Question

Prove the following
$\sf \dfrac{tan \theta}{1- cot \theta} + \dfrac{cot \theta}{1 -tan \theta} = 1 + \tan \theta + \cot \theta$

Answer 1

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Prove:-}}}}}}}$

$\sf \dfrac{tan ∅}{1- cot ∅} + \dfrac{cot ∅}{1 -tan ∅} = 1 + tan ∅ + cot ∅$

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Proof:-}}}}}}}$

• Taking LHS,

$\sf \dfrac{tan ∅}{1- cot ∅} + \dfrac{cot ∅}{1 -tan ∅}$

• This implies that,

$\sf \dfrac{tan ∅}{1 \: - \: cot ∅} \: - \: \Big(\dfrac{cot ∅}{tan ∅ \: + \: 1}\Big)$

$\sf \dfrac{tan ∅}{1 \: - \: cot ∅} \: - \: \dfrac{cot ∅}{tan ∅ \: - \: 1}$

• We know that,

$\sf cot ∅ \: = \: \dfrac{1}{tan ∅}$

• Substituting the values,

$\sf \dfrac{tan ∅}{1 \: - \: \dfrac{1}{tan ∅}} \: - \: \dfrac{\dfrac{1}{tan ∅}}{tan ∅ \: - \: 1}$

$\sf \dfrac{tan ∅}{\dfrac{tan ∅ \: - \: 1}{tan ∅}} \: - \: \dfrac{1}{tan ∅ \: (tan ∅ \: - \: 1)}$

$\sf \dfrac{tan^2 ∅}{tan ∅ \: - \: 1} \: - \: \dfrac{1}{tan ∅ \: (tan ∅ \: - \: 1)}$

• Taking $\sf \dfrac{1}{tan ∅ \: - \: 1}$ out,

$\sf \dfrac{1}{tan ∅ \: - \: 1} \: \Big(\dfrac{tan^2 ∅}{1} \: - \: \dfrac{1}{tan ∅}\Big)$

• Taking LCM,

$\sf \dfrac{1}{tan ∅ \: - \: 1} \: \Big(\dfrac{tan^3 ∅ \: - \: 1}{tan ∅}\Big)$

• We know that,

• a³ - b³ = (a - b)(a² + ab + b²)

$\sf \dfrac{1}{tan ∅ \: - \: 1} \: \bigg(\dfrac{(tan ∅ \: - \: 1)(tan^2 ∅ \: + \: tan ∅ \: + \: 1^2)}{tan ∅}\bigg)$

$\sf \dfrac{1}{\cancel{tan ∅ \: - \: 1}} \: \bigg(\dfrac{\cancel{(tan ∅ \: - \: 1)}(tan ^2 ∅ \: + \: tan ∅ \: + \: 1^2)}{tan ∅}\bigg)$

$\sf 1 \: \bigg(\dfrac{tan^2 ∅ \: + \: tan ∅ \: + \: 1^2}{tan ∅}\bigg)$

$\sf \dfrac{tan^2 ∅ \: + \: tan ∅ \: + \: 1}{tan ∅}$

$\sf \dfrac{tan^2 ∅}{tan ∅} \: + \: \dfrac{tan ∅}{tan ∅} \: + \: \dfrac{1}{tan ∅}$

$\sf \dfrac{tan^{\cancel{2}} ∅}{\cancel{tan ∅}} \: + \: \dfrac{\cancel{tan ∅}}{\cancel{tan ∅}} \: + \: \dfrac{1}{tan ∅}$

$\sf tan ∅ \: + \: 1 \: + \: cot ∅$

• LHS = RHS

• Hence, proved.

$\Large{\bf{\purple{\mathfrak{\dag{\underline{\underline{Important \: events \: of \: the \: answer:-}}}}}}}$

• $\sf Making \: the \: equation \: \dfrac{tan ∅}{1 \: - \: cot ∅} \: - \: \dfrac{cot ∅}{tan ∅ \: - \: 1}$

• Substituting $\sf \dfrac{1}{tan ∅}$ in place of cot ∅

• Taking out $\sf \dfrac{1}{tan ∅ \: - \: 1}$ common.

