If Tan ⁡ 22 ∘ = T \tan 22^{\circ}=t Tan 2 2 ∘ = T , Write The Following In Terms Of T T T .(a) Tan ⁡ 202 ∘ \tan 202^{\circ} Tan 20 2 ∘ (b) Tan ⁡ 338 ∘ \tan 338^{\circ} Tan 33 8 ∘ (c) \tan \left(-22^{\circ}\right ] (d) Tan ⁡ 518 ∘ \tan 518^{\circ} Tan 51 8 ∘ (e) $\cos

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Expressing Trigonometric Functions in Terms of tt

In this article, we will explore the relationship between the tangent function and the given angle 2222^{\circ}. We will express various trigonometric functions in terms of tt, where t=tan22t = \tan 22^{\circ}. This will involve using trigonometric identities and properties to simplify the expressions.

Expressing tan202\tan 202^{\circ} in Terms of tt

To express tan202\tan 202^{\circ} in terms of tt, we can use the fact that the tangent function has a period of 180180^{\circ}. This means that tan202=tan(202180)=tan22=t\tan 202^{\circ} = \tan (202^{\circ} - 180^{\circ}) = \tan 22^{\circ} = t.

Expressing tan338\tan 338^{\circ} in Terms of tt

Similarly, to express tan338\tan 338^{\circ} in terms of tt, we can use the fact that the tangent function has a period of 180180^{\circ}. This means that tan338=tan(338180)=tan158\tan 338^{\circ} = \tan (338^{\circ} - 180^{\circ}) = \tan 158^{\circ}. We can further simplify this expression by using the fact that tan158=tan(18022)=tan22=t\tan 158^{\circ} = \tan (180^{\circ} - 22^{\circ}) = -\tan 22^{\circ} = -t.

Expressing tan(22)\tan \left(-22^{\circ}\right) in Terms of tt

To express tan(22)\tan \left(-22^{\circ}\right) in terms of tt, we can use the fact that the tangent function is an odd function. This means that tan(22)=tan22=t\tan \left(-22^{\circ}\right) = -\tan 22^{\circ} = -t.

Expressing tan518\tan 518^{\circ} in Terms of tt

To express tan518\tan 518^{\circ} in terms of tt, we can use the fact that the tangent function has a period of 180180^{\circ}. This means that tan518=tan(518180)=tan338=t\tan 518^{\circ} = \tan (518^{\circ} - 180^{\circ}) = \tan 338^{\circ} = -t.

Expressing cos202\cos 202^{\circ} in Terms of tt

To express cos202\cos 202^{\circ} in terms of tt, we can use the fact that cos202=cos(180+22)=cos22\cos 202^{\circ} = \cos (180^{\circ} + 22^{\circ}) = -\cos 22^{\circ}. We can further simplify this expression by using the fact that cos22=1sec22=11+tan222=11+t2\cos 22^{\circ} = \frac{1}{\sec 22^{\circ}} = \frac{1}{\sqrt{1 + \tan^2 22^{\circ}}} = \frac{1}{\sqrt{1 + t^2}}. Therefore, cos202=11+t2\cos 202^{\circ} = -\frac{1}{\sqrt{1 + t^2}}.

Expressing cos338\cos 338^{\circ} in Terms of tt

To express cos338\cos 338^{\circ} in terms of tt, we can use the fact that cos338=cos(180+158)=cos158\cos 338^{\circ} = \cos (180^{\circ} + 158^{\circ}) = -\cos 158^{\circ}. We can further simplify this expression by using the fact that cos158=cos(18022)=cos22=11+t2\cos 158^{\circ} = \cos (180^{\circ} - 22^{\circ}) = -\cos 22^{\circ} = \frac{1}{\sqrt{1 + t^2}}. Therefore, cos338=11+t2\cos 338^{\circ} = -\frac{1}{\sqrt{1 + t^2}}.

Expressing cos(22)\cos \left(-22^{\circ}\right) in Terms of tt

To express cos(22)\cos \left(-22^{\circ}\right) in terms of tt, we can use the fact that the cosine function is an even function. This means that cos(22)=cos22=11+t2\cos \left(-22^{\circ}\right) = \cos 22^{\circ} = \frac{1}{\sqrt{1 + t^2}}.

Expressing cos518\cos 518^{\circ} in Terms of tt

To express cos518\cos 518^{\circ} in terms of tt, we can use the fact that the cosine function has a period of 180180^{\circ}. This means that cos518=cos(518180)=cos338=11+t2\cos 518^{\circ} = \cos (518^{\circ} - 180^{\circ}) = \cos 338^{\circ} = -\frac{1}{\sqrt{1 + t^2}}.

In this article, we have expressed various trigonometric functions in terms of tt, where t=tan22t = \tan 22^{\circ}. We have used trigonometric identities and properties to simplify the expressions. The results are as follows:

  • tan202=t\tan 202^{\circ} = t
  • tan338=t\tan 338^{\circ} = -t
  • tan(22)=t\tan \left(-22^{\circ}\right) = -t
  • tan518=t\tan 518^{\circ} = -t
  • cos202=11+t2\cos 202^{\circ} = -\frac{1}{\sqrt{1 + t^2}}
  • cos338=11+t2\cos 338^{\circ} = -\frac{1}{\sqrt{1 + t^2}}
  • cos(22)=11+t2\cos \left(-22^{\circ}\right) = \frac{1}{\sqrt{1 + t^2}}
  • cos518=11+t2\cos 518^{\circ} = -\frac{1}{\sqrt{1 + t^2}}

In our previous article, we explored the relationship between the tangent function and the given angle 2222^{\circ}. We expressed various trigonometric functions in terms of tt, where t=tan22t = \tan 22^{\circ}. In this article, we will answer some frequently asked questions related to this topic.

