[tex]\text{Given that,}\\\\1-\cos A = \dfrac 12\implies \cos A = 1-\dfrac 12 = \dfrac 12\\\\\text{L.H.S}\\\\=\tan^2 A + 4 \sin^2 A\\\\=\dfrac{\sin^2 A}{\cos^2 A}+4(1-\cos^2 A)\\\\\\=\dfrac{1-\cos^2 A}{\cos^2 A} +4-4\cos^2 A\\\\\\=\dfrac{1-\left( \dfrac 12 \right)^2 }{\left(\dfrac 12 \right)^2 } +4 -4 \left( \dfrac 12 \right)^2\\\\\\=\dfrac{1- \dfrac 14}{\dfrac 14}+4-4\cdot \dfrac 14\\\\\\=\dfrac{\tfrac 34}{\tfrac 14}+4-1\\\\\\=\dfrac 34 \cdot 4 +3\\\\=3+3\\\\=6\\\\=\text{R.H.S}[/tex]
