Question

1) а)sin^2a+cos^2a+ctg^2a б) sin^2(1+tg^2) в) 1-1/sin^2a г) 4-tga*ctga

Answer 1

Ответ:

Применяем основные тригонометрические тождества .

$1)\ \ \underbrace{sin^2a+cos^2a}_{1}+ctg^2a=1+ctg^2a=\dfrac{1}{sin^2a}\\\\\\2)\ \ sin^2a\, (\underbrace{1+tg^2a}_{1/cos^2a})=sin^2a\cdot \dfrac{1}{cos^2a}=\dfrac{sin^2a}{cos^2a}=tg^2a\\\\\\3)\ \ 1-\dfrac{1}{sin^2a}=\dfrac{sin^2a-1}{sin^2a}=\dfrac{-(1-sin^2a)}{sin^2a}=\dfrac{-cos^2a}{sin^2a}=-ctg^2a\\\\\\4)\ \ 4-\underbrace{tga\cdot ctga}_{1}=4-1=3$  

Answer 2

Let's simplify each expression one by one:

a) sin^2(a) + cos^2(a) + ctg^2(a): The identity sin^2(a) + cos^2(a) = 1 holds true for any angle a. Therefore, we can simplify the expression as follows: sin^2(a) + cos^2(a) + ctg^2(a) = 1 + ctg^2(a).

b) sin^2(1 + tg^2): It seems like there may be some missing information or a typo in this expression. If you could provide the complete expression or clarify what you meant, I would be able to assist you better.

c) 1 - 1/sin^2(a): To simplify this expression, we'll start by finding a common denominator for the fraction: 1 - 1/sin^2(a) = (sin^2(a) - 1)/sin^2(a).

Now, we can factor the numerator: sin^2(a) - 1 = (sin(a) - 1)(sin(a) + 1).

Therefore, the simplified expression becomes: 1 - 1/sin^2(a) = (sin(a) - 1)(sin(a) + 1)/sin^2(a).

d) 4 - tg(a) * ctg(a): The tangent (tg) and cotangent (ctg) functions are reciprocals of each other. Thus, tg(a) * ctg(a) is equivalent to 1.

Therefore, the expression simplifies to: 4 - tg(a) * ctg(a) = 4 - 1 = 3.

In summary: a) sin^2(a) + cos^2(a) + ctg^2(a) simplifies to 1 + ctg^2(a). b) Please provide additional information or clarify the expression sin^2(1 + tg^2). c) 1 - 1/sin^2(a) simplifies to (sin(a) - 1)(sin(a) + 1)/sin^2(a). d) 4 - tg(a) * ctg(a) simplifies to 3.