Question

Tg(180°+a)tg(90°+B)+tg(90°-a)ctg(270°-a)

Answer 1

Решение

(tg(180°-a)cos(180°-a)tg(90°-a))/(sin(90°+a)ctg(90°+a)tg(90°+a)) =  

= [- tgα * (- cosα) * ctgα] / [ cosα * (- tgα) * (- ctgα)] = 1

Answer 2

To simplify the given expression, we can use trigonometric identities. Let's break down the expression step by step:

1. First, let's deal with the tangent and cotangent functions using their periodicity properties:

   • tg(180° + a) = tg(a) (since tangent has a period of 180°)
   • ctg(270° - a) = ctg(90° + a) = -tg(a) (since cotangent has a period of 180°)

2. Next, let's use the sum and difference identities for tangent:

   • tg(90° + B) = -cot(B) (1)

3. We can simplify the other terms as follows:

   • tg(90° - a) = cot(a) (2)

Now, we can rewrite the expression:

Tg(180° + a) * tg(90° + B) + tg(90° - a) * ctg(270° - a)

Using the identities derived above, we get:

tg(a) * (-cot(B)) + cot(a) * (-tg(a))

Now, notice that both terms have a common factor of -1, so we can factor it out:

-1 * [tg(a) * cot(B) + cot(a) * tg(a)]

Finally, we can apply the identity: tg(a) * cot(B) = 1 (valid when a and B are acute angles):

-1 * [1 + 1]

And the simplified expression is:

-1 * 2 = -2

So, the simplified form of the expression is -2.