Prove that :
[tex]\sf{\dfrac{1+cos\theta+sin\theta}{1+cos\theta-sin\theta} = \dfrac{1+sin\theta}{cos\theta}[/tex]

January 4, 2022

By Vivian

1 thought on “Prove that :<br /> [tex]\sf{\dfrac{1+cos\theta+sin\theta}{1+cos\theta-sin\theta} = \dfrac{1+sin\theta}{cos\theta}[/tex]<br /><br /”

1. Answer:

   Step-by-step explanation:

   To Prove :-

   [tex]\sf \dfrac{1+cos\theta+sin\theta}{1+cos\theta-sin\theta}=\dfrac{1+sin\theta}{cos\theta}[/tex]

   Solution :-

   Taking L.H.S :-

   [tex]=\sf \dfrac{1+cos\theta+sin\theta}{1+cos\theta-sin\theta}[/tex]

   [tex]\sf=\dfrac{1+cos\theta+sin\theta}{1+cos\theta-sin\theta}\times \dfrac{1+cos\theta+sin\theta}{1+cos\theta+sin\theta}[/tex]

   [tex]\sf=\dfrac{((1+cos\theta)+sin\theta)^2}{(1+cos\theta)^2-(sin\theta)^2}[/tex]

   [tex]\sf=\dfrac{(1+cos\theta)^2+sin^2\theta+2(1+cos\theta)(sin\theta)}{1+2cos\theta+cos^2\theta-sin^2\theta}[/tex]

   [tex]\sf=\dfrac{1+2cos\theta+cos^2\theta+sin^2\theta+2sin\theta(1+cos\theta)}{2cos\theta+cos^2\theta+cos^2\theta}[/tex]

   {since, 1 – sin^2A = cos^2A]

   [tex]\sf=\dfrac{1+2cos\theta+1+2sin\theta(1+cos\theta)}{2cos^2\theta+2cos\theta}[/tex]

   [tex]\sf=\dfrac{2+2cos\theta+2sin\theta(1+cos\theta)}{2(cos^2\theta+cos\theta)}[/tex]

   [tex]\sf=\dfrac{2(1+cos\theta)+2sin\theta(1+cos\theta)}{2(cos^2\theta+cos\theta)}[/tex]

   [tex]\sf=\dfrac{2(1+cos\theta)(1+sin\theta)}{2cos\theta(cos\theta+1)}[/tex]

   [tex]\sf=\dfrac{1+sin\theta}{cos\theta}[/tex]

   Reply