Question

Comprove a Identidade Trigonométrica:

a)cossec²x+tg²x=sec²x+cotg²x

b) (tgx-senx)/sen³x=secx/1+cosx

Poderiam explicar também?Obrigado!

Answer 1

Vou chamar sen x de s e cos x de c para facilitar.

É importante lembrar que:
s² + c² = 1, então   c² = 1 - s²   e   s² = 1 - c²
tg x = s/c
cotg x = c/s
sec x = 1/c
cossec x = 1/s

a)
$cossec^2 x + tg^2x \\ \\ = \dfrac{1}{s^2}+\dfrac{s^2}{c^2} \\ \\ Multiplicando \ \ tudo \ \ por \ \ c^2 + s^2 = 1: \\ \\ = \dfrac{(c^2+s^2)}{s^2}+\dfrac{s^2(c^2+s^2)}{c^2} \\ \\ = \dfrac{c^2}{s^2}+\dfrac{s^2}{s^2}+\dfrac{s^2(c^2+s^2)}{c^2} \\ \\ = cotg^2x+\dfrac{s^2}{s^2}+\dfrac{s^2(c^2+s^2)}{c^2} \\ \\ = cotg^2x+\dfrac{s^2}{s^2}+\dfrac{s^2}{c^2} \\ \\ = cotg^2x+\dfrac{c^2}{c^2}+\dfrac{s^2}{c^2} \\ \\ = cotg^2x+\dfrac{c^2+s^2}{c^2} \\ \\ = cotg^2x+\dfrac{c^2+1-c^2}{c^2}$
$= cotg^2x+\dfrac{1}{c^2} \\ \\ = cotg^2x+sec^2x$

b)
$\dfrac{(tg x - s)}{s^3} \\ \\ = \dfrac{\left(\dfrac{s}{c} - s\right)}{s^3} \\ \\ = \dfrac{\left(\dfrac{s}{c} - \dfrac{sc}{c}\right)}{s^3} \\ \\ = \dfrac{\left(\dfrac{s-sc}{c}\right)}{s^3} \\ \\ = \dfrac{\left(\dfrac{1}{c}(s-sc)\right)}{s^3} \\ \\ = \dfrac{\left(sec \ x \ (s-sc)\right)}{s^3} \\ \\ = \dfrac{(sec \ x \ [s(1-c)]}{s^3} \\ \\ = \dfrac{sec \ x \ (1-c)}{s^2} \\ \\ = \dfrac{sec \ x \ (1-c)}{1-c^2} \\ \\ = \dfrac{sec \ x \ (1-c)}{(1-c)(1+c)} \\ \\ = \dfrac{sec \ x }{(1+c)}$