Prove that 2sin² π/6cos²π/3=3/2prove that..I need Quality answer p

1.	Prove that 2sin² π/6cos²π/3=3/2prove that..I need Quality answer please don't post unnecessary Answer and don't spam..
Answer» ━━━━━━━━━━━━━━━━━━━━━━━━━ $\huge\bf\red{\underline{\underline{Solution}}}$ ⠀⠀⠀⠀⠀⠀⠀⠀ TAKING L.H.S $\bold{2sin²}$ $\dfrac{π}{6}$ $\bold{+\:cosec²\:}$ $\dfrac{7π}{6}$ $\bold{cos²}$ $\dfrac{π}{3}$ ⠀⠀⠀⠀⠀⠀⠀⠀ putting π = 180° ⠀⠀⠀⠀⠀⠀⠀⠀ = $\bold{2sin²}$ $\dfrac{180}{6}$ $\bold{+\:cosec²}$ $\dfrac{7\:×\:180}{6}$ $\bold{cos²}$ ] $\dfrac{180}{3}$ = $\bold{2sin² \:30° \:+ cosec² \:210°\:cos² \:60}$ = $\bold{2(<klux>SIN</klux>\:30°)²\:+\:(cosec\:210°) \: (cos\:60°)²}$ ⠀⠀⠀⠀⠀⠀⠀⠀ Here, $\bold{sin\:30°\:=}$ $\dfrac{<klux>1</klux>}{2}$ & $\bold{sin\:cos\:60°=}$ $\dfrac{1}{2}$ ⠀⠀⠀⠀ ⠀ For cosec 210° ⠀⠀⠀⠀⠀⠀let's first calculate sin 210° ⠀⠀⠀⠀⠀⠀sin 210° = $\bold{sin\:(180+30)}$ ⠀⠀⠀⠀⠀⠀= $\bold{- \:sin 30°}$ ⠀⠀⠀⠀⠀⠀= $\dfrac{-1}{2}$ ⠀⠀⠀⠀⠀ So, cosec 210° = $\dfrac{1}{sin\:210°}$ = $\dfrac{1}{-1}$ = $\dfrac{2}{-1}$ = - 2 $\text{\large\underline{\red{putting\: values\: in \:equation}}}$ ⠀⠀⠀⠀⠀⠀⠀⠀ $\bold{=\:2(sin\:30°)2 \:+ \:(cosec\:210°)? \:(cos\: 60°)²}$ ⠀⠀⠀⠀⠀= 2 $\dfrac{1²}{2}$ + $\bold{(-2)²}$ $\dfrac{1²}{2}$ ⠀⠀⠀⠀⠀⠀⠀ $\bold{=\:2\:×}$ $\dfrac{1}{2}$ + 4 × $\dfrac{1}{2}$ ⠀⠀⠀⠀⠀⠀= $\dfrac{1}{2}$ + 1 ⠀⠀⠀⠀⠀⠀= $\dfrac{1}{2}$ + 1 ⠀⠀⠀⠀⠀⠀= $\dfrac{3}{2}$ ⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀ $\bold{= \:R.H.S}$ ⠀⠀⠀⠀⠀⠀ $\text{\large\underline{\red{Hence\:proved}}}$ ━━━━━━━━━━━━━━━━━━━━━━━━━