If cosecA-sinA=x^3 and secA-cosA=y^3, then show that x^2y^2{x^2+y^2}=1

Friday, April 4, 2014

If cosecA-sinA=x^3 and secA-cosA=y^3, then show that x^2y^2{x^2+y^2}=1

$\displaystyle{c}{\sec{{A}}}-{\sin{{A}}}={{x}}^{{3}}$

$\displaystyle=\Rightarrow\frac{{1}}{{\sin{{A}}}}-{\sin{{A}}}=$
x^3

$\displaystyle=\Rightarrow\frac{{{1}-{{\sin}}^{{2}}{A}}}{{\sin{{A}}}}=$
x^3

$\displaystyle=\Rightarrow\frac{{{{\cos}}^{{2}}{A}}}{{\sin{{A}}}}={{x}}^{{3}}$

$\displaystyle=\Rightarrow{x}=$
((cos^2A)/sinA)^(1/3)...............(1)

$\displaystyle{\sec{{A}}}-{\cos{{A}}}=$
y^3

$\displaystyle=\Rightarrow\frac{{1}}{{\cos{{A}}}}-{\cos{{A}}}=$
y^3

$\displaystyle=\Rightarrow\frac{{{1}-{{\cos}}^{{2}}{A}}}{{\cos{{A}}}}=$
y^3

$\displaystyle=\Rightarrow\frac{{{{\sin}}^{{2}}{A}}}{{\cos{{A}}}}=$
y^3

$\displaystyle=\Rightarrow{y}={\left({\left({{\sin}}^{{2}}\right.}\right.}$
A)/cosA)^(1/3)......................(2)

Now we will
substitute into the equations:

$\displaystyle{\left({{x}}^{{2}}{{y}}^{{2}}\right)}{\left({{x}}^{{2}}+{{y}}^{{2}}\right)}=$
1

$\displaystyle{{\left(\frac{{{{\cos}}^{{2}}{A}}}{{\sin{{A}}}}\right)}}^{{\frac{{2}}{{3}}}}{{\left(\frac{{{{\sin}}^{{2}}{A}}}{{\cos{{A}}}}\right)}}^{{\frac{{2}}{{3}}}}{\left({\left({\left({{\cos}}^{{2}}\right.}\right.}\right.}$
A)/sinA)^(2/3) + ((sin^2 A)/cosA)^(2/3)) = 1

$\displaystyle{\left({\left({\cos{{A}}}\right.}\right.}$
sinA)^(4/3) )/ (sinA cosA)^(2/3) (((cos^(4/3) A)(cos^ (2/3) A)+(sin^(4/3) A) (sin^(2/3)
A))/(sinA cosA)^(2/3))= 1

$\displaystyle=\Rightarrow{{\left({\cos{{A}}}{\sin{{A}}}\right)}}^{{\frac{{2}}{{3}}}}{\left({\left({(}\right.}\right.}$
cos^2 A) + (sin^2 A))/ (sinA cosA)^(2/3))= 1

Reduce
similar terms.

==> $\displaystyle{{\cos}}^{{2}}{A}+{{\sin}}^{{2}}{A}=$
1

$\displaystyle=\Rightarrow{1}={1}$

Then,
we have proved that if $\displaystyle{\sec{{A}}}-{\cos{{x}}}={{y}}^{{3}}$ and $\displaystyle{c}{\sec{{A}}}-{\sin{{A}}}={{x}}^{{3}}$ Then, $\displaystyle{\left({{x}}^{{2}}{{y}}^{{2}}\right)}$
(x^2 +y^2) =
1

Help Notes

Friday, April 4, 2014