Q: What is the period of the tangent function?

A: The period of the tangent function is 180180^{\circ}. This means that the tangent function repeats itself every 180180^{\circ}.

Q: How do you express tan202\tan 202^{\circ} in terms of tt?

A: To express tan202\tan 202^{\circ} in terms of tt, we can use the fact that the tangent function has a period of 180180^{\circ}. This means that tan202=tan(202180)=tan22=t\tan 202^{\circ} = \tan (202^{\circ} - 180^{\circ}) = \tan 22^{\circ} = t.

Q: How do you express tan338\tan 338^{\circ} in terms of tt?

A: To express tan338\tan 338^{\circ} in terms of tt, we can use the fact that the tangent function has a period of 180180^{\circ}. This means that tan338=tan(338180)=tan158\tan 338^{\circ} = \tan (338^{\circ} - 180^{\circ}) = \tan 158^{\circ}. We can further simplify this expression by using the fact that tan158=tan(18022)=tan22=t\tan 158^{\circ} = \tan (180^{\circ} - 22^{\circ}) = -\tan 22^{\circ} = -t.

Q: How do you express tan(22)\tan \left(-22^{\circ}\right) in terms of tt?

A: To express tan(22)\tan \left(-22^{\circ}\right) in terms of tt, we can use the fact that the tangent function is an odd function. This means that tan(22)=tan22=t\tan \left(-22^{\circ}\right) = -\tan 22^{\circ} = -t.

Q: How do you express tan518\tan 518^{\circ} in terms of tt?

A: To express tan518\tan 518^{\circ} in terms of tt, we can use the fact that the tangent function has a period of 180180^{\circ}. This means that tan518=tan(518180)=tan338=t\tan 518^{\circ} = \tan (518^{\circ} - 180^{\circ}) = \tan 338^{\circ} = -t.

Q: How do you express cos202\cos 202^{\circ} in terms of tt?

A: To express cos202\cos 202^{\circ} in terms of tt, we can use the fact that cos202=cos(180+22)=cos22\cos 202^{\circ} = \cos (180^{\circ} + 22^{\circ}) = -\cos 22^{\circ}. We can further simplify this expression by using the fact that cos22=1sec22=11+tan222=11+t2\cos 22^{\circ} = \frac{1}{\sec 22^{\circ}} = \frac{1}{\sqrt{1 + \tan^2 22^{\circ}}} = \frac{1}{\sqrt{1 + t^2}}. Therefore, cos202=11+t2\cos 202^{\circ} = -\frac{1}{\sqrt{1 + t^2}}.

Q: How do you express cos338\cos 338^{\circ} in terms of tt?

A: To express cos338\cos 338^{\circ} in terms of tt, we can use the fact that cos338=cos(180+158)=cos158\cos 338^{\circ} = \cos (180^{\circ} + 158^{\circ}) = -\cos 158^{\circ}. We can further simplify this expression by using the fact that cos158=cos(18022)=cos22=11+t2\cos 158^{\circ} = \cos (180^{\circ} - 22^{\circ}) = -\cos 22^{\circ} = \frac{1}{\sqrt{1 + t^2}}. Therefore, cos338=11+t2\cos 338^{\circ} = -\frac{1}{\sqrt{1 + t^2}}.

Q: How do you express cos(22)\cos \left(-22^{\circ}\right) in terms of tt?

A: To express cos(22)\cos \left(-22^{\circ}\right) in terms of tt, we can use the fact that the cosine function is an even function. This means that cos(22)=cos22=11+t2\cos \left(-22^{\circ}\right) = \cos 22^{\circ} = \frac{1}{\sqrt{1 + t^2}}.

Q: How do you express cos518\cos 518^{\circ} in terms of tt?

A: To express cos518\cos 518^{\circ} in terms of tt, we can use the fact that the cosine function has a period of 180180^{\circ}. This means that cos518=cos(518180)=cos338=11+t2\cos 518^{\circ} = \cos (518^{\circ} - 180^{\circ}) = \cos 338^{\circ} = -\frac{1}{\sqrt{1 + t^2}}.

In this article, we have answered some frequently asked questions related to expressing trigonometric functions in terms of tt. We have used trigonometric identities and properties to simplify the expressions. The results are as follows:

  • tan202=t\tan 202^{\circ} = t
  • tan338=t\tan 338^{\circ} = -t
  • tan(22)=t\tan \left(-22^{\circ}\right) = -t
  • tan518=t\tan 518^{\circ} = -t
  • cos202=11+t2\cos 202^{\circ} = -\frac{1}{\sqrt{1 + t^2}}
  • cos338=11+t2\cos 338^{\circ} = -\frac{1}{\sqrt{1 + t^2}}
  • cos(22)=11+t2\cos \left(-22^{\circ}\right) = \frac{1}{\sqrt{1 + t^2}}
  • cos518=11+t2\cos 518^{\circ} = -\frac{1}{\sqrt{1 + t^2}}

These results demonstrate the relationship between the tangent and cosine functions and the given angle 2222^{\circ}. They also illustrate the use of trigonometric identities and properties to simplify expressions.