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SnowyKíng

Given :

[tex] \tt cotA = \sqrt{ \dfrac{1}{3}}[/tex]

[tex] \tt \implies cotA = \dfrac{1}{\sqrt{3}}[/tex]

To Find :

All other trigonometric ratios, which are :

  • sinA
  • cosA
  • tanA
  • cosecA
  • secA

Solution :

Let's make a diagram of right angled triangle ABC.

Now, From point A,

AC = Hypotenuse

BC = Perpendicular

AB = Base

[tex] \tt We \: are \: given, \: cotA = \dfrac{1}{\sqrt{3}}[/tex]

[tex] \tt We \: know \: that \: cot \theta = \dfrac{base}{perpendicular}[/tex]

[tex] \tt \implies \dfrac{base}{perpendicular} = \dfrac{1}{\sqrt{3}}[/tex]

[tex] \tt \implies \dfrac{AB}{BC} = \dfrac{1}{\sqrt{3}}[/tex]

[tex] \tt \implies AB = 1x \: ; \: BC = \sqrt{3}x \: (x \: is \: positive)[/tex]

Now, by Pythagoras' theorem, we have

AC² = AB² + BC²

[tex] \tt \implies AC^{2} = (1x)^{2} + (\sqrt{3}x)^{2}[/tex]

[tex] \tt \implies AC^{2} = 1x^{2} + 3x^{2}[/tex]

[tex] \tt \implies AC^{2} = 4x^{2}[/tex]

[tex] \tt \implies AC = \sqrt{4x^{2}}[/tex]

[tex] \tt \implies AC = 2x[/tex]

Now,

[tex] \tt sin \theta = \dfrac{perpendicular}{hypotenuse}[/tex]

[tex] \tt \implies sinA = \dfrac{BC}{AC} [/tex]

[tex] \tt \implies sinA = \dfrac{\sqrt{3}x}{2x} [/tex]

[tex] \tt \implies sinA = \dfrac{\sqrt{3}}{2} [/tex]

[tex] \Large \boxed{\tt sinA = \dfrac{\sqrt{3}}{2}} [/tex]

[tex] \tt cos \theta = \dfrac{base}{hypotenuse}[/tex]

[tex] \tt \implies cosA = \dfrac{AB}{AC} [/tex]

[tex] \tt \implies cosA = \dfrac{1x}{2x} [/tex]

[tex] \tt \implies cosA = \dfrac{1}{2} [/tex]

[tex] \Large \boxed{\tt cosA = \dfrac{1}{2}} [/tex]

[tex] \tt tan \theta = \dfrac{perpendicular}{base}[/tex]

[tex] \tt \implies tanA = \dfrac{BC}{AB} [/tex]

[tex] \tt \implies tanA = \dfrac{\sqrt{3}x}{1x} [/tex]

[tex] \tt \implies tanA = \sqrt{3}[/tex]

[tex] \Large \boxed{\tt tanA = \sqrt{3}}[/tex]

[tex] \tt cosec \theta = \dfrac{hypotenuse}{perpendicular}[/tex]

[tex] \tt \implies cosecA = \dfrac{AC}{BC} [/tex]

[tex] \tt \implies cosecA = \dfrac{2x}{\sqrt{3}x} [/tex]

[tex] \tt \implies cosecA = \dfrac{2}{\sqrt{3}}[/tex]

[tex] \Large \boxed{\tt cosecA = \dfrac{2}{\sqrt{3}}}[/tex]

[tex] \tt sec \theta = \dfrac{hypotenuse}{base}[/tex]

[tex] \tt \implies secA = \dfrac{AC}{AB} [/tex]

[tex] \tt \implies secA = \dfrac{2x}{1x} [/tex]

[tex] \tt \implies secA = 2[/tex]

[tex] \Large \boxed{\tt secA = 2}[/tex]

Ver imagen SnowyKíng

Diagram :-

[tex]\setlength{\unitlength}{2mm}\begin{picture}(0,0)\thicklines\put(0,0){\line(3,0){2.5cm}}\put(0,0){\line(0,3){2.5cm}}\qbezier(12.4,0)(6.6,5)(0,12.4)\put(-2,13){\sf A}\put(13,-2){\sf C}\put(-2,-2){\sf B}\put(-3,6){\sf 1}\put(6,-3){\sf \sqrt3$}\put(7,7){\sf 2}\end{picture}[/tex]

Solution :-

Given ,

  • cotA = [tex]\sf \sqrt{\dfrac{1}{3}}=\dfrac{1}{\sqrt3}[/tex]

We need to find ,

  • All the trigonometric identities

First finding the other side of the triangle using Pythagoras theorem .

Hypotenuse² = Base² + Height²

[tex]\to\sf Hypotenuse^2 = (1)^2 + (\sqrt3)^2[/tex]

[tex]\to\sf Hypotenuse^2 = 1 + 3 [/tex]

[tex]\to \sf Hypotenuse = \sqrt4[/tex]

[tex]\to\bf Hypotenuse = 2[/tex]

Now ,

  • [tex]\rm sinA = \dfrac{opposite}{hypotenuse}=\sf\dfrac{\sqrt3}{2}[/tex]

  • [tex]\rm cosA = \dfrac{adjacent}{hypotenuse}=\sf\dfrac{1}{2}[/tex]

  • [tex]\rm tanA = \dfrac{opposite}{adjacent}=\sf\dfrac{\sqrt3}{1}[/tex]

  • [tex]\rm cosecA=\dfrac{hypotenuse}{adjacent}=\sf\dfrac{2}{\sqrt3}[/tex]

  • [tex]\rm secA = \dfrac{Hypotenuse}{adjacent}=\sf\dfrac{2}{1}[/tex]

  • [tex]\rm cotA = Already\; given =\sf \dfrac{1}{\sqrt3}[/tex]

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