Question

`Given \: cotA = \sqrt{(1)/(3)}`

Find all other trigonometric ratios. ​

Answer 1

Diagram :-

`\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}`

Solution :-

Given ,

• cotA =
  `\sf \sqrt{(1)/(3)}=(1)/(\sqrt3)`

We need to find ,

• All the trigonometric identities

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

Hypotenuse² = Base² + Height²

`\to\sf Hypotenuse^2 = (1)^2 + (\sqrt3)^2`

`\to\sf Hypotenuse^2 = 1 + 3`

`\to \sf Hypotenuse = \sqrt4`

`\to\bf Hypotenuse = 2`

Now ,

• 
  `\rm sinA = (opposite)/(hypotenuse)=\sf(\sqrt3)/(2)`

• 
  `\rm cosA = (adjacent)/(hypotenuse)=\sf(1)/(2)`

• 
  `\rm tanA = (opposite)/(adjacent)=\sf(\sqrt3)/(1)`

• 
  `\rm cosecA=(hypotenuse)/(adjacent)=\sf(2)/(\sqrt3)`

• 
  `\rm secA = (Hypotenuse)/(adjacent)=\sf(2)/(1)`

• 
  `\rm cotA = Already\; given =\sf (1)/(\sqrt3)`

Answer 2

Given :

`\tt cotA = \sqrt{ (1)/(3)}`

`\tt \implies cotA = (1)/(√(3))`

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

`\tt We \: are \: given, \: cotA = (1)/(√(3))`

`\tt We \: know \: that \: cot \theta = (base)/(perpendicular)`

`\tt \implies (base)/(perpendicular) = (1)/(√(3))`

`\tt \implies (AB)/(BC) = (1)/(√(3))`

`\tt \implies AB = 1x \: ; \: BC = √(3)x \: (x \: is \: positive)`

Now, by Pythagoras' theorem, we have

AC² = AB² + BC²

`\tt \implies AC^(2) = (1x)^(2) + (√(3)x)^(2)`

`\tt \implies AC^(2) = 1x^(2) + 3x^(2)`

`\tt \implies AC^(2) = 4x^(2)`

`\tt \implies AC = \sqrt{4x^(2)}`

`\tt \implies AC = 2x`

Now,

`\tt sin \theta = (perpendicular)/(hypotenuse)`

`\tt \implies sinA = (BC)/(AC)`

`\tt \implies sinA = (√(3)x)/(2x)`

`\tt \implies sinA = (√(3))/(2)`

`\Large \boxed{\tt sinA = (√(3))/(2)}`

`\tt cos \theta = (base)/(hypotenuse)`

`\tt \implies cosA = (AB)/(AC)`

`\tt \implies cosA = (1x)/(2x)`

`\tt \implies cosA = (1)/(2)`

`\Large \boxed{\tt cosA = (1)/(2)}`

`\tt tan \theta = (perpendicular)/(base)`

`\tt \implies tanA = (BC)/(AB)`

`\tt \implies tanA = (√(3)x)/(1x)`

`\tt \implies tanA = √(3)`

`\Large \boxed{\tt tanA = √(3)}`

`\tt cosec \theta = (hypotenuse)/(perpendicular)`

`\tt \implies cosecA = (AC)/(BC)`

`\tt \implies cosecA = (2x)/(√(3)x)`

`\tt \implies cosecA = (2)/(√(3))`

`\Large \boxed{\tt cosecA = (2)/(√(3))}`

`\tt sec \theta = (hypotenuse)/(base)`

`\tt \implies secA = (AC)/(AB)`

`\tt \implies secA = (2x)/(1x)`

`\tt \implies secA = 2`

`\Large \boxed{\tt secA = 2}`