Question

solve this as fast as possible​

Answer 1

TO PROVE :–

$\\ \sf \implies \dfrac{ \cos( \theta) }{1 - \sin( \theta) } = \tan \left( \frac{\pi}{4} + \frac{ \theta}{2} \right)$

SOLUTION :–

• Let's take L.H.S. –

$\\ \sf \: \: = \: \: \dfrac{ \cos( \theta) }{1 - \sin( \theta) }$

• We know that –

$\\ \sf \longrightarrow \: \cos( \theta) = \sin \left( \dfrac{ \pi}{2} - \theta \right)$

$\\ \sf \longrightarrow \: \sin( \theta) = \cos \left( \dfrac{ \pi}{2} - \theta \right)$

• So that –

$\\ \sf \: \: = \: \: \dfrac{ \sin \left( \dfrac{ \pi}{2} - \theta \right)}{1 - \cos\left( \dfrac{ \pi}{2} - \theta \right)}$

• Using identity –

$\\ \sf \longrightarrow \: \cos( \theta) =1 - 2 \sin^{2} \left( \dfrac{ \theta}{2} \right)$

$\\ \sf \longrightarrow \: \sin( \theta) = 2 \sin \left(\dfrac{ \theta}{2} \right) \cos \left( \dfrac{ \theta}{2} \right)$

• So –

$\\ \sf \: \: = \: \: \dfrac{2 \sin \left( \dfrac{ \pi}{4} - \dfrac{\theta }{2} \right) \cos \left( \dfrac{ \pi}{4} - \dfrac{\theta }{2} \right) }{1 - \left[ 1 - 2 \sin^{2} \left( \dfrac{\pi}{4} - \dfrac{ \theta}{2} \right) \right]}$

$\\ \sf \: \: = \: \: \dfrac{2 \sin \left( \dfrac{ \pi}{4} - \dfrac{\theta }{2} \right) \cos \left( \dfrac{ \pi}{4} - \dfrac{\theta }{2} \right) }{1 - 1 + 2 \sin^{2} \left( \dfrac{\pi}{4} - \dfrac{ \theta}{2} \right) }$

$\\ \sf \: \: = \: \: \dfrac{2 \sin \left( \dfrac{ \pi}{4} - \dfrac{\theta }{2} \right) \cos \left( \dfrac{ \pi}{4} - \dfrac{\theta }{2} \right) }{ 2 \sin^{2} \left( \dfrac{\pi}{4} - \dfrac{ \theta}{2} \right) }$

$\\ \sf \: \: = \: \: \dfrac{ \sin \left( \dfrac{ \pi}{4} - \dfrac{\theta }{2} \right) \cos \left( \dfrac{ \pi}{4} - \dfrac{\theta }{2} \right) }{ \sin^{2} \left( \dfrac{\pi}{4} - \dfrac{ \theta}{2} \right) }$

$\\ \sf \: \: = \: \: \dfrac{ \cos \left( \dfrac{ \pi}{4} - \dfrac{\theta }{2} \right) }{ \sin \left( \dfrac{\pi}{4} - \dfrac{ \theta}{2} \right) }$

$\\ \sf \: \: = \: \: \cot \left( \dfrac{ \pi}{4} - \dfrac{\theta }{2} \right) \: \: \: \left[ \: \: \because \: \: \dfrac{ \cos( \theta) }{ \sin( \theta) } = \cot( \theta) \right]$

• We also know that –

$\\ \sf \: \longrightarrow \: \: \cot ( \theta) = \tan\left( \dfrac{ \pi}{2} - \theta \right)$

• So that –

$\\ \sf \: \: = \: \: \tan \left \{ \dfrac{\pi}{2} - \left( \dfrac{ \pi}{4} - \dfrac{\theta }{2} \right) \right \}$

$\\ \sf \: \: = \: \: \tan \left \{ \dfrac{\pi}{2} - \dfrac{ \pi}{4} + \dfrac{\theta }{2} \right \}$

