Question

identidades trigonométricas​

Answer 1

Hola!

Explicación paso a paso:

■》Solución 01 :

$2(1 - { \cos }^{2} \alpha ) + { \cos}^{2} \alpha = 1 + { \sin}^{2} \alpha \\2( { \sin}^{2} \alpha ) + { \cos}^{2} \alpha = 1 + { \sin}^{2} \alpha \\ 2 { \sin}^{2} \alpha - { \sin}^{2} \alpha + { \cos }^{2} \alpha = 1 \\ { \sin}^{2} \alpha + { \cos }^{2} \alpha = 1 \\ 1 = 1$

■》Solución 02:

$(1 + \sin \alpha )(1 - \sin \alpha ) = { \cos }^{2} \alpha \\ {1}^{2} - {( \sin \alpha )}^{2} = { \cos }^{2} \alpha \\ 1 - { \sin}^{2} \alpha = { \cos }^{2} \alpha \\ 1 = { \cos }^{2} \alpha + { \sin}^{2} \alpha \\ { \sin }^{2} \alpha + { \cos }^{2} \alpha = 1 \\ 1 = 1$

■》Solución 03:

$\frac{ \cos \alpha }{ \cot \alpha } = \sin \alpha \\ \\ \frac{ \cos \alpha }{ \sin \alpha } = \cot \alpha \\ \\ \cot \alpha = \cot \alpha \\ \\ 1 = 1$

■》Solución 04:

$\frac{ \tan \alpha }{ \sin \alpha } - \sec \alpha = 0 \\ \\ \frac{ \tan \alpha }{ \sin \alpha } = \sec \alpha \\ \\ \tan \alpha = \sec \alpha \times \sin \alpha \\ \\ \frac{ \sin \alpha }{ \cos \alpha } = \frac{1}{ \cos \alpha } \times \sin \alpha \\ \\ \frac{\cos \alpha }{\cos \alpha } = \frac{ \sin \alpha }{ \sin \alpha } \\ \\ 1 = 1$

■》Solución 05:

$\frac{ \sin \alpha }{ \csc \alpha } + \frac{ \cos \alpha }{ \sec \alpha } = 1 \\ \\ \sin \alpha \times \sec \alpha + \cos \alpha \times \csc \alpha = \sec \alpha \times \csc \alpha \\ \\ \sin \alpha \times \frac{1}{ \cos \alpha } + \cos \alpha \times \frac{1}{ \sin \alpha } = \frac{1}{ \cos \alpha } \times \frac{1}{ \sin \alpha } \\ \\ \frac{ \sin \alpha }{ \cos \alpha } + \frac{ \cos \alpha }{ \sin \alpha } = \frac{1}{\cos \alpha \times\sin \alpha } \\ \\ { \sin}^{2} \alpha + { \cos }^{2} \alpha = \frac{\cos \alpha \times\sin \alpha}{\cos \alpha \times\sin \alpha} \\ \\ 1 = 1$