Question

if tan teta =3/4 then 1-cos teta / 1+Cos teta​

Answer 1

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf 3}\put(2.8,.3){\large\bf 4}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf \Theta }\end{picture}$

Given:

$tan \theta = \dfrac{3}{4} \implies \dfrac{opposite side}{adjacent side} = \dfrac{3}{4}$

To find:

$\dfrac{1 - cos \theta}{1 + cos \theta}$

Method:

By Pythagoras Theorem;

AC² = AB² + BC²

AC² = 3² + 4²

AC² = 9 + 16 = 25

AC = √25

AC = ±5

Side cannot be negative.

So, AC = 5 cm

$\bf{cos \theta = \dfrac{adjacent \: side}{hypotenuse} = \dfrac{4}{5}}$

$1 - cos \theta = 1 - \dfrac{4}{5} = \dfrac{5 - 4}{5} = \dfrac{1}{5} \\\\ \implies 1 + cos \theta = 1 + \dfrac{4}{5} = \dfrac{5 + 4}{5} = \dfrac{9}{5}\\\\\\\ \dfrac{1 - cos \theta}{1 + cos \theta} = \dfrac{\dfrac{1}{5}}{\dfrac{9}{5} } = \dfrac{1}{9}$

Answer:

$\large \boxed{{\bf{\dfrac{1 - cos \theta}{1 + cos \theta} = \dfrac{1}{9}}}}$