Question

en \: cotA = \sqrt{\dfrac{1}{3}}" alt="Given \: cotA = \sqrt{\dfrac{1}{3}}" align="absmiddle" class="latex-formula">

Find all other trigonometric ratios. ​

Answer 1

Given :

$\tt cotA = \sqrt{ \dfrac{1}{3}}$

$\tt \implies cotA = \dfrac{1}{\sqrt{3}}$

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

$\tt We \: are \: given, \: cotA = \dfrac{1}{\sqrt{3}}$

$\tt We \: know \: that \: cot \theta = \dfrac{base}{perpendicular}$

$\tt \implies \dfrac{base}{perpendicular} = \dfrac{1}{\sqrt{3}}$

$\tt \implies \dfrac{AB}{BC} = \dfrac{1}{\sqrt{3}}$

$\tt \implies AB = 1x \: ; \: BC = \sqrt{3}x \: (x \: is \: positive)$

Now, by Pythagoras' theorem, we have

AC² = AB² + BC²

$\tt \implies AC^{2} = (1x)^{2} + (\sqrt{3}x)^{2}$

$\tt \implies AC^{2} = 1x^{2} + 3x^{2}$

$\tt \implies AC^{2} = 4x^{2}$

$\tt \implies AC = \sqrt{4x^{2}}$

$\tt \implies AC = 2x$

Now,

$\tt sin \theta = \dfrac{perpendicular}{hypotenuse}$

$\tt \implies sinA = \dfrac{BC}{AC}$

$\tt \implies sinA = \dfrac{\sqrt{3}x}{2x}$

$\tt \implies sinA = \dfrac{\sqrt{3}}{2}$

$\Large \boxed{\tt sinA = \dfrac{\sqrt{3}}{2}}$

$\tt cos \theta = \dfrac{base}{hypotenuse}$

$\tt \implies cosA = \dfrac{AB}{AC}$

$\tt \implies cosA = \dfrac{1x}{2x}$

$\tt \implies cosA = \dfrac{1}{2}$

$\Large \boxed{\tt cosA = \dfrac{1}{2}}$

$\tt tan \theta = \dfrac{perpendicular}{base}$

$\tt \implies tanA = \dfrac{BC}{AB}$

$\tt \implies tanA = \dfrac{\sqrt{3}x}{1x}$

$\tt \implies tanA = \sqrt{3}$

$\Large \boxed{\tt tanA = \sqrt{3}}$

$\tt cosec \theta = \dfrac{hypotenuse}{perpendicular}$

$\tt \implies cosecA = \dfrac{AC}{BC}$

$\tt \implies cosecA = \dfrac{2x}{\sqrt{3}x}$

$\tt \implies cosecA = \dfrac{2}{\sqrt{3}}$

$\Large \boxed{\tt cosecA = \dfrac{2}{\sqrt{3}}}$

$\tt sec \theta = \dfrac{hypotenuse}{base}$

$\tt \implies secA = \dfrac{AC}{AB}$

$\tt \implies secA = \dfrac{2x}{1x}$

$\tt \implies secA = 2$

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

Answer 2

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 = $\sf \sqrt{\dfrac{1}{3}}=\dfrac{1}{\sqrt3}$

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 ,

• $\rm sinA = \dfrac{opposite}{hypotenuse}=\sf\dfrac{\sqrt3}{2}$

• $\rm cosA = \dfrac{adjacent}{hypotenuse}=\sf\dfrac{1}{2}$

• $\rm tanA = \dfrac{opposite}{adjacent}=\sf\dfrac{\sqrt3}{1}$

• $\rm cosecA=\dfrac{hypotenuse}{adjacent}=\sf\dfrac{2}{\sqrt3}$

• $\rm secA = \dfrac{Hypotenuse}{adjacent}=\sf\dfrac{2}{1}$

• $\rm cotA = Already\; given =\sf \dfrac{1}{\sqrt3}$