Question

Найдите tg a, если cos a = -10/√101 и a принадлежит (π;3π/2).

Answer 1

Из тождества $1+tg^2 \alpha=\frac{1}{cos^2 \alpha}$

$tg^2 \alpha=\frac{1}{cos^2 \alpha}-1=\frac{1}{(-\frac{10}{\sqrt{101} })^2 }-1=\frac{101}{100} -1=\frac{1}{100}}$

Так как α лежит в III четверти, то $tg\alpha=\frac{1}{10}$

Answer 2

To find the value of tangent (tg) of angle a, given that cos a = -10/√101 and a belongs to the interval (π, 3π/2), we can use the trigonometric identity:

tg a = sin a / cos a

Since we already have the value of cos a, we need to find the value of sin a to calculate tg a. We can use the Pythagorean identity to find sin a:

sin^2 a + cos^2 a = 1

sin^2 a = 1 - cos^2 a sin a = ± √(1 - cos^2 a)

Now, let's proceed with the calculation:

cos a = -10/√101

sin a = ± √(1 - cos^2 a) sin a = ± √(1 - (-10/√101)^2) sin a = ± √(1 - 100/101) sin a = ± √(1/101) = ± 1/√101

However, we know that a is in the interval (π, 3π/2), which corresponds to the third quadrant on the unit circle. In the third quadrant, both sine and cosine are negative. Therefore, we have:

sin a = -1/√101 cos a = -10/√101

Now we can find tg a:

tg a = sin a / cos a tg a = (-1/√101) / (-10/√101) tg a = 1/10

So, tg a = 1/10.