Question

Choose the correct option for the following.

The value of $\frac{\cos 3A-2\cos 4A}{\sin 3A+2\sin 4A}$ is $\frac{1-\sqrt{2}}{1+\sqrt{6}}$, when $A={15}^0$.

• A. True
• B. False
• C. Ambiguous
• D. Data insufficient

Answer 1

The correct option is A True

Given, $A={15}^{\circ }$

Thus, $\frac{\cos 3A-2\cos 4A}{\sin 3A+2\sin 4A}$

Put $A={15}^{\circ }$

= $\frac{\cos {45}^0-2\cos {60}^0}{\sin {45}^0+2\sin {60}^0}$

= $\frac{\frac{1}{\sqrt{2}}-2\left(\frac{1}{2}\right)}{\frac{1}{\sqrt{2}}+2\left(\frac{\sqrt{3}}{2}\right)}$

= $\frac{\frac{1}{\sqrt{2}}-1}{\frac{1}{\sqrt{2}}+\sqrt{3}}$

= $\frac{\frac{1-\sqrt{2}}{\sqrt{2}}}{\frac{1}{\sqrt{2}}+\sqrt{3}}$

= $\frac{1-\sqrt{2}}{1+\sqrt{6}}$