Math, asked by Queendishaa, 6 months ago

Solve it fast my friends​

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Answered by MoodyCloud
5

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

Answered by ahervandan39
0

see here is u answer

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