Question

Find the value of $\frac{\cos {75}^{\circ }.\sin {12}^{\circ }.\cos {18}^{\circ }}{\sin {15}^{\circ }.\cos {78}^{\circ }.\sin {72}^{\circ }}$

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

Answer: (B)

$1$

Given trigonometric expression is $\frac{\cos 75°\sin 12°\cos 18°}{\sin 15°\cos 78°\sin 22°}$

We can write, $\cos 75°$ as $\cos (90°-15°),\sin 12°$ as $\sin (90°-78°)$ and $\cos 18°$ as $\cos (90°-72°)$

Hence,

$\frac{\cos 75°\sin 12°\cos 18°}{\sin 15°\cos 78°\sin 22°}=\frac{\cos (90°-15°)\sin (90°-78°)\cos (90°-72°)}{\sin 15°\cos 78°\sin 22°}$

We know that, $\cos (90°-\theta )=\sin \theta$ and $\sin (90°-\theta )=\cos \theta$

Hence, we get:

$\frac{\sin 15°\cos 78°\sin 22°}{\sin 15°\cos 78°\sin 22°}$

$=1$

Hence, the answer is $\frac{\cos 75°\sin 12°\cos 18°}{\sin 15°\cos 78°\sin 22°}=1$.