Question

If \[cosec x - \sin x = a^3 , \sec x - \cos x = b^3\], then prove that \[a^2 b^2 \left( a^2 + b^2 \right) = 1\]

Answer 1

\[cosec x - \sin x = a^3 \]
\[ \therefore \frac{1}{\sin x} - \sin = a^3 \]
\[ \Rightarrow \frac{1 - \sin^2 x}{\sin x} = a^3 \]
\[ \Rightarrow \frac{\cos^2 x}{\sin x} = a^3 \]
\[ \Rightarrow a = \left( \frac{\cos^2 x}{\sin x} \right)^\frac{1}{3} . . . . (i)\]
\[\text{ Also, }\sec x - \cos x = b^3 \]
\[ \Rightarrow \frac{1}{\cos x} - \cos = b^3 \]
\[ \Rightarrow \frac{1 - \cos^2 x}{\cos x} = b^3 \]
\[ \Rightarrow \frac{\sin^2 x}{\cos x} = b^3 \]
\[ \Rightarrow b = \left( \frac{\sin^2 x}{\cos x} \right)^\frac{1}{3} . . . . . (ii)\]
\[\text{ Now, LHS }= a^2 b^2 \left( a^2 + b^2 \right) = \left( ab \right)^2 \left( a^2 + b^2 \right)\]
\[ = \left[ \left( \frac{\cos^2 x}{\sin x} \right)^\frac{1}{3} \left( \frac{\sin^2 x}{\cos x} \right)^\frac{1}{3} \right]^2 \left[ \left( \left( \frac{\cos^2 x}{\sin x} \right)^\frac{1}{3} \right)^2 + \left( \left( \frac{\sin^2 x}{\cos x} \right)^\frac{1}{3} \right)^2 \right]\]
\[ = \left( \sin x \cos x \right)^\frac{2}{3} \left[ \frac{\left( \cos^2 x \right)^\frac{2}{3}}{\left( \sin x \right)^\frac{2}{3}} + \frac{\left( \sin^2 x \right)^\frac{2}{3}}{\left( \cos x \right)^\frac{2}{3}} \right]\]
\[ = \left( \sin x \cos x \right)^\frac{2}{3} \left[ \frac{\left( \cos^3 x \right)^\frac{2}{3} + \left( \sin^3 x \right)^\frac{2}{3}}{\left( \sin x \right)^\frac{2}{3} \left( \cos x \right)^\frac{2}{3}} \right]\]
\[ = \left( \sin x \cos x \right)^\frac{2}{3} \left[ \frac{\cos^2 x + \sin^2 x}{\left( \sin x \cos x \right)^\frac{2}{3}} \right]\]
 = 1 = RHS