Tricky Tangent Trigo

Brief explanation of my method:

You can rewrite the expression as $\dfrac{\prod \limits_{k=1}^{24} \cos \frac{k\pi}{50}}{\left(\prod \limits_{k=1}^{12} \cos \frac{k\pi}{25}\right)^2}$

You can get these products by using vieta's on the polynomial with roots in the form $\cos \frac{2\pi k}{n}$ where $k$ ranges from $0$ to $n-1$ , and then taking advantage of the symmetry of the product of the roots of the polynomial.

The polynomial $P(x)$ with the roots in the form $\cos \frac{2\pi k}{n}$ satisfies $P(\cos(\theta))=\cos(n\theta)-1$ . You can express $\cos(n\theta)$ in terms of $\cos(\theta)$ by using De Moivre's and expanding the RHS of the equation $\cos(n\theta)+i\sin(n\theta)=(\cos(\theta)+i\sin(\theta))^n$ , and replacing $\sin^2(\theta)$ with $1-\cos^2(\theta)$ . You don't need to find the value of each coefficient of every term in the polynomial, just the first and last non-zero coefficients (since you want to use vieta's on the product of the roots).

You should get $\prod \limits_{k=0,k\nmid25}^{99} \cos \frac{k\pi}{50}=\dfrac{\binom{100}{2}+50}{2^{99}}=\dfrac{625}{2^{96}}=\left(\prod \limits_{k=1}^{24} \cos \frac{k\pi}{50}\right)^4$

and $\prod \limits_{k=0,k\nmid25}^{49} \cos \frac{k\pi}{25}=\dfrac{1}{2^{48}}=\left(\prod \limits_{k=1}^{12} \cos \frac{k\pi}{25}\right)^4$

so the answer is $\dfrac{\frac{5}{2^{24}}}{\frac{1}{2^{24}}}=\boxed{5}$ .

Angle not in degree and Excel shall give fastest calculation.