Question

If $cot\Theta=\frac{1}{\sqrt{3} }$, show that $\frac{1-cosx^{2}\Theta }{2-sinx^{2}\Theta } =\frac{3}{5}$.

Answer 1

SOLUTION :  

Given: cot θ = 1/√3

cot θ = 1/√3 = B/P = AB /BC

Draw a right ∆ABC, ∠B = 90°

Base(AB) = 1 unit & Perpendicular(BC) = √3 unit

In ∆ABC,  

AC² = AB² + BC²

[By using Pythagoras theorem]

AC² = 1² + (√3)²

AC² = 1 + 3

AC² = 4

AC =√4

AC = 2  

Now, sin θ = P/H = BC /AC = √3/2

Cos θ = B/H = AB/AC = ½

L.H.S  = 1- cos²θ / 2 - sin²θ  

= 1 -(½)² / 2 - (√3/2)²

= 1 - ¼ / 2 -¾

=( 4 -1)/4 / (8 - 3)/4

= ¾ / 5/4

= ⅗  

= R.H.S

Hence ,  1- cos²θ / 2 - sin²θ = ⅗

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

$in \:right \: triangle \: abc \: \\ by \: pythagorous \: theoram \\ \\ {(ac)}^{2} = {(ab)}^{2} + {(bc)}^{2} \\ hypotenuse = \sqrt{ { (\sqrt{3}) }^{2} + {(1)}^{2} } \\ hypotenuse = 2$
Now,

$\sin(θ) = \frac{ \sqrt{3} }{2} \\ \cos(θ) = \frac{1}{2} \\ now \: according \: to \: question \\ \\ = \frac{1 - {cos}^{2} θ}{2 - {sin}^{2} θ} \\ = \frac{1 - ({ \frac{1}{2} )}^{2} }{2 - {( \frac{ \sqrt{3} }{2} )}^{2} } \\ = \frac{1 - \frac{1}{4} }{2 - \frac{3}{4} } \\ = \frac{ \frac{3}{4} }{ \frac{5}{4} } \\ = \frac{3}{5} = rhs$
L.H.S=R.H.S

Hence proved✔️