Question

Решить(tg(2x-п/6)+1) / (1-tg(2x-п/6))

Answer 1

$\frac{tga+1}{1-tga}=\frac{\frac{sina}{cosa}+1}{1-\frac{sina}{cosa}}=\frac{sina+cosa}{cosa-sina}=\frac{\sqrt2\cdot (\frac{1}{\sqrt2}\, sina+\frac{1}{\sqrt2}\, cosa)}{\sqrt2\cdot (\frac{1}{\sqrt2}\, cosa-\frac{1}{\sqrt2}\, sina)}=\\\\\\=\frac{sina\cdot cos\frac{\pi}{4} +cosa\cdot sin\frac{\pi}{4}}{cosa\cdot cos\frac{\pi}{4}-sina\cdot sin\frac{\pi}{4}}=\frac{sin(a+\frac{\pi}{4})}{cos(a+\frac{\pi}{4})}=tg(a+\frac{\pi}{4})\; ;$

$a=2x-\frac{\pi}{6}\; \; \Rightarrow \; \; \; \frac{tg(2x-\frac{\pi}{6})+1}{1-tg(2x-\frac{\pi}{6})}=tg((2x-\frac{\pi}{6})+\frac{\pi}{4})=tg(2x+\frac{\pi}{12})$

$P.S.\qquad \frac{1}{\sqrt2}=\frac{\sqrt2}{2}=sin\frac{\pi}{4}=cos\frac{\pi}{4}$

Answer 2

To solve the expression, we need to simplify it first. Let's work through the steps:

Step 1: Use the trigonometric identity for tangent sum: tg(A + B) = (tg(A) + tg(B)) / (1 - tg(A) * tg(B))

Step 2: Let A = 2x and B = π/6

Step 3: Substitute A and B in the identity: tg(2x + π/6) = (tg(2x) + tg(π/6)) / (1 - tg(2x) * tg(π/6))

Step 4: Use the trigonometric value of tg(π/6): tg(π/6) = √3

Step 5: Substitute the value of tg(π/6) into the equation: tg(2x + π/6) = (tg(2x) + √3) / (1 - tg(2x) * √3)

Step 6: Now we have the expression (tg(2x + π/6)) in terms of tg(2x). Let's find tg(2x) by isolating it:

tg(2x + π/6) = (tg(2x) + √3) / (1 - tg(2x) * √3)

Cross-multiply to eliminate the fraction:

(tg(2x + π/6)) * (1 - tg(2x) * √3) = tg(2x) + √3

Expand the left side:

tg(2x + π/6) - tg(2x) * √3 = tg(2x) + √3

Move all terms involving tg(2x) to one side:

tg(2x + π/6) - tg(2x) = tg(2x) * √3 + tg(2x) + √3

Factor out tg(2x) from the right side:

tg(2x + π/6) - tg(2x) = tg(2x) * (1 + √3) + √3

Now, isolate tg(2x):

tg(2x) = (tg(2x + π/6) - √3) / (1 + √3)

And the expression is simplified to:

tg(2x) = (tg(2x + π/6) - √3) / (1 + √3)

This is the simplified form of the expression (tg(2x - π/6) + 1) / (1 - tg(2x - π/6)).