Question

if tan theta = 4/5 then 1-cos theta/1+cos theta​

Answer 1

Answer:

$\frac{1-cos\theta}{1+cos\theta}=\frac{66-10\sqrt{41}}{16}$

Step-by-step explanation:

We are given that

$tan\theta=\frac{4}{5}$

We have to find the value of $\frac{1-cos\theta}{1+cos\theta}$

We know that $tan\theta=\frac{perpendicular\;arm}{Base}$

Compare with the given values of tangent then we get

Perpendicular arm of triangle =4 units

Base of triangle =5 units

Using Pythogorous theorem

$(Hypotenuse)^2=(Base)^2+(Perpendicular arm)^2$

Therefore, $(Hypotenuse)^2=(4)^2+(5)^2$

$(Hypotenuse)^2=16+25=41$

$Hypotenuse=\sqrt{41}$

$cos\theta=\frac{Base}{hypotenuse}$

$cos\theta=\frac{5}{\sqrt{41}}$

Substitute the value then we get

$\frac{1-cos\theta}{1+cos\theta}=\frac{1-\frac{5}{\sqrt{41}}}{1+\frac{5}{\sqrt{41}}}$

$\frac{1-cos\theta}{1+cos\theta}=\frac{\sqrt{41}-5}{\sqrt{41}+5}$

Using rationalization property

$\frac{a-b}{a+b}=\frac{(a-b)(a-b)}{(a+b)(a-b)}$

$\frac{1-cos\theta}{1+cos\theta}=\frac{(\sqrt{41}-5)(\sqrt{41}-5)}{(\sqrt{41}-5)(\sqrt{41}+5)}$

$\frac{1-cos\theta}{1+cos\theta}=\frac{41+25-10\sqrt{41}}{41-25}$

Using property :$(a-b)^2=a^2+b^2-2 ab$

$a^2-b^2=(a+b)(a-b)$

$\frac{1-cos\theta}{1+cos\theta}=\frac{66-10\sqrt{41}}{16}$