dfrac{\pi}{4} - \alpha\right)\right] \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{1}{\sin \left(\dfrac{\pi}{4} - \alpha\right) \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{2}{2 \sin \left(\dfrac{\pi}{4} - \alpha\right) \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{2}{\sin \left[2 \left(\dfrac{\pi}{4} - \alpha\right)\right]}$

$= \dfrac{2}{\sin \left(\dfrac{\pi}{2} - 2 \alpha\right)}$

$= \dfrac{2}{\cos 2 \alpha}$

$= 2 \sec 2 \alpha = $ RHS