Question

Найдите sin a -10 cos a / 2 sin a+ cos a, если tg a=3

Answer 1

Ответ: -1

Пошаговое объяснение:

разделим числитель изнаменатель  почленно на cosa,  тогда sina/cosa=tga=3,   cosa/cosa=1,  получим   (tga-10)/2tga+1=(3-10)/(2*3+1)=-1

Answer 2

To find the value of the expression, we first need to find the values of sin(a) and cos(a) using the given information that tan(a) = 3.

We know that: tan(a) = sin(a) / cos(a)

Given that tan(a) = 3, we can write: 3 = sin(a) / cos(a)

Now, we can solve for sin(a) and cos(a). Let's assume that a is an acute angle in the first quadrant, as the tangent is positive (tan(a) = 3).

Step 1: Find cos(a): cos(a) = 1 / sqrt(1 + tan^2(a)) cos(a) = 1 / sqrt(1 + 3^2) cos(a) = 1 / sqrt(1 + 9) cos(a) = 1 / sqrt(10)

Step 2: Find sin(a): sin(a) = tan(a) * cos(a) sin(a) = 3 * (1 / sqrt(10)) sin(a) = 3 / sqrt(10)

Now, we can substitute sin(a) and cos(a) back into the given expression and simplify:

(sin(a) - 10cos(a)) / (2sin(a) + cos(a)) = ((3/sqrt(10)) - 10 * (1/sqrt(10))) / (2 * (3/sqrt(10)) + (1/sqrt(10))) = (3 - 10) / (2 * 3 + 1) = -7 / 7 = -1

Therefore, sin(a) - 10cos(a) / (2sin(a) + cos(a)) = -1.