If A = 30°; show that :

$\dfrac{\text{1 + sin 2 A + cos 2 A}}{\text{sin A + cos A}}$ = 2 cos A

$\dfrac{\text{1 + sin 2 A + cos 2 A}}{\text{sin A + cos A}}$ = 2 cos A

$\text{L.H.S.} = \dfrac{\text{1 + sin 2 A + cos 2 A}}{\text{sin A + cos A}}\\[1em] = \dfrac{\text{1 + sin (2 x 30°) + cos (2 x 30°)}}{\text{sin 30° + cos 30°}}\\[1em] = \dfrac{\text{1 + sin 60° + cos 60°}}{\text{sin 30° + cos 30°}}\\[1em] = \dfrac{1 + \dfrac{\sqrt3}{2} + \dfrac{1}{2}}{\dfrac{1}{2} + \dfrac{\sqrt3}{2}}\\[1em] = \dfrac{\dfrac{2}{2} + \dfrac{\sqrt3}{2} + \dfrac{1}{2}}{\dfrac{1}{2} + \dfrac{\sqrt3}{2}}\\[1em] = \dfrac{\dfrac{2 + \sqrt3 + 1}{2}}{\dfrac{1 + \sqrt3}{2}}\\[1em] = \dfrac{\dfrac{3 + \sqrt3}{2}}{\dfrac{1 + \sqrt3}{2}}\\[1em] = \dfrac{\dfrac{3 + \sqrt3}{\cancel2}}{\dfrac{1 + \sqrt3}{\cancel2}}\\[1em] = \dfrac{3 + \sqrt3}{1 + \sqrt3}\\[1em] = \dfrac{(3 + \sqrt3) \times (1 - \sqrt3)}{(1 + \sqrt3) \times (1 - \sqrt3)}\\[1em] = \dfrac{3 - 3\sqrt3 + \sqrt3 - 3}{1 - 3}\\[1em] = \dfrac{- 2\sqrt3}{- 2}\\[1em] = \sqrt3$

R.H.S. = 2 cos A

= 2 cos 30°

= $2 \times \dfrac{\sqrt3}{2}$

= $\sqrt3$

∴ L.H.S. = R.H.S.

Hence, $\dfrac{\text{1 + sin 2 A + cos 2 A}}{\text{sin A + cos A}}$ = 2 cos A.