Question

Solve it fast my friends​

Answer 1

To prove:-

$\bigstar \sf \cfrac{(1 + cot \: A + tan \: A)(sin \:A - cos \: A}{ {sec}^{3} \: A - {cosec}^{3} \: A)} = {sin}^{2} \: A \: {cos}^{2} \:A$

Solution:-

Take L.H.S

$\star \sf \cfrac{(1 + cot \: A + tan \: A)(sin \:A - cos \: A}{ {sec}^{3} \: A - {cosec}^{3} \: A}$

$\implies \sf \cfrac{[1 + \cfrac{cos \:A}{sin \: A} \ + \cfrac{sin \:A }{cos \: A} ](sin \: A - cos \:A )}{[ \cfrac{1}{ {cos}^{3} \: A} - \cfrac{1}{ {sin}^{3}A } ]}$

$\implies \sf \cfrac{[1 + \dfrac{ {cos}^{2} \: A + {sin}^{2} \: A}{sin \:A \: cos \: A} ](sin \: A - cos \: A)}{[ \dfrac{ {sin}^{3} \: A - {cos}^{3} \:A }{ {sin}^{3} \: A \: {cos}^{3} \: A } ]}$

$\implies \sf \cfrac{[1 + \dfrac{1}{sin \:A \: cos \:A} ](sin \:A - cos \: A) \: {sin}^{3} \: A \: cos^{3} \: A}{( {sin}^{3} - cos \: ^{3} \: A}$

$\implies \sf \cfrac{(sin \:A \: cos \: A + 1)(sin \: A - cos \: A) {sin}^{2}A \: {cos}^{2} \: A }{(sin \:A - cos \: A)( {sin}^{2} \: A + {cos}^{2} \: A + sin \:A \: cos \: A)}$

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

$\implies \sf \cfrac{(sin \:A \: cos \: A + 1) {sin}^{2} A \: {cos}^{2} \:A }{(1 + sin \: A \: cos \: A)}$

$\implies \sf {sin}^{2} \: A \: {cos}^{2} \: A$

Therefore,

L.H.S = R.H.S

Answer 2

see here is u answer

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