Question

prove that :
tan (theta) / 1-cot (theta) +cot (theta) / 1-tan (theta) = 1+ sec (theta) cosec (theta)

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Answer 1

$\large \bf \: Prove \: that \: :$ ↴

$\sf \dfrac{ \tan \theta }{1 - \cot\theta } \: - \: \dfrac{ \cot \theta }{1 - \cot\theta } \: = \: 1 \: + \: \sec\theta \: cosec\theta$

$\: \large\: \bf \: So \: Let's \: begin \: :$ ↴

$\sf \: Solving \: LHS$ ↴

$\large\: \;\tt{ \;\;{L.H.S.\;=\;\bigg[\dfrac{\tan \theta }{(1\:-\:\cot \theta)}\bigg]\;+\;\bigg[\dfrac{\cot \theta}{(1\:-\:\tan \theta)}\bigg]}}$

$\large\: \;\tt{\longrightarrow\dfrac{\dfrac{\sin \theta}{\cos \theta}}{ 1 \:-\:\dfrac{\cos \theta}{\sin \theta}}\;+\;\dfrac{\dfrac{\cos \theta}{\sin \theta}}{ 1 \:-\:\dfrac{\sin \theta}{\cos \theta}}}$

$\large\: \;\tt {\longrightarrow\dfrac{\dfrac{\sin \theta}{\cos \theta}}{ \dfrac{1}{1} \:-\:\dfrac{\cos \theta}{\sin \theta}}\;+\;\dfrac{\dfrac{\cos \theta}{\sin \theta}}{ \dfrac{1}{1} \:-\:\dfrac{\sin \theta}{\cos \theta}}}$

$\large\: \;\tt{\longrightarrow\dfrac{\dfrac{\sin \theta}{\cos \theta}}{ \:\dfrac{ \sin \theta \: - \: \cos \theta}{\sin \theta}}\;+\;\dfrac{\dfrac{\cos \theta}{\sin \theta}}{ \:\dfrac{ \cos \theta \: - \: \sin \theta}{\cos \theta}}}$

$\large\: \sf \dfrac{\sin^{2} \theta}{cos\theta( \sin\theta - \cos\theta) } \: \: + \: \dfrac{\ \cos^{2} \theta}{ \sin \theta( \ \cos \theta - \ \sin \theta) }$

$\large\: \sf \: \: Taking \: - \: as \: common.$

$\large\: \sf \dfrac{\sin^{2} \theta}{cos\theta( \sin\theta - \cos\theta) } \: \: + \: \dfrac{\ \cos^{2} \theta}{ - \sin \theta( \ \ \sin\theta - \ \ \cos \theta) }$

$\large\: \sf \dfrac{\sin^{2} \theta}{cos\theta( \sin\theta - \cos\theta) } \: \: - \: \dfrac{\ \cos^{2} \theta}{ \sin \theta( \ \ \sin\theta - \ \ \cos \theta) }$

$\large\: \dfrac{1}{ \sin \theta \: - \cos\theta } \: \bigg[ \: \dfrac{ \sin ^{2} \theta}{ \cos \theta} \: - \: \dfrac{\cos^{2} \theta}{\sin \theta} \bigg]$

$\large\: \sf \: Taking \: LCM, \: we \: get$

$\large\: \dfrac{1}{ \sin \theta \: - \cos\theta } \: \bigg[ \: \dfrac{ \sin ^{3} \theta - \: \cos ^{3} \theta}{sin \theta \: cos \theta} \bigg]$

$\large\: \longrightarrow \sf \: \: Formula \: used :$ ↴

$\large\: \boxed { \sf \: a ^{3} \: - \:b ^{3} \: = \: (a \: - \: b) \: (a^{2} \: + b ^{2} \: + \: ab )}$

$\large\: \dfrac{1}{ \sin \theta \: - \cos\theta } \: \dfrac{(\sin\theta - \cos\theta)(sin ^{2} + \cos^{2} + \sin\theta \: cos\theta) }{sin\theta\cos\theta}$

$\large\: \dfrac{1}{ \cancel{\sin \theta \: - \cos\theta }} \: \dfrac{( \cancel{\sin\theta - \cos\theta})(sin ^{2} + \cos^{2} + \sin\theta \: cos\theta) }{sin\theta\cos\theta}$

$\large\: \dfrac{(sin ^{2} + \cos^{2} + \sin\theta \: cos\theta) }{sin\theta\cos\theta}$

$\large\: \sf{ We \: know \: that (sin ^{2} + \cos^{2} + \sin\theta \: cos\theta) =1}$

$\sf \: \large\: \dfrac{1 \: + \sin \theta \: \cos\theta }{\sin \theta \: \cos\theta}$

$\sf \: \large\: \dfrac{1}{ \sin \theta \: \cos \theta } \: + \: \dfrac{ \sin\theta \cos\theta }{ \sin \theta \: \cos \theta }$

$\sf \: \large \:\dfrac{1}{ \sin \theta \: \cos \theta } \: + \: \dfrac{ \cancel{ \sin\theta \cos\theta }}{\cancel{ \sin \theta \: \cos \theta } }$

$\sf \: \large\: \dfrac{1}{ \sin \theta } \: \times \dfrac{1}{ \cos \theta } \: + \: 1$

$\sf \:\large \: LHS = \sec \theta \: \: \cos \theta+1$

$\sf \: \large \:RHS = \: 1 \: + \: \sec \theta \: \: \cos \theta$

$\sf LHS \: = \:RHS$

$\sf\:Hence\: proved$