Respuesta:
Queda verificado
Explicación paso a paso:
∆) Solución:
[tex] { \tan }^{2} \theta = {csc}^{2} \theta \: \: { \tan }^{2} \theta - 1 \\ 1 = {csc}^{2} \theta \: \: { \tan }^{2} \theta - { \tan }^{2} \theta \\ 1 = \frac{1}{{ \sin}^{2} \theta}. \frac{{ \sin}^{2} \theta}{{ \cos}^{2} \theta} - \frac{{ \sin}^{2} \theta}{{ \cos}^{2} \theta} \\ 1 = \frac{1}{{ \cos}^{2} \theta} - \frac{{ \sin}^{2} \theta}{{ \cos}^{2} \theta} \\ 1 = \frac{1 - { \sin}^{2} \theta}{{ \cos}^{2} \theta} \\ 1 = \frac{{ \cos}^{2} \theta}{{ \cos}^{2} \theta} \\ 1 = 1[/tex]
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Respuesta:
Queda verificado
Explicación paso a paso:
∆) Solución:
[tex] { \tan }^{2} \theta = {csc}^{2} \theta \: \: { \tan }^{2} \theta - 1 \\ 1 = {csc}^{2} \theta \: \: { \tan }^{2} \theta - { \tan }^{2} \theta \\ 1 = \frac{1}{{ \sin}^{2} \theta}. \frac{{ \sin}^{2} \theta}{{ \cos}^{2} \theta} - \frac{{ \sin}^{2} \theta}{{ \cos}^{2} \theta} \\ 1 = \frac{1}{{ \cos}^{2} \theta} - \frac{{ \sin}^{2} \theta}{{ \cos}^{2} \theta} \\ 1 = \frac{1 - { \sin}^{2} \theta}{{ \cos}^{2} \theta} \\ 1 = \frac{{ \cos}^{2} \theta}{{ \cos}^{2} \theta} \\ 1 = 1[/tex]