Question

Given cot tita =7/8 then evaluate (I) (1+sin tita) (1 -sin tita)/ (1 + cos tita)(1-cos tita)​

Answer 1

Answer:

$\rm \dfrac{49}{64}$

Step-by-step explanation:

$\sf \large\red{Given:-}$

$\rm \cot θ = \dfrac{7}{8}$

$\sf \large \red{To \: find:-}$

$\rm \dfrac{(1+ \sin θ)(1- \sin θ)}{(1+ \cos θ)(1-\cos θ)}$

$\sf \large \red{Solution:-}$

$\rm \cot θ = \dfrac{7}{8}$

We know that,

$\boxed{\rm \cot ^2 θ = \cosec ^2 θ -1 }$

$\cot ^2 θ = \dfrac{49}{64}$

$\cot ^2 θ +1 = \dfrac{49}{64}+1$

$\cosec ^2 θ = \dfrac{49+64}{64}$

$\cosec ^2 θ = \dfrac{113}{64}$

$\cosec θ = \sqrt{\dfrac{113}{64}}$

$\cosec θ = \dfrac{\sqrt{113}}{8}$

$\sin θ = \dfrac{8}{\sqrt{113}}$

$\sin ^2 θ = \bigg( \dfrac{8}{\sqrt{113}} \bigg) ^2$

$\sin ^2 θ = \dfrac{64}{113}$

$1- \sin ^2 θ = 1- \dfrac{64}{113}$

$\cos ^2 θ = \dfrac{113-64}{113}$

$\cos ^2 θ = \dfrac{49}{113}$

$\cos θ = \dfrac{7}{\sqrt{113}}$

$\rm \dfrac{(1+ \sin θ)(1- \sin θ)}{(1+ \cos θ)(1-\cos θ)}$

$\rm \dfrac{ \bigg(1+ \dfrac{8}{\sqrt{113}} \bigg) \bigg(1- \dfrac{8}{\sqrt{113}} \bigg)}{ \bigg(1+ \dfrac{7}{\sqrt{113}} \bigg) \bigg(1-\dfrac{7}{\sqrt{113}} \bigg)}$

$\rm \dfrac{ \bigg( \dfrac{\sqrt{113}+8}{\sqrt{113}} \bigg) \bigg( \dfrac{\sqrt{113}-8}{\sqrt{113}} \bigg)}{ \bigg( \dfrac{\sqrt{113}+7}{\sqrt{113}} \bigg) \bigg(\dfrac{\sqrt{113}-7}{\sqrt{113}} \bigg)}$

$\rm \dfrac{ \dfrac{113-64}{113}}{\dfrac{113-49}{113}}$

$\rm \dfrac{49}{64}$