Question

prove that: (see the attachment)
solve step by step​

Answer 1

⠀⠀ || ✪✪ Prove.✪✪  ||

Take L.H.S.

tan A /(1-cot A) + cot A /(1-tan A)

( change quantity )

★ Sin A / Cos A = tan A

★ Cos A / Sin A = Cot A

= (sin A / Cos A )/(1-cos A/Sin A) + (cos A/Sin A )/(1- sin A/ cos A)

= sin² A /cos A(Sin A - Cos A) - Cos² A/sin A(sin A - cos A)

= (Sin³ A - Cos³ A)/sin A Cos A(Sin A - Cos A)

= ( Sin A - Cos A)(Sin² A + Cos² A + Sin A Cos A)/sin A Cos A(Sin A - Cos A)

( eliminate Sin A - Cos A from Denominator and numerator )

= ( sin² A + Cos² A + Sin A Cos A )/ Sin A cos A

= (1 + Sin A Cos A)/ Sin A Cos A

= (1/Sin A Cos A) + (Sin A Cos A/ Sin A Cos A)

= (1/Sin A ) × (1/ Cos A) + (Sin A Cos A/ Sin A Cos A)

= sec A Cosec A + 1

= R.H.S.

That's proved.

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Use Formula

★ (a³-b³) = (a-b)(a²+b²+ab)

★ ( 1/ Sinx ) = Cosec x

★ ( 1/Cos x ) = Sec x

★ sin² x + Cos² x = 1

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Answer 2

HEY MATE YOUR ANSWER IS HERE...

BY TAKING LHS

$\frac{ \tan \: a }{1 - \cot \: a } + \frac{ \cot \: a}{1 - \tan \: a }$

NOW BY TRIGNOMATERIC RATIOS

$tan \: a \: = \frac{sin \: a}{cos \: a} \\ \\ cot \: a \: = \frac{cos \: a}{sin \: a}$

HENCE ,

$\frac{ \frac{ \sin \: a }{ \cos \: a } }{1 - \frac{cos \: a}{sin \: a} } + \frac{ \frac{cos \: a}{sin \: a} }{1 - \frac{sin \: a}{cos \: a} }$

NOW

$\frac{ \frac{sin \: a}{cos \: a}}{ \frac{cos \: a - sin \: a \: }{cos \: a} } + \frac{ \frac{cos \: a}{sina} }{ \frac{sin \: a \: - cos \: a}{sin \: a} }$

BY SOLVING IT FURTHER WE GET

$\frac{ {sin}^{2}a }{cos \: a(sin \: a - \: cos \: a)} + \frac{ {cos}^{2}a }{sin \: a(cos \: a \: - sin \: a)}$

THEN ,

BY STANDARDIZED FORM ( taking - as common)

$\frac{ {sin}^{2}a }{cos \: a(sin \: a - \: cos \: a)} - \frac{ {cos}^{2}a }{sin \: a(sin \: a \: - \: cos \: a)}$

Now

$\frac{ {sin}^{3}a - {cos}^{3}a }{sin \: a \times \: cos \: a \: (sin \: a \: - \: cos \: a \: )}$

NOW BY IDENTITY

a³ - b³ = ( a - b ) ( a² + b² + ab )

let a = sin a

let b = cos a

by applying identity

$\frac{( \: sin \: a \: - \: cos \: a \: )( \: {sin}^{2}a \: + { \: cos}^{2} a + sin \: a \: cos \: a) }{sin \: a \: cos \: a( \: sin \: a \: - \: cos \: a)}$

AFTER CANCEL OUTING TERMS

$\frac{( \: {sin}^{2}a \: + { \: cos}^{2} a + sin \: a \: cos \: a) }{sin \: a \: cos \: a}$

THEN ,

BY TRIGNOMATERIC IDENTITY

SIN²A + COS²A = 1

$\frac{( \: 1 + sin \: a \: cos \: a) }{sin \: a \: cos \: a}$

NOW ,

$= \frac{1}{sin \: a \: cos \: a} + 1$

NOW

1/ sin A = cosec A

1/ COS A = SEC A

HENCE ,

$sec \: a \: cosec \: a \: + 1$

HENCE PROVED

THANKS FOR YOUR QUESTION HOPE THIS HELPS...

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