Question

(tga+ctga)^2+(sina*cosa)^2=

Answer 1

Ответ:

$= {1}^{2} + { \sin}^{2} \alpha + 2 \sin \alpha \cos \alpha + { \cos }^{2} \alpha = 1 + 1 + \sin2 \alpha = \sin2 \alpha + 2$

Answer 2

To simplify the expression $(tga+ctga)^2 + (sina*cosa)^2$, we can start by expanding and simplifying each term separately.

1. Expand $(tga+ctga)^2$: $(tga+ctga)^2 = t^2g^2a^2 + 2tcg^2a^3 + c^2t^2g^2a^2$.

2. Expand $(sina*cosa)^2$: $(sina*cosa)^2 = sin^2a*cos^2a$.

Now, let's add the two simplified expressions together:

$(tga+ctga)^2 + (sina*cosa)^2 = t^2g^2a^2 + 2tcg^2a^3 + c^2t^2g^2a^2 + sin^2a*cos^2a$.

We can factor out $a^2$ from the first and third terms:

$a^2(t^2g^2 + c^2t^2g^2) + 2tcg^2a^3 + sin^2a*cos^2a$.

Now, let's factor out $a^2$ from the entire expression:

$a^2(t^2g^2 + c^2t^2g^2 + 2tcg^2a + sin^2a*cos^2a)$.

So, the simplified expression is $a^2(t^2g^2 + c^2t^2g^2 + 2tcg^2a + sin^2a*cos^2a)$.