Trigonometric Identities and Values

$\sqrt {2}\sec 45^{\circ} - \tan 30^{\circ} = \sqrt {2}\times \sqrt {2} - \dfrac {1}{\sqrt {3}} = \dfrac {2\sqrt {3} - 1}{\sqrt {3}}$.

The roots of the equation $x^{2n + 1}- 1 = 0$ are
$1, \displaystyle cos \frac{2 \pi}{2n + 1} + i  sin  \frac{2 \pi}{2n + 1}, cos \frac{4 \pi}{2n + 1} + i  sin \frac{4 \pi}{2n + 1}, ........, cos \frac{4 n \pi}{2n + 1} + i  sin  \frac{4n  \pi}{2n + 1}$
Therefore $\displaystyle x^{2n+1} - 1 = (x - 1) \left ( x - cos \frac{2 \pi}{2n + 1} - i  sin \frac{2 \pi}{2n + 1} \right ) \left ( x - cos \frac{4 \pi}{2n + 1} - i  sin \frac{4 \pi}{2n + 1} \right ) ....... \left ( x - cos \frac{4 n\pi}{2n + 1} - i  sin \frac{4n \pi}{2n + 1} \right )$
Further since
$\displaystyle cos \left ( \frac{(2n + 1) - r}{2n + 1} \right ) 2\pi = cos \frac{2 r \pi}{2n + 1}$
and $\displaystyle sin \left ( \frac{(2n + 1) - r}{2n + 1} \right ) 2\pi = -sin \frac{2 r \pi}{2n + 1}$
it follows that
$\displaystyle \left ( x - cos \frac{2 \pi}{2n + 1} - i  sin \frac{2 \pi}{2n + 1}\right ) \left ( x - cos \frac{4 \pi}{2n + 1} - i  sin \frac{4 \pi}{2n + 1}\right )$
$= x^2 - 2x  cos \displaystyle \frac{2 \pi}{2n + 1} + 1$
$\left ( x - cos \frac{4 \pi}{2n + 1} - i  sin \frac{4 \pi}{2n + 1}\right )\left ( x - cos \frac{(4n - 2) \pi}{2n + 1} - i  sin \frac{(4n - 2) \pi}{2n + 1}\right )$
$=x^2 - 2x  cos \displaystyle \frac{4 \pi}{2n + 1} + 1$
$\left ( x - cos \frac{2 n\pi}{2n + 1} - i  sin \frac{2 n\pi}{2n + 1}\right )\left ( x - cos \frac{(2n + 2) \pi}{2n + 1} - i  sin \frac{(2n + 2)}{(2n + 1)} \pi\right )$
$= x^2 - 2x   cos \frac{2 n \pi}{2n + 1} + 1$
Thus the polynomial $x^{2n + 1} - 1$ can be rewritten thus
$x^{2n + 1} - 1 = (x - 1) \displaystyle \left ( x^2 - 2x  cos  \frac{2 \pi}{2n + 1} + 1\right ) \left ( x^2 - 2x  cos  \frac{4 \pi}{2n + 1} + 1\right )........ \left ( x^2 - 2x  cos  \frac{2 n\pi}{2n + 1} + 1\right ) $
or $\displaystyle \frac{x^{2n + 1} - 1}{x - 1} = \left ( x^2 - 2x  cos  \frac{2 \pi}{2n + 1} + 1\right ) \left ( x^2 - 2x  cos  \frac{4 \pi}{2n + 1} + 1\right ) ......... \left ( x^2 - 2x  cos  \frac{2 n\pi}{2n + 1} + 1\right ) $
Taking $\displaystyle \lim _{x \rightarrow 1}$ on both sides
$(2n + 1) = 2^{2n} sin^2 \displaystyle \frac{\pi}{2n + 1} sin^2 \frac{2 \pi}{2n + 1} ..... sin^2 \frac{n \pi}{2n + 1}$
Hence, $\displaystyle sin \frac{\pi}{2n + 1} sin \frac{2 \pi}{2n + 1} ...... sin \frac{n \pi}{2n + 1} = \frac{\sqrt{(2n + 1)}}{2^n}$

We know,

As we know that 

This is a telescoping product. Each term (1 - cos(60+x)/cos(x)) simplifies to (cos x - cos(60+x))/cos x = (2 sin(60/2 + x) sin(60/2))/cos x. The product eventually cancels out to -1.

Using the identity tan(A)tan(B)tan(C) = tan(A+B+C) - (tan A + tan B + tan C) or expanding the terms, the expression simplifies to 0.

Express sin(75 degrees) as sin(45 degrees + 30 degrees). Using the sine addition formula, sin(45)cos(30) + cos(45)sin(30) = (1/sqrt(2))(sqrt(3)/2) + (1/sqrt(2))(1/2) = (sqrt(3) + 1) / (2 * sqrt(2)).

Using the identity sin(3x) = 3 sin x - 4 sin^3 x and product-to-sum formulas, the expression simplifies to 1.

Using cos^2 A = (1 + cos 2A)/2, the expression becomes (1 + cos 146)/2 + (1 + cos 94)/2 + (cos 146 + cos 94)/2. This simplifies to 1 + (cos 146 + cos 94)/2 + (cos 146 + cos 94)/2 = 1 + cos 146 + cos 94. Using sum-to-product, this results in 3/4.

$\dfrac{\cos 13 - \sin 13}{\cos 13 + \sin 13} + \dfrac{1}{\cot 148}$

Pair the terms using the property (1 + tan(theta))(1 + tan(45 - theta)) = 2. Since there are terms from 1 to 44 degrees pairing up to give 2^22, and the term (1 + tan(45)) = 2, the total product is 2^22 * 2 = 2^23, so n = 23.

As we know that