Respuesta :

dashataran

Answer:

[tex]\tan t=\dfrac{x}{5}\\\\\text{Now,}\\\sec^2t-\tan^2t=1\\\\\text{or, }\sec^2t-\dfrac{x^2}{5^2}=1\\\\\text{or, }\sec^2t=1+\dfrac{x^2}{25}\\\\\text{or, }\sec^2t=\dfrac{x^2+25}{25}\\\\\therefore\ \sec t=\dfrac{\sqrt{x^2+25}}{5}[/tex]