Question

Сtg(n/2 +a) + tg (2n+a)

Answer 1

Используем формулы приведения сtg(n/2 +a) + tg (2n+a)= -tg a + tg a=0

Answer 2

Resheniye na fotografiye

Answer 3

To simplify the expression $\tan\left(\frac{n}{2} + a\right) + \tan(2n + a)$, we can use the trigonometric identity for the sum of tangents:

$\tan(x + y) = \frac{\tan(x) + \tan(y)}{1 - \tan(x) \tan(y)}.$

Let's set $x = \frac{n}{2}$ and $y = a$ in the above identity and then apply it to the given expression:

$\tan\left(\frac{n}{2} + a\right) = \frac{\tan\left(\frac{n}{2}\right) + \tan(a)}{1 - \tan\left(\frac{n}{2}\right) \tan(a)}.$

Now, let's set $x = 2n$ and $y = a$ in the same identity:

$\tan(2n + a) = \frac{\tan(2n) + \tan(a)}{1 - \tan(2n) \tan(a)}.$

Since $\tan(2n) = \frac{2\tan(n)}{1 - \tan^2(n)}$, we can substitute this into the expression for $\tan(2n + a)$:

$\tan(2n + a) = \frac{\frac{2\tan(n)}{1 - \tan^2(n)} + \tan(a)}{1 - \frac{2\tan(n)}{1 - \tan^2(n)} \tan(a)}.$

Now, we can rewrite the original expression $\tan\left(\frac{n}{2} + a\right) + \tan(2n + a)$ as:

$\frac{\tan\left(\frac{n}{2}\right) + \tan(a)}{1 - \tan\left(\frac{n}{2}\right) \tan(a)} + \frac{\frac{2\tan(n)}{1 - \tan^2(n)} + \tan(a)}{1 - \frac{2\tan(n)}{1 - \tan^2(n)} \tan(a)}.$

To further simplify, we can find a common denominator for the two fractions:

$\text{Common Denominator} = \left(1 - \tan\left(\frac{n}{2}\right) \tan(a)\right) \left(1 - \frac{2\tan(n)}{1 - \tan^2(n)} \tan(a)\right).$

Now, we can combine the fractions over the common denominator:

$\frac{\left(\tan\left(\frac{n}{2}\right) + \tan(a)\right)\left(1 - \frac{2\tan(n)}{1 - \tan^2(n)} \tan(a)\right) + \left(\frac{2\tan(n)}{1 - \tan^2(n)} + \tan(a)\right)\left(1 - \tan\left(\frac{n}{2}\right) \tan(a)\right)}{\text{Common Denominator}}.$