Question

Please answer this question

Answer 1

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

`\\=\frac{9-16} 3 - \frac 42 \\\\\\=(-7)/(3)-2\\\\\\=\frac{-7-6}3\\\\\\=-\frac{13}3\\\\\\=-4\frac 13 \\\\\\=\text{R.H.S}\\\\\\\text{Proved.}`