Respuesta :
For the answer to the question above, I'll provide a solution to my answer below
sin²θ + cos²θ = 1
so, cos²θ = 1 - (5/6)² => 11/36
=> cosθ = ±√11/6
Also, tanθ = sinθ/cosθ => (5/6)/(±√11/6)
i.e. tanθ = ±5/√11
I hope my answer helped you. Feel free to ask more questions. Have a nice day!
sin²θ + cos²θ = 1
so, cos²θ = 1 - (5/6)² => 11/36
=> cosθ = ±√11/6
Also, tanθ = sinθ/cosθ => (5/6)/(±√11/6)
i.e. tanθ = ±5/√11
I hope my answer helped you. Feel free to ask more questions. Have a nice day!
Answer: The required values are
[tex]\cos\theta=\pm\dfrac{\sqrt{11}}{6},\\\\\\\tan\theta=\pm\dfrac{5\sqrt{11}}{11}.[/tex]
Step-by-step explanation: We are given the following value :
[tex]\sin\theta=\dfrac{5}{6}[/tex]
We are to find the values of [tex]\cos\theta[/tex] and [tex]\tan\theta.[/tex]
We will be using the following formulas :
[tex](i)\sin^2x+\cos^2x=1,\\\\\\(ii)\tan x=\dfrac{\sin x}{\cos x}.[/tex]
We have
[tex]\cos\theta\\\\\\=\pm\sqrt{1-\sin^2\theta}\\\\\\=\pm\sqrt{1-\dfrac{25}{36}}\\\\\\=\pm\sqrt{\dfrac{11}{36}}\\\\\\=\pm\dfrac{\sqrt{11}}{6}[/tex]
and
[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\frac{5}{6}}{\pm\frac{\sqrt{11}}{6}}=\pm\dfrac{5}{\sqrt{11}}=\pm\dfrac{5\sqrt{11}}{11}.[/tex]
Thus, the required values are
[tex]\cos\theta=\pm\dfrac{\sqrt{11}}{6},\\\\\\\tan\theta=\pm\dfrac{5\sqrt{11}}{11}.[/tex]