Question

Доказать, что: tg(45+a) - tg(45-a) = 2tg2a

Answer 1

Доказательство во вложении.

Answer 2

To prove the trigonometric identity:

tg(45° + a) - tg(45° - a) = 2tg(2a)

We'll start with the left-hand side (LHS) and manipulate it step by step to obtain the right-hand side (RHS).

Step 1: Express the tangent of a sum of angles formula

tg(α + β) = (tgα + tgβ) / (1 - tgα * tgβ)

Step 2: Apply the formula to the first term of LHS (tg(45° + a))

tg(45° + a) = (tg45° + tga) / (1 - tg45° * tga)

Step 3: Simplify tg45°

tg45° = 1

Step 4: Substitute back into Step 2

tg(45° + a) = (1 + tga) / (1 - tga)

Step 5: Apply the formula to the second term of LHS (tg(45° - a))

tg(45° - a) = (tg45° - tga) / (1 + tg45° * tga)

Step 6: Simplify tg45°

tg45° = 1

Step 7: Substitute back into Step 5

tg(45° - a) = (1 - tga) / (1 + tga)

Step 8: Now we can rewrite the LHS

LHS = (1 + tga) / (1 - tga) - (1 - tga) / (1 + tga)

Step 9: Find a common denominator for both terms

Common denominator = (1 - tga)(1 + tga)

Step 10: Rewrite the fractions with the common denominator

LHS = [(1 + tga)(1 + tga) - (1 - tga)(1 - tga)] / [(1 - tga)(1 + tga)]

Step 11: Expand the numerator

LHS = (1 + 2tga + tga^2 - 1 + 2tga - tga^2) / [(1 - tga)(1 + tga)]

Step 12: Simplify the numerator

LHS = (4tga) / [(1 - tga)(1 + tga)]

Step 13: Factor out 2tg(α)

LHS = 2tga / [(1 - tga)(1 + tga)]

Step 14: Apply the identity: tg2α = 2tgα / (1 - tg²α)

RHS = 2tg2a

Since LHS and RHS are the same, we have proven the trigonometric identity:

tg(45° + a) - tg(45° - a) = 2tg(2a)