Question

Cosec-sin=m3 and sec-cos=n3 then prove that m4n2+m2n4=1

Answer 1

As we know that
$cosec \: x = \frac{1}{sin \: x} \\ \\ sec \: x = \frac{1}{cos \: x}$
Given
$cosec \: x - sin \: x = {m}^{3} \\ \\ \frac{1}{sin \: x} - sin \: x = {m}^{3} \\ \\ \frac{1 - {sin}^{2}x }{sin \: x} = {m}^{3} \\ \\ \frac{ {cos}^{2}x }{sin \: x} = {m}^{3} ...eq1$
By the same way
$sec \: x - cos \: x = {n}^{3} \\ \\ \frac{1}{cos \: x} - cos \: x = {n}^{3} \\ \\ \frac{1 - {cos}^{2}x }{cos \: x} = {n}^{3} \\ \\ \frac{ {sin}^{2}x }{cos \: x} = {n}^{3} ...eq2$

Squaring both side eq1
$\frac{ {sin}^{4}x }{ {cos}^{2} \: x} = {n}^{6} \\ \\ put \: value \: of \: {cos}^{2} x \: from \: second \: equation \\ \\ \frac{ {sin}^{4}x }{ {m}^{3} sin\: x } = {n}^{6} \\ \\ {sin}^{3} x = {m}^{3} {n}^{6} \\ \\ {sin}^{3}\: x = ( {m {n}^{2} })^{3} \\ \\ compare \: base \: sin \: x = m {n}^{2}$

Thus put the value of sin x in any equation and find cos x,
$cos \: x = n {m}^{2}$
Now we know that

${sin}^{2} x + {cos}^{2} x = 1 \\ \\ ({m {n}^{2} })^{2}+ {(n {m}^{2})}^{2} = 1 \\ \\ {m}^{2} {n}^{4} + {n}^{2} {m}^{4} = 1$
Hence proved