Question

If $L=\left(4\tan 1^{\circ }\right)\left(4\tan 2^{\circ }\right)\left(4\tan 3^{\circ }\right)\dots \left(4\tan {10}^{\circ }\right),$$M=\left(\tan 2^{\circ }\right)\left(\tan 4^{\circ }\right)\left(\tan 6^{\circ }\right)\dots \left(\tan {20}^{\circ }\right),$ and $N=(1+\tan 1^{\circ })(1-\tan 1^{\circ })(1+\tan 2^{\circ })(1-\tan 2^{\circ })\dots (1+\tan {10}^{\circ })(1-\tan {10}^{\circ })$, then the value of $\frac{L}{128MN}$ is

Answer 1

Given
$L=\left(4\tan 1^{\circ }\right)\left(4\tan 2^{\circ }\right)\left(4\tan 3^{\circ }\right)\dots \left(4\tan {10}^{\circ }\right)M=\left(\tan 2^{\circ }\right)\left(\tan 4^{\circ }\right)\left(\tan 6^{\circ }\right)\dots \left(\tan {20}^{\circ }\right)N=(1+\tan 1^{\circ })(1-\tan 1^{\circ })(1+\tan 2^{\circ })(1-\tan 2^{\circ })\dots (1+\tan {10}^{\circ })(1-\tan {10}^{\circ })\Rightarrow N=(1-{\tan }^2{1}^{\circ })(1-{\tan }^2{2}^{\circ })\dots (1-{\tan }^2{10}^{\circ })$

We know that
$\tan 2^{\circ }=\frac{2\tan 1^{\circ }}{1-{\tan }^2{1}^{\circ }}$
So,
$MN=2^{10}\tan 1^{\circ }\tan 2^{\circ }\tan 3^{\circ }\dots \tan {10}^{\circ }\Rightarrow \frac{L}{MN}=\frac{4^{10}}{2^{10}}\Rightarrow \frac{L}{MN}=2^{10}\therefore \frac{L}{128MN}=\frac{2^{10}}{2^7}=8$