Question

demuestra las siguientes identidades
a) csc θ - sen θ = cot θ * cos θ
b) sen x + cos x * cot x = csc x
c) cos θ (tan θ + 1) = sen θ + cos θ
d) cos x ( sec x _ sen x tan x) = cos^{2}x

Answer 1

¡Buenas!

$\boldsymbol{a)}\\ \\ csc(\alpha) - sen(\alpha) = cot(\alpha) \cdot cos(\alpha)\\ \\ \dfrac{1}{sen(\alpha)} - sen(\alpha) = \dfrac{cos(\alpha)}{sen(\alpha)} \cdot cos(\alpha) \\ \\ \\ \dfrac{1-sen^{2}(\alpha)}{sen(\alpha)} = \dfrac{cos^{2}(\alpha)}{sen(\alpha)}\\ \\ \\ \boxed{cos^{2}(\alpha)+sen^{2}(\alpha) = 1} \\ \\ \\ \dfrac{cos^{2}(\alpha)}{sen(\alpha)} = \dfrac{cos^{2}(\alpha)}{sen(\alpha)}\\ \\ \\ 1 = 1\ \ \ \ \checkmark \\ \\ \\ \textrm{l.q.q.d}$

$\boldsymbol{b)}\\ \\ sen(\alpha) + cos(\alpha) \cdot cot(\alpha) = csc(\alpha) \\ \\ sen(\alpha) + cos(\alpha) \cdot \dfrac{cos(\alpha)}{sen(\alpha)} = \dfrac{1}{sen(\alpha)}\\ \\ \\ sen(\alpha) + \dfrac{cos^{2}(\alpha)}{sen(\alpha)} = \dfrac{1}{sen(\alpha)} \\ \\ \\ \dfrac{sen^{2}(\alpha ) + cos^{2}(\alpha )}{sen(\alpha )} = \dfrac{1}{sen(\alpha)} \\ \\ \\ \dfrac{1}{sen(\alpha)} = \dfrac{1}{sen(\alpha)} \\ \\ \\ 1=1\ \ \ \ \checkmark \\ \\ \\ \textrm{l.q.q.d}$

$\boldsymbol{c)}\\ \\ cos(\alpha)(tan(\alpha)+1)= sen(\alpha)+cos(\alpha)\\ \\ cos(\alpha) \cdot tan(\alpha) + cos(\alpha) = sen(\alpha) + cos(\alpha)\\ \\ cos(\alpha) \cdot \dfrac{sen(\alpha)}{cos(\alpha)} + cos(\alpha) = sen(\alpha) + cos(\alpha)\\ \\ sen(\alpha) + cos(\alpha) = sen(\alpha) + cos(\alpha) \\ \\ 0= 0\ \ \ \ \checkmark \\ \\ \\ \textrm{l.q.q.d}$

$\boldsymbol{d)}\\ \\ cos(\alpha) (sec(\alpha) - sen(\alpha) \cdot tan(\alpha))= cos^{2}(\alpha)\\ \\ cos(\alpha) \cdot sec(\alpha) - cos(\alpha) \cdot sen(\alpha) \cdot tan(\alpha) = cos^{2}(\alpha)\\ \\ cos(\alpha) \cdot \dfrac{1}{cos(\alpha)} - cos(\alpha) \cdot sen(\alpha) \cdot \dfrac{sen(\alpha)}{cos(\alpha)} = cos^{2}(\alpha)\\ \\ \\ 1-sen^{2}(\alpha) = cos^{2}(\alpha)\\ \\ cos^{2}(\alpha) = cos^{2}(\alpha)\\ \\ 1=1\ \ \ \ \checkmark \\ \\ \\ \textrm{l.q.q.d}$