$\\ \sf \: \: = \: \: \tan \left ( \dfrac{ \pi}{4} + \dfrac{\theta }{2} \right)$

$\\ \sf \: \: = \: \: R.H.S .$

$\\ \: \: \: \: \: \: \: \: \: \: \: \:{ \underbrace{ \sf \: Hence \: \: proved}}$

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\dagger \: {\sf{ \frac{sin \theta}{1 - sin \theta} = tan \bigg( \frac{\pi}{4} + \frac{ \theta}{2} \bigg) }}$

${\bf{\blue{\underline{Now:}}}}$

Take L.H.S

${\sf{ L.H.S = \frac{sin \theta}{1 - sin \theta} }}$

${\sf{ = \frac{sin \bigg( \frac{\pi}{2} - \theta \bigg) }{1 - cos \bigg( \frac{\pi}{2} - \theta \bigg)} }}$

$\dagger \: \boxed{\sf{ sin2 \theta = 2sin \theta \: cos \theta}}$

$\dagger \: \boxed{\sf{1 - 2{sin}^{2} \theta = {cos}2 \theta }}$

${\sf{ = \frac{2sin \bigg( \frac{\pi}{4} - \frac{ \theta}{2} \bigg)cos \bigg( \frac{\pi}{4} - \frac{ \theta }{2} \bigg) }{1 - 1 - 2 {sin}^{2} \theta \bigg( \frac{\pi}{4} - \frac{ \theta}{2} \bigg) } }}$

${\sf{ = \frac{2sin \bigg( \frac{\pi}{4} - \frac{ \theta}{2} \bigg)cos \bigg( \frac{\pi}{4} - \frac{ \theta }{2} \bigg) }{ 2 {sin}^{2} \theta \bigg( \frac{\pi}{4} - \frac{ \theta}{2} \bigg) } }}$

${\sf{ = \frac{ \cancel2sin \bigg( \frac{\pi}{4} - \frac{ \theta}{2} \bigg)cos \bigg( \frac{\pi}{4} - \frac{ \theta }{2} \bigg) }{ \cancel2 {sin}^{2} \theta \bigg( \frac{\pi}{4} - \frac{ \theta}{ 2} \bigg) } }}$

${\sf{ = \frac{sin \bigg( \frac{\pi}{4} - \frac{ \theta}{2} \bigg)cos \bigg( \frac{\pi}{4} - \frac{ \theta }{2} \bigg) }{ {sin}^{2} \theta \bigg( \frac{\pi}{4} - \frac{ \theta}{ 2} \bigg) } }}$

${\sf{ = \frac{sin \bigg( \frac{\pi}{4} - \frac{ \theta}{2} \bigg)cos \bigg( \frac{\pi}{4} - \frac{ \theta }{2} \bigg) }{ {sin} \bigg( \frac{\pi}{4} - \frac{ \theta}{ 2} \bigg) {sin} \bigg( \frac{\pi}{4} - \frac{ \theta}{ 2} \bigg) } }}$

${\sf{ = \frac{ \cancel{sin \bigg( \frac{\pi}{4} - \frac{ \theta}{2} \bigg)}cos \bigg( \frac{\pi}{4} - \frac{ \theta }{2} \bigg) }{ \cancel{{sin} \bigg( \frac{\pi}{4} - \frac{ \theta}{ 2} \bigg) } {sin} \bigg( \frac{\pi}{4} - \frac{ \theta}{ 2} \bigg) } }}$

${\sf{ = \frac{cos \bigg( \frac{\pi}{4} - \frac{ \theta }{2} \bigg) }{ {sin} \bigg( \frac{\pi}{4} - \frac{ \theta}{ 2} \bigg) } }}$

${\sf{ = cot \bigg( \frac{\pi}{4} - \frac{ \theta }{2} \bigg) }}$

${\sf{ = tan\bigg( \frac{\pi}{2} - \theta \bigg) }}$

${\sf{ = tan\bigg( \frac{\pi}{2} - \frac{\pi}{4} - \frac{ \theta}{2} \bigg) }}$

${\sf{ = tan\bigg( \frac{\pi}{4} + \frac{ \theta}{2} \bigg) }}$

Hence proved!!

L.H.S = R.HS