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!

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]

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