If \[ \cosec x-\sin x=a^3,\qquad \sec x-\cos x=b^3 \] Prove that \[ a^2b^2(a^2+b^2)=1 \]

Solution:

\[ \cosec x-\sin x = \frac{1}{\sin x}-\sin x = \frac{1-\sin^2 x}{\sin x} = \frac{\cos^2 x}{\sin x} \]

\[ a^3=\frac{\cos^2 x}{\sin x} \]

Similarly,

\[ \sec x-\cos x = \frac{1}{\cos x}-\cos x = \frac{1-\cos^2 x}{\cos x} = \frac{\sin^2 x}{\cos x} \]

\[ b^3=\frac{\sin^2 x}{\cos x} \]

Multiplying,

\[ a^3b^3 = \sin x\cos x \]

\[ (ab)^3=\sin x\cos x \]

Also,

\[ a^2 = \left(\frac{\cos^2 x}{\sin x}\right)^{\frac23}, \qquad b^2 = \left(\frac{\sin^2 x}{\cos x}\right)^{\frac23} \]

\[ a^2+b^2 = \frac{\cos^{4/3}x}{\sin^{2/3}x} + \frac{\sin^{4/3}x}{\cos^{2/3}x} \]

Taking LCM,

\[ = \frac{\cos^2 x+\sin^2 x} {\sin^{2/3}x\cos^{2/3}x} \]

\[ = \frac{1} {(\sin x\cos x)^{2/3}} \]

Since \[ (ab)^3=\sin x\cos x \]

\[ a^2+b^2 = \frac{1}{a^2b^2} \]

Therefore,

\[ a^2b^2(a^2+b^2)=1 \]

Hence proved.