Trigo .. again!

Claim 1: $\frac{7\cos (b-c)}{\cos (a-d)}=2$ Proof: $\;\;$ Using the Sum and Difference Formulas for Cosine, $\cos (b-c)=\cos b\cos c+\sin b\sin c\;$ $\;\cos(a-d)=\cos a\cos d+\sin a\sin d$ In order to match the given equations with the above equations, we rewrite it as, $\sin a-8\sin d=4\sin c-7\sin b\;\implies \sin^2 a-16\sin a \sin d+64\sin^2 d=16\sin^2 c-56\sin c\sin b+49\sin ^2b \tag{1}$ $\;\;\;\cos a -8\cos d=4\cos c-7\cos b\implies \cos^2 a-16\cos a\cos d+64\cos^2 d=16\cos^2 c-56\cos c\cos b+49\cos ^2b\tag{2}$ Adding $(1)$ and $(2)$ , $\begin{aligned} \sin^2a+\cos^2a-16(\sin a \sin d+\cos a\cos d)+64(\sin^2 d+\cos^2 d)&=16(\sin^2 c+\cos ^2c)-56(\sin c\sin b+\cos c\cos b)+49(\sin ^2b+\cos ^2b) \\ 1-16\cos (a-d)+64&=16-56\cos (b-c)+49 \\ 2\cos (a-d)&=7\cos (b-c) \\ \therefore \frac{7\cos (b-c)}{\cos (a-d)}&=2 \end{aligned}$

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Claim 2: $(1+\tan 1^{\circ})(1+\tan 2^{\circ})\cdots (1+\tan 45^{\circ})=2^{23}$ Proof: $\;\;$ Using the Sum and Difference Formulas for Tangent, when the two angles add up to $45^{\circ}$ , $\begin{aligned} \tan (x+y)&=\frac{\tan x+\tan y}{1-\tan x\tan y} \\ 1-\tan x\tan y&=\tan x+\tan y \\ \implies 2&=(1+\tan x)(1+\tan y) \end{aligned}$ Note, $\begin{aligned} (1+\tan 1^{\circ})&(1+\tan 44^{\circ})=2 \\ (1+\tan 2^{\circ})&(1+\tan 43^{\circ})=2 \\ (1+\tan 3^{\circ})&(1+\tan 42^{\circ})=2 \\ &. \\ &. \\ &. \\ (1+\tan 22^{\circ})&(1+\tan 23^{\circ})=2 \\ &\;\;\;1+\tan 45^{\circ}=2 \end{aligned}$ Multiplying the above equations, $(1+\tan 1^{\circ})(1+\tan 2^{\circ})\cdots (1+\tan 45^{\circ})=2^{23}$

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So, $\boxed{m=2}$ and $\boxed{n=23}$ and the only true statement is: $m$ is a prime number.