Question

(AYUDA!)verifica las siguientes identidades trigonométricas.

Answer 1

a)
$\frac{senx}{cosx} = \frac{1}{cotx} \\ \frac{senx}{cosx} = \frac{1}{ \frac{cosx}{senx} } \\ \frac{senx}{cosx} = \frac{senx}{cosx}$

b)
${cot}^{2} x + 1 = \frac{1}{ {sen}^{2} x} \\ \frac{ {cos}^{2}x }{ {sen}^{2}x } + 1 = \frac{1}{ {sen}^{2}x } \\ \frac{ {cos}^{2} x + {sen}^{2} x}{ {sen}^{2}x } = \frac{1}{ {sen}^{2}x } \\ \frac{1}{ {sen}^{2}x } = \frac{1}{ {sen}^{2}x }$

c)
$1 + {tan}^{2} x = {sec}^{2} x \\ 1 + \frac{ {sen}^{2} x}{ {cos}^{2}x } = {sec}^{2} x \\ \frac{ {cos}^{2} x + {sen}^{2} x}{ {cos}^{2} x} = {sec}^{2} x \\ \frac{1}{ {cos}^{2}x } = {sec}^{2} x \\ {sec}^{2} x = {sec}^{2} x$

d)
$\frac{senx}{1 - cosx} = \frac{1 + cosx}{senx} \\ ( \frac{senx}{1 - cosx} )( \frac{1 + cosx}{1 + cosx}) = \frac{1 + cosx}{senx} \\ \frac{senx(1 + cosx)}{1 - {cos}^{2} x} = \frac{1 + cosx}{senx} \\ \frac{senx(1 + cosx)}{ {sen}^{2}x } = \frac{1 + cosx}{senx} \\ \frac{1 + cosx}{senx} = \frac{1 + cosx}{senx}$