Question

If $x=$ ${20}^{\circ }$, evaluate : $12\sin \frac{3x}{2}\cos 3x+5\tan (2x+5^{\circ })-3{\cot }^{2}3x$.

• A. $7$
• B. $3$
• C. $9$
• D. $10$

Answer 1

The correct option is A $7$

$12\sin \frac{3x}{2}\cos 3x+5\tan (2x+5^{\circ })-3{\cot }^{2}3x$
Put $x={20}^{\circ }$
= $12\sin {30}^{\circ }\cos {60}^{\circ }+5\tan {45}^{\circ }-3{\cot }^2{60}^{\circ }$
= $12\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+5\left(1\right)-3{\left(\frac{1}{\sqrt{3}}\right)}^2$
= $3+5-1$
= $7$