Question

If $x={15}^0$, evaluate:
$12\sin 3x\cos 3x+5\tan (2x+{15}^0)-3{\cot }^{2}3x$

Answer 1

Given, $x={15}^{\circ }$
Then, $12\sin 3x\cos 3x+5\tan (2x+{15}^0)-3{\cot }^{2}3x$
= $12\sin {45}^0\cos {45}^0+5\tan ({30}^0+{15}^0)-3{\cot }^2{45}^0$
= $12\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)+5\left(1\right)-3(1{)}^2$
= $6+5-3$
= $8$