Question

prove that tanA /(1-cotA )+cotA (1-tanA )=1+SecAcosecA ​

Answer 1

To Prove :

$\sf \bullet\ \; \dfrac{tanA}{1-cotA}+\dfrac{cotA}{1-tanA}=1+secA.cscA$

Proof :

LHS

$\to \sf \dfrac{tanA}{1-cotA}+\dfrac{cotA}{1-tanA}$

• cot A = 1 / tan A

$\to \sf \dfrac{tanA}{1-\frac{1}{tanA}}+\dfrac{cotA}{1-tanA}\\\\\to \sf \dfrac{tanA.tanA}{tanA-1}+\dfrac{cotA}{1-tanA}\\\\\to \sf \dfrac{tan^2A}{tanA-1}+\dfrac{cotA}{1-tanA}\\\\\to \sf \dfrac{cotA-tan^2A}{1-tanA}\\\\\to \sf \dfrac{\frac{1}{tanA}-tan^2A}{1-tanA}\\\\\to \sf \dfrac{1-tan^3A}{tanA(1-tanA)}$

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

$\\ \to \sf \dfrac{\cancel{(1-tanA)}(1+tanA+tan^2A)}{tanA\cancel{(1-tanA)}}\\\\\to \sf \dfrac{1+tanA+tan^2A}{tanA}$

• sec²A = 1 + tan²A

$\to \sf \dfrac{tanA+sec^2A}{tanA}\\\\\to \sf \dfrac{tanA}{tanA}+\dfrac{sec^2A}{tanA}$

• sec A = 1 / cos A  ,  tan A = sin A / cos A

$\to \sf 1+\dfrac{\frac{1}{cos^2A}}{\frac{sinA}{cosA}}$

$\to \sf 1+\dfrac{cosA}{cos^2A.sinA}$

$\to \sf 1+\dfrac{1}{cosA}.\dfrac{1}{sinA}$

$\to \sf 1+secA.cscA$

RHS

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Alternate Method :

LHS

$\to \sf \dfrac{tanA}{1-cotA}+\dfrac{cotA}{1-tanA}\\\\\to \sf \dfrac{\frac{sinA}{cosA}}{1-\frac{cosA}{sinA}}+\dfrac{\frac{cosA}{sinA}}{1-\frac{sinA}{cosA}}\\\\\to \sf \dfrac{\frac{sinA}{cosA}}{\frac{sinA-cosA}{sinA}}+\dfrac{\frac{cosA}{sinA}}{\frac{cosA-sinA}{cosA}}$

$\to \sf \dfrac{sin^2A}{cosA(sinA-cosA)}+\dfrac{cos^2A}{sinA(cosA-sinA)}\\\\\to \sf \dfrac{sin^2A}{cosA(sinA-cosA)}-\dfrac{cos^2A}{sinA(sinA-cosA)}$

$\to \sf \dfrac{1}{sinA-cosA}\bigg(\dfrac{sin^2A}{cosA}-\dfrac{cos^2A}{sinA}\bigg)\\\\\to \sf \dfrac{1}{sinA-cosA}\bigg(\dfrac{sin^3A-cos^3A}{sinA.cosA}\bigg)$

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

$\to \sf \dfrac{\cancel{(sinA-cosA)}(sin^2A+sinA.cosA+cos^2A)}{\cancel{(sinA-cosA)}(sinA.cosA)}$

$\to \sf \dfrac{sin^2A+cos^2A+sinA.cosA}{sinA.cosA}$

• sin²A + cos²A = 1

$\to \sf \dfrac{1+sinA.cosA}{sinA.cosA}$

$\to \sf \dfrac{1}{sinA.cosA}+\dfrac{sinA.cosA}{sinA.cosA}$

$\to \sf \dfrac{1}{sinA}.\dfrac{1}{cosA}+1$

• csc A = 1 / sin A , sec A = 1 / cos A

$\to \sf cscA.secA+1\\\\\leadsto \sf \pink{1+secA.cscA}\ \; \bigstar$

RHS

Answer 2

Answer:

LHS

$:\implies \sf \dfrac{\tan(A)}{ 1 - \cos(A) }+ \dfrac{\cot(A)}{ 1 - \tan(A) }$

$:\implies \sf \dfrac{ \dfrac{\sin(A)}{ \cos(A) }}{1 - \dfrac{ \cos(A) }{ \sin(A) }} + \dfrac{ \dfrac{\cos(A)}{ \sin(A) }}{1 - \dfrac{ \sin(A) }{ \cos(A) }}$

$:\implies \sf \dfrac{ \dfrac{\sin(A)}{ \cos(A) }}{\dfrac{ \sin(A) - \cos(A) }{ \sin(A) }} + \dfrac{ \dfrac{\cos(A)}{ \sin(A) }}{ \dfrac{ \cos(A) - \sin(A) }{ \cos(A) }}$

$:\implies \sf \dfrac{\sin(A)}{ \cos(A) } \times \dfrac{\sin(A)}{ \sin(A) - \cos(A) } - \dfrac{\cos(A)}{ \sin(A) } \times \dfrac{\cos(A)}{ \sin(A) - \cos(A) }$

$:\implies \sf \dfrac{\sin^{2} (A)}{ \cos(A) \bigg( \sin(A) - \cos(A) \bigg)} - \dfrac{\cos^{2} (A)}{ \sin(A) \bigg( \sin(A) - \cos(A) \bigg)}$

$:\implies \sf \dfrac{\sin^{3} (A) - \cos^{3} (A)}{ \sin(A)\cos(A) \bigg( \sin(A) - \cos(A) \bigg)}$

$:\implies \sf \dfrac{ \bigg(\sin (A) - \cos (A) \bigg)\bigg(\sin^{2} (A) + \cos^{2} (A) + \sin(A)\cos(A)\bigg)}{ \sin(A)\cos(A) \bigg( \sin(A) - \cos(A) \bigg)}$

$:\implies \sf \dfrac{ 1 + \sin(A).\cos(A)}{ \sin(A)\cos(A) }$

$:\implies \sf \dfrac{1}{ \sin(A) } \times \dfrac{1}{ \cos(A) } + \dfrac{\sin(A)\cos(A)}{ \sin(A)\cos(A) }$

$:\implies \sf \cosec(A) \sec(A) + 1$

$:\implies \textsf{\textbf{ 1 + cosec(A) sec(A) }}$

RHS