Question

Tg (π/4 + x) + tg (π/4 - x)

Answer 1

Ответ:

Пошаговое объяснение:

Tg (π/4 + x) + tg (π/4 - x))=син(π/4 + x +π/4 - x)/кос(π/4 + x)кос(π/4 - x)=

син(π/2 )/кос(π/4 + x)кос(π/4 - x)=

1/(1/2(кос ((π/4 + x)-(π/4 - x))+кос ((π/4 + x)+(π/4 - x)))=

2 / кос (π/4 + x-π/4+ x)+кос (π/4 + x+π/4 - x)=2/кос 2х

Answer 2

To simplify the expression $\tan\left(\frac{\pi}{4} + x\right) + \tan\left(\frac{\pi}{4} - x\right)$, you can use the trigonometric identity for the sum of tangents:

$\tan(A + B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A) \tan(B)}$

In this case, $A = \frac{\pi}{4}$ and $B = x$, so:

$\tan\left(\frac{\pi}{4} + x\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan(x)}{1 - \tan\left(\frac{\pi}{4}\right) \tan(x)}$

Now, we can calculate $\tan\left(\frac{\pi}{4}\right)$:

$\tan\left(\frac{\pi}{4}\right) = 1$

So the expression becomes:

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

Now, to add these fractions, we need a common denominator, which is $(1 - \tan(x))(1 + \tan(x))$:

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

Now, combine the fractions:

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

Now, simplify the numerator:

$\frac{(1 + \tan(x))^2 + (1 - \tan(x))^2}{(1 - \tan(x))(1 + \tan(x))}$

Expand and simplify the squares:

$\frac{1 + 2\tan(x) + \tan^2(x) + 1 - 2\tan(x) + \tan^2(x)}{1 - \tan^2(x)}$

Combine like terms:

$\frac{2 + 2\tan^2(x)}{1 - \tan^2(x)}$

Now, notice that $1 - \tan^2(x)$ is equal to $\sec^2(x)$ (a trigonometric identity), so we can replace it:

$\frac{2 + 2\tan^2(x)}{\sec^2(x)}$

Finally, since $\tan^2(x) + 1 = \sec^2(x)$ (another trigonometric identity), we can simplify further:

$\frac{2\sec^2(x)}{\sec^2(x)}$

Now, the $\sec^2(x)$ terms cancel out:

$\frac{2}{1} = 2$

So, the simplified expression for $\tan\left(\frac{\pi}{4} + x\right) + \tan\left(\frac{\pi}{4} - x\right)$