Question

If $\sin 4A-\cos 2A=\cos 4A-\sin 2A$ (where, $0<A<\frac{\mathrm{\pi }}{4}$) then the value of $\tan 4A$ is:

• A. $1$
• B. $\frac{1}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
• E. $\frac{\sqrt{3}+1}{\sqrt{3}-1}$

Answer 1

The correct option is C

$\sqrt{3}$

Explanation for the correct option.

Given that, $\sin 4A-\cos 2A=\cos 4A-\sin 2A$.

$\sin 4A+\sin 2A=\cos 4A+\cos 2A$

Now using identity, $\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)\mathrm{and}\mathrm{cosa}+\cos b=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$.

$$\begin{gathered} \Rightarrow 2\sin \left(\frac{4A+2A}{2}\right)\cos \left(\frac{4A-2A}{2}\right)=2\cos \left(\frac{4A+2A}{2}\right)\cos \left(\frac{4A-2A}{2}\right) \\ \Rightarrow 2\sin \left(3A\right)\cos \left(A\right)=2\cos \left(3A\right)\cos \left(A\right) \\ \Rightarrow Sin3A=\cos 3A \\ \Rightarrow \tan 3A=1 \\ \Rightarrow 3A=\frac{\mathrm{\pi }}{4} \\ \Rightarrow A=\frac{\mathrm{\pi }}{12} \end{gathered}$$

So, the value of:

$$\begin{gathered} \therefore \tan 4A=\tan \frac{\mathrm{\pi }}{3} \\ =\sqrt{3} \end{gathered}$$

Hence, option C is correct.