Question

Помогите доказать тождество1) ctgt/(tgt+ctgt)=cos^2t2) (1+tgt)/(1+ctgt)=tgt

Answer 1

Вот пожалуйста тебе решение))))

Answer 2

Proof of Identity 1: ctg(t)/(tg(t)+ctg(t)) = cos^2(t)

To prove the given identity, we can start by simplifying the left-hand side (LHS) expression and showing that it is equal to the right-hand side (RHS) expression.

Let's rewrite the LHS expression using trigonometric identities:

ctg(t) / (tg(t) + ctg(t))

Since ctg(t) is the reciprocal of tg(t), we can substitute tg(t) with 1/ctg(t) in the denominator:

ctg(t) / (1/ctg(t) + ctg(t))

Now, we need to find a common denominator for the two fractions in the denominator:

ctg(t) / ((1 + ctg^2(t)) / ctg(t))

Next, we can simplify the expression further by multiplying the numerator and denominator by ctg(t):

ctg^2(t) / (1 + ctg^2(t))

Using the Pythagorean identity sin^2(t) + cos^2(t) = 1, we can replace ctg^2(t) with 1 - cos^2(t):

(1 - cos^2(t)) / (1 + (1 - cos^2(t)))

Now, we can simplify the expression by combining like terms:

(1 - cos^2(t)) / (2 - cos^2(t))

Finally, we can simplify further by canceling out common factors:

1 - cos^2(t) = sin^2(t)

Therefore, the LHS expression is equal to sin^2(t), which is equivalent to the RHS expression. Hence, we have proved the given identity.

Proof of Identity 2: (1 + tg(t))/(1 + ctg(t)) = tg(t)

To prove the second identity, we can start by simplifying the LHS expression and showing that it is equal to the RHS expression.

Let's rewrite the LHS expression using trigonometric identities:

(1 + tg(t)) / (1 + ctg(t))

Next, we can multiply the numerator and denominator by ctg(t):

(1 + tg(t)) * ctg(t) / (1 + ctg(t)) * ctg(t)

Expanding the numerator and denominator:

(ctg(t) + tg(t) * ctg(t)) / (ctg(t) + ctg^2(t))

Using the fact that tg(t) * ctg(t) = 1, we can simplify the expression:

(ctg(t) + 1) / (ctg(t) + ctg^2(t))

Now, we can factor out ctg(t) from the denominator:

(ctg(t) + 1) / (ctg(t) * (1 + ctg(t)))

Canceling out the common factor ctg(t):

(1 + ctg(t)) / (1 + ctg(t))

The expression simplifies to 1, which is equivalent to the RHS expression. Hence, we have proved the given identity.

Therefore, both identities have been proven to be true.