Respuesta :

Answer:

[tex]\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}[/tex]

Step by step explanation:

[tex]\text{Given that,}\\\\~~~~~~\sin A = \dfrac 38 \\\\\implies \sin^2 A = \dfrac 9{64}\\\\\implies 1 - \cos^2 A = \dfrac{9}{64}\\\\\implies \cos ^2 A = 1 - \dfrac 9{64}\\\\\implies \cos^2 A = \dfrac{55}{64}\\\\\implies \cos A =\pm\sqrt{\dfrac{55}{64}}\\ \\\implies \cos A = \pm\dfrac{\sqrt{55}}8\\\\[/tex]

[tex]\implies \dfrac 1{\cos A} = \pm\dfrac{8}{\sqrt{55}}[/tex]

[tex]\text{Now,}\\\\\csc A - \sec A\\\\=\dfrac{1}{\sin A}- \dfrac{1}{\cos A}\\\\=\dfrac 83 -\left(\pm \dfrac 8{\sqrt {55}} \right)\\ \\\text{Hence,}\\\\\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}[/tex]