• Using a³ - b³ = (a² + ab + b²)(a - b)

• Cancelling tan ∅ - 1

Answer 2

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

Exigency To Prove : $\sf \dfrac{tan \theta}{1- cot \theta} + \dfrac{cot \theta}{1 -tan \theta} = 1 + \tan \theta + \cot \theta$ [ L.H.S = R.H.S ]

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad \dag\:\:\bigg\lgroup \sf{ \sf \dfrac{tan \theta}{1- cot \theta} + \dfrac{cot \theta}{1 -tan \theta} = 1 + \tan \theta + \cot \theta }\bigg\rgroup$

Here ,

• $\bf L.H.S = \:\sf \dfrac{tan \theta}{1- cot \theta} + \dfrac{cot \theta}{1 -tan \theta}$
• $\bf R.H.S = \sf 1 + \tan \theta + \cot \theta$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Solving \: the \: L.H.S \: \::}}$

$\qquad :\implies \bf L.H.S = \:\sf \dfrac{tan \theta}{1- cot \theta} + \dfrac{cot \theta}{1 -tan \theta}$

$\qquad :\implies \sf \dfrac{tan \theta}{1- cot \theta} + \dfrac{cot \theta}{1 -tan \theta}$

$\qquad :\implies \sf \dfrac{tan \theta}{1- cot \theta} - \dfrac{cot \theta}{tan \theta-1}$

$\dag\:\:\it{ As,\:We\:know\:that\::}$

• $\sf cot \theta = \dfrac {1}{cot \theta}$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: this \::}}$

$\qquad :\implies \sf \dfrac{tan \theta}{1- cot \theta} - \dfrac{cot \theta}{tan \theta-1}$

$\qquad :\implies \sf \dfrac{tan \theta}{1- \dfrac{1}{tan \theta}} - \dfrac{\dfrac{1}{tan \theta }}{tan \theta-1}$

$\qquad :\implies \sf \dfrac{tan \theta}{ \dfrac{tan \theta - 1}{tan \theta}} - \dfrac{\dfrac{1}{tan \theta }}{tan \theta-1}$

$\qquad :\implies \sf \dfrac{tan \theta}{ \dfrac{tan \theta - 1}{tan \theta}} - \dfrac{1}{tan \theta (tan \theta-1)}$

$\qquad :\implies \sf \dfrac{tan^2 \theta}{ tan \theta - 1} - \dfrac{1}{tan \theta (tan \theta-1)}$

• Taking $\bf \dfrac {1}{tan \theta -1 }$ as common :

$\qquad :\implies \sf \dfrac{tan^2 \theta}{ tan \theta - 1} - \dfrac{1}{tan \theta (tan \theta-1)}$

$\qquad :\implies \sf \dfrac{1}{tan \theta - 1} \bigg[ tan^2 \theta - \dfrac{1}{tan \theta } \bigg]$

$\qquad :\implies \sf \dfrac{1}{tan \theta - 1} \bigg[ \dfrac{tan^3 \theta - 1 }{tan \theta } \bigg]$

$\dag\:\:\it{ As,\:We\:know\:that\::}$

• Algebraic Indentity : a³ - b³ = (a² + ab + b² ) (a - b)

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Applying \: this \:Indentity \::}}$

$\qquad :\implies \sf \dfrac{1}{tan \theta - 1} \bigg[ \dfrac{tan^3 \theta - 1 }{tan \theta } \bigg]$

$\qquad :\implies \sf \dfrac{1}{tan \theta - 1} \bigg[ \dfrac{(tan^2 \theta + tan \theta + 1^2 ) ( tan \theta - 1 ) }{tan \theta } \bigg]$

$\qquad :\implies \sf \dfrac{1}{tan \theta - 1} \times \dfrac {(tan^2 \theta + tan \theta + 1^2 ) ( tan \theta - 1 ) }{tan \theta }$

$\qquad :\implies \sf \dfrac{1}{\cancel {tan \theta - 1}} \times \dfrac {(tan^2 \theta + tan \theta + 1^2 )\cancel {( tan \theta - 1 )} }{tan \theta }$

$\qquad :\implies \sf \ \dfrac {tan^2 \theta + tan \theta + 1^2 }{tan \theta }$

$\qquad :\implies \sf \ \dfrac {tan^2 \theta + tan \theta + 1 }{tan \theta }$

$\qquad :\implies \sf \ \dfrac {tan^2 \theta}{tan \theta } + \cancel {\dfrac{ tan \theta}{tan \theta }}+\dfrac{ 1 }{tan \theta }$

$\qquad :\implies \sf \ \cancel {\dfrac {tan^2 \theta}{tan \theta }} +1 + \dfrac{ 1 }{tan \theta }$

$\qquad :\implies \sf \ tan \theta +1 + \dfrac{ 1 }{tan \theta }$

$\qquad :\implies \sf \ tan \theta +1 + cot \theta$

$\qquad :\implies \bf L.H.S \: : \sf 1 + tan \theta + cot \theta$

Therefore,

• $\qquad :\implies \bf L.H.S \:= \sf 1 + tan \theta + cot \theta$

• $\qquad :\implies \bf R.H.S \: = \sf 1 + tan \theta + cot \theta$

⠀⠀⠀⠀⠀━━ As , We can see that here L.H.S = R.HS .

⠀⠀⠀⠀⠀$\therefore {\underline {\bf{ Hence, \:Proved\:}}}$

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