Question

(sin²a+ctg²a+cos²a) sin²a+tg²a

Answer 1

Ответ:

We can start by simplifying the terms inside the first set of parentheses:

sin²a + ctg²a + cos²a = sin²a + (1/sin²a) + cos²a (using the identity ctg(a) = 1/tan(a) = cos(a)/sin(a))

= (sin⁴a + 1 + cos⁴a) / (sin²a)

= (1 + sin²a cos²a) / (sin²a)

Now, substituting this expression into the initial equation and simplifying, we get:

(sin²a + ctg²a + cos²a) sin²a + tg²a

= (1 + sin²a cos²a) / (sin²a) * sin²a + (sin²a / cos²a)

= 1 + sin²a cos²a + sin²a

= sin²a (1 + cos²a) + 1

= sin²a sin²a + cos²a sin²a + 1

= sin⁴a + (1 - sin²a) sin²a + 1

= sin⁴a + sin²a - sin⁴a + 1

= sin²a + 1

Therefore, (sin²a+ctg²a+cos²a) sin²a+tg²a simplifies to sin²a + 1.

Answer 2

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