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prove that sec^2thia = cos^2thita can never be less than 2 xcothita-ysinthita=a xsinthita+ycosthita=b then prove that x^2+y^2=a^2+b^2

  1. prove that sec^2thia = cos^2thita can never be less than 2
  2. xcothita-ysinthita=a xsinthita+ycosthita=b then prove that x^2+y^2=a^2+b^2

Grade:10

1 Answers

Y RAJYALAKSHMI
45 Points
7 years ago
I think you need to check these  problems.  
1.check the sign for sec^2thia = cos^2thita ( may not be =)
2. The second problem should be xcos thita – ysinthita = a & xsin thita + ycos thita = b.  Then square these equations and add.
 
 
Pl. check
 

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