Question

Please answer this question

Answer 1

$\text{L.H.S}\\\\=\cot^2 \dfrac{\pi}6 -2\cos^2 \dfrac{\pi}{3}-\dfrac 34 \sec^2 \dfrac{\pi}4-4\sec^2 \dfrac{\pi}6\\\\\\=\cot^2 30^\circ -2 \cos^2 60^\circ- \dfrac 34 \sec^2 45^{\circ} -4 \sec^2 30^{\circ}~~~~~~~;[\pi= 180^{\circ}]\\\\\\=\left(\sqrt 3 \right)^2 -2 \left( \dfrac 12 \right)^2 -\dfrac 34 \left(\sqrt 2 \right)^2 -4 \left(\dfrac 2{\sqrt 3} \right)^2\\\\\\=3-2 \cdot \dfrac 14 - \dfrac 34 \cdot 2 -4 \cdot \dfrac 43\\ \\\\=3-\dfrac 12-\dfrac 32 - \dfrac{16}3$

$\\=\dfrac{9-16} 3 - \dfrac 42 \\\\\\=\dfrac{-7}{3}-2\\\\\\=\dfrac{-7-6}3\\\\\\=-\dfrac{13}3\\\\\\=-4\dfrac 13 \\\\\\=\text{R.H.S}\\\\\\\text{Proved.}$