Question

prove the following:

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

Answer 1

Given:

$\sf\boxed{\bullet} \: \sqrt{\dfrac{1- \cos\theta}{1+ \cos\theta}} = \dfrac{sin\theta }{1+ \cos\theta}$

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Prove That:

$\sf\boxed{\bullet} \: L.H.S = R.H.S$

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Solution:

L.H.S,

$\sf \Longrightarrow \sqrt{\dfrac{1- \cos\theta}{1+ \cos\theta}}$

Multiply, both numerator and denominator with 1 - cos θ

$\sf \Longrightarrow \sqrt{\dfrac{1- \cos\theta}{1+ \cos\theta} \times \dfrac{1- \cos\theta}{1- \cos\theta} }$

Using, (a + b) (a - b) = a² - b²

$\sf \Longrightarrow \sqrt{\dfrac{{(1- \cos\theta)}^{2} }{1^{2} - \cos^{2} \theta}}$

$\sf \Longrightarrow \sqrt{\dfrac{{(1- \cos\theta)}^{2} }{1 - \cos^{2} \theta}}$

Using, 1 - cos² θ = sin² θ

$\sf \Longrightarrow \sqrt{\dfrac{{(1- \cos\theta)}^{2} }{\sin^{2} \theta}}$

$\sf \Longrightarrow \sqrt{ {\bigg\lgroup\dfrac{1- \cos\theta}{\sin \theta}\bigg\rgroup}^{2}}$

$\sf \Longrightarrow \dfrac{1- \cos\theta}{\sin \theta}$

Multiply, both numerator and denominator with 1 + cos θ

$\sf \Longrightarrow\dfrac{(1- \cos\theta)(1+\cos \theta)}{\sin \theta(1+\cos \theta)}$

Using, (a + b) (a - b) = a² - b²

$\sf \Longrightarrow\dfrac{1^2- \cos^{2} \theta}{\sin \theta(1+\cos \theta)}$

$\sf \Longrightarrow\dfrac{1- \cos^{2} \theta}{\sin \theta(1+\cos \theta)}$

Using, 1 - cos² θ = sin² θ

$\sf \Longrightarrow\dfrac{\sin^{2} \theta}{\sin \theta(1+\cos \theta)}$

$\sf \Longrightarrow\dfrac{\sin \theta}{(1+\cos \theta)}$

$\sf \Longrightarrow\dfrac{\sin \theta}{1+\cos \theta}$

$\sf \Longrightarrow\dfrac{\sin \theta}{1+\cos \theta}= R.H.S$

Hence, Proved

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Extra Shots:

$\boxed{\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}}$