Question
If $(\text{cosec }\theta+\sin\theta)=\text{a}^3$ and $(\sec\theta-\cos\theta)=\text{b}^3,$ prove that $\big(\text{a}^2\text{b}^2\big)\big(\text{a}^2+\text{b}^2\big)=1.$
✓
Answer
We have $(\text{cosec }\theta-\sin\theta)=\text{a}^3$ and $(\sec\theta-\cos\theta)=\text{n}$
$\Rightarrow\text{a}^3=\Big(\frac{1}{\sin\theta}-\sin\theta\Big)$
$\Rightarrow\text{a}^3=\frac{\big(1-\sin^2\theta\big)}{\sin\theta}$
$=\frac{\cos^2\theta}{\sin\theta}$
$\therefore\ \text{a}=\frac{\cos^{\frac23}\theta}{\sin^\frac13\theta}$
Again, $(\sec\theta-\cos\theta)=\text{b}^3$
$\Rightarrow\text{b}^3=\Big(\frac{1}{\cos\theta}-\cos\theta\Big)$
$=\frac{\big(1-\cos^2\theta\big)}{\cos\theta}$
$=\frac{\sin^2\theta}{\cos\theta}$
$\therefore\ \text{b}=\frac{\sin^\frac{2}{3}\theta}{\cos^\frac13\theta}$
Now, $\text{LHS}=\text{a}^2\text{b}^2\big(\text{a}^2+\text{b}^2\big)$
$=\text{a}^4\text{b}+\text{a}^2\text{b}^4$
$=\text{a}^3\big(\text{ab}^2\big)+\big(\text{a}^2\text{b}\big)\text{b}^3$
$=\frac{\cos^3\theta}{\sin\theta}\times\Bigg[\frac{\cos^\frac23\theta}{\sin^\frac13\theta}\times\frac{\sin^\frac43\theta}{\cos^\frac23\theta}\Bigg]\\ \ +\frac{\cos^3\theta}{\sin\theta}\times\Bigg[\frac{\cos^\frac23\theta}{\sin^\frac13\theta}\times\frac{\sin^\frac43\theta}{\cos^\frac23\theta}\Bigg]\times\frac{\sin^2\theta}{\cos\theta}$
$=\frac{\cos^2\theta}{\sin\theta}\times\sin\theta+\cos\theta\times\frac{\sin^2\theta}{\cos\theta}$
$=\cos^2\theta+\sin^2\theta=1$
$=\text{RHS}$
Hence proved.
$\Rightarrow\text{a}^3=\Big(\frac{1}{\sin\theta}-\sin\theta\Big)$
$\Rightarrow\text{a}^3=\frac{\big(1-\sin^2\theta\big)}{\sin\theta}$
$=\frac{\cos^2\theta}{\sin\theta}$
$\therefore\ \text{a}=\frac{\cos^{\frac23}\theta}{\sin^\frac13\theta}$
Again, $(\sec\theta-\cos\theta)=\text{b}^3$
$\Rightarrow\text{b}^3=\Big(\frac{1}{\cos\theta}-\cos\theta\Big)$
$=\frac{\big(1-\cos^2\theta\big)}{\cos\theta}$
$=\frac{\sin^2\theta}{\cos\theta}$
$\therefore\ \text{b}=\frac{\sin^\frac{2}{3}\theta}{\cos^\frac13\theta}$
Now, $\text{LHS}=\text{a}^2\text{b}^2\big(\text{a}^2+\text{b}^2\big)$
$=\text{a}^4\text{b}+\text{a}^2\text{b}^4$
$=\text{a}^3\big(\text{ab}^2\big)+\big(\text{a}^2\text{b}\big)\text{b}^3$
$=\frac{\cos^3\theta}{\sin\theta}\times\Bigg[\frac{\cos^\frac23\theta}{\sin^\frac13\theta}\times\frac{\sin^\frac43\theta}{\cos^\frac23\theta}\Bigg]\\ \ +\frac{\cos^3\theta}{\sin\theta}\times\Bigg[\frac{\cos^\frac23\theta}{\sin^\frac13\theta}\times\frac{\sin^\frac43\theta}{\cos^\frac23\theta}\Bigg]\times\frac{\sin^2\theta}{\cos\theta}$
$=\frac{\cos^2\theta}{\sin\theta}\times\sin\theta+\cos\theta\times\frac{\sin^2\theta}{\cos\theta}$
$=\cos^2\theta+\sin^2\theta=1$
$=\text{RHS}$
Hence proved.
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