Question

do it fasttt kal exam hai​

Answer 1

Given:-

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

__________________________

Prove That:-

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

__________________________

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!

____________________________

Extra Info:-

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

Note:-

• kindly see the answer from brainly.in website

Answer 2

$\huge\boxed{\fcolorbox{purple}{ink}{Answer}}$

Given that -

$\rm \dfrac{cos \theta}{(1-tan \theta)}+ \dfrac{sin \theta}{(1- cot \theta)} = (cos \theta + sin \theta) </p><p>(1−tanθ)$

Consider LHS :

$⇒ \rm \dfrac{cos \theta}{(1-tan \theta)}+ \dfrac{sin \theta}{(1- cot \theta)}$

We know that -

• tanθ = sinθ/cosθ
• cotθ = cosθ/sinθ

$⇒ \begin{gathered}\rm \dfrac{cos \theta}{(1- \dfrac{\sin \theta}{\cos \theta})}+ \dfrac{sin \theta}{(1- \dfrac{\cos \theta}{\sin \theta})} \\\end{gathered}$

$⇒ \rm \dfrac{cos \theta}{(\dfrac{\cos \theta - \sin \theta}{\cos \theta})}+ \dfrac{sin \theta}{( \dfrac{\sin \theta - \cos \theta}{\sin \theta})}$

$⇒ \rm \dfrac{\cos^2 \theta}{(\cos \theta - \sin \theta)} + \dfrac{\sin^2 \theta}{(\sin \theta - \cos \theta)}$

$⇒ \rm \dfrac{\cos^2 \theta}{(\cos \theta - \sin \theta)} - \dfrac{\sin^2 \theta}{(\cos \theta - \sin \theta)}$

$⇒ \rm \dfrac{\cos^2 \theta - \sin^2 \theta }{(\cos \theta - \sin \theta)}$

$⇒ \rm \dfrac{(\cos \theta - \sin \theta)(\cos \theta + \sin \theta)}{(\cos \theta - \sin \theta)}$

$⇒ \cos \theta + \sin \theta = \text{RHS}$

∴ LHS = RHS

Hence, it is proved.