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

Amogh , 10 Years ago
Grade 10
anser 1 Answers
Y RAJYALAKSHMI

Last Activity: 10 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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