Question

If cosecA-sinA=l and secA-cosA=m.prove that l²m²(l²+m²+3)=1.​

Answer 1

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2 answers · Mathematics 

 Best Answer

cosecA-sinA=cos^2A/sinA = l 
secA - cosA=sin^2A/cosA = m 

l*l*m*m(l*l+m*m+3) 
l^2*m^2(l^2+m^2+3) 
sub the values of l and m we get 
=cos^2A * sin^2A ( cos^6A+sin^6A+3 * sin^2Acos^2A ) / (sin^2A * cos^2A) 
= cos^6A+sin^6A+3 * sin^2Acos^2A 
=(cos^2A+sin^2A)^3 
=1^3 we know cos^2A+sin^2A=1 
=1 


is the above ans and we need to prove that(this isnot obvious) 

= cos^6A+sin^6A+3 * sin^2Acos^2A 
=(cos^2A+sin^2A)^3 

we know (cos^2 A + sin ^2 A)^3 
= cos^6A + sin ^6 A + 3 cos^2 A sin ^2 A ( cos^2 A + sin ^2 A) 

using (a+b)^3 = a^3 + b^3 + 3ab(a^2+b^2) 
= cos^6A+sin^6A+3 * sin^2Acos^2A as ( cos^2 A + sin ^2 A) = 1

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