If - =a^3, - =b^3, than proved that a^2b^2 (a^2 + b^2) = 1.

Question

If $\cos\text{x}-\sin\text{x}=\text{a}^3, \sec\text{x}-\cos\text{x}=\text{b}^3,$ than proved that $a^2b^2 (a^2 + b^2) = 1.$

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Answer

Given: $\text{cosec }\text{x}-\sin\text{x}=\text{a}^3,\sec\text{x}-\cos\text{x}=\text{b}^2$
To show: $\text{a}^2\text{b}^2(\text{a}^2+\text{b}^2)=1$
Since, $\text{cosec }\text{x}-\sin\text{x}=\text{a}^3$
$\Rightarrow\frac{1}{\sin\text{x}}-\sin\text{x}=\text{a}^3$ $\Big(\because\text{cosec }\text{x}=\frac{1}{\sin\text{x}}\Big)$
$\Rightarrow\frac{1-\sin^2\text{x}}{\sin\text{x}}=\text{a}^3$
$\Rightarrow\frac{\cos^2\text{x}}{\sin\text{x}}=\text{a}^3$ $(\because1-\sin^2\text{x}=\cos^2\text{x})$
$\Rightarrow\text{a}=\frac{\cos\frac{2}{3}\text{x}}{\sin\frac{1}{3}\text{x}}$
Since, $\frac{1}{\cos\text{x}}-\cos\text{x}=\text{b}^3$ $\Big(\because\sec\text{x}=\frac{1}{\cos\text{x}}\Big)$
$\Rightarrow\frac{1-\cos^2\text{x}}{\cos\text{x}}=\text{b}^3$
$\Rightarrow\frac{\sin^2\text{x}}{\cos\text{x}}=\text{b}^3$ $(\because1-\cos^2\text{x}=\sin^2\text{x})$
$\Rightarrow\text{b}=\frac{\sin\frac{2}{3}\text{x}}{\cos\frac{1}{3}\text{x}}$
Now, $\text{a}^2\text{b}^2 \text{(a}^2 +\text{ b}^2)$
$=\frac{\cos\frac{4}{3}\text{x}}{\sin\frac{2}{3}\text{x}}\times\frac{\sin\frac{4}{3}\text{x}}{\cos\frac{2}{3}\text{x}}\Bigg(\frac{\cos\frac{4}{3}\text{x}}{\sin\frac{2}{3}\text{x}}+\frac{\sin\frac{4}{3}\text{x}}{\cos\frac{2}{3}\text{x}}\Bigg)$
$=\cos\frac{2}{3}\text{x}\times\sin\frac{2}{3}\text{x}\frac{\Big(\cos\frac{6}{3}\text{x}+\sin\frac{6}{3}\text{x}\Big)}{\sin\frac{2}{3}\text{x}.\cos\frac{2}{3}\text{x}}$
$=\cos^2\text{x}+\sin^2\text{x}$
$=1$
$\text{Proved}$

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