Subject:
MathAuthor:
guinness52Created:
1 year agoAnswer:
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2[tex]\displaystyle \sf{ \frac{{cos}^{2} \:A }{ cos\:A - sin \:A} + \frac{sin\:A}{1 - cot\:A} = sin\:A + cos\:A} \: \: is \: \bf proved[/tex]
Given :
[tex]\displaystyle \sf{ \frac{{cos}^{2} \:A }{ cos\:A - sin \:A} + \frac{sin\:A}{1 - cot\:A} = sin\:A + cos\:A}[/tex]
To find :
To prove the expression
Solution :
Step 1 of 2 :
Write down the given expression to prove
Here the given expression is
[tex]\displaystyle \sf{ \frac{{cos}^{2} \:A }{ cos\:A - sin \:A} + \frac{sin\:A}{1 - cot\:A} = sin\:A + cos\:A}[/tex]
Step 2 of 2 :
Prove the expression
LHS
[tex]\displaystyle \sf{ = \frac{{cos}^{2} \:A }{ cos\:A - sin \:A} + \frac{sin\:A}{1 - cot\:A} }[/tex]
[tex]\displaystyle \sf{ = \frac{{cos}^{2} \:A }{ cos\:A - sin \:A} + \frac{sin\:A}{1 - \dfrac{cos\:A}{sin\:A} } }[/tex]
[tex]\displaystyle \sf{ = \frac{{cos}^{2} \:A }{ cos\:A - sin \:A} + \frac{sin\:A}{ \dfrac{sin\:A - cos\:A}{sin\:A} } }[/tex]
[tex]\displaystyle \sf{ = \frac{{cos}^{2} \:A }{ cos\:A - sin \:A} + \frac{{sin}^{2} \:A}{sin\:A - cos\:A} }[/tex]
[tex]\displaystyle \sf{ = \frac{{cos}^{2} \:A }{ cos\:A - sin \:A} + \frac{{sin}^{2} \:A}{ - (cos\:A - sin \:A)} }[/tex]
[tex]\displaystyle \sf{ = \frac{{cos}^{2} \:A }{ cos\:A - sin \:A} - \frac{{sin}^{2} \:A}{ cos\:A - sin \:A} }[/tex]
[tex]\displaystyle \sf{ = \frac{{cos}^{2} \:A - {sin}^{2} \:A}{ cos\:A - sin \:A} }[/tex]
[tex]\displaystyle \sf{ = \frac{(cos\:A + sin \:A)(cos\:A - sin \:A)}{ cos\:A - sin \:A} }[/tex]
[tex]\displaystyle \sf{ = (sin\:A + cos \:A) }[/tex]
= RHS
Hence the proof follows
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