Question
If $(\cot\theta+\tan\theta)=\text{m}$ and $(\sec\theta-\cos\theta)=\text{n},$ prove that $\big(\text{m}^2\text{n}\big)^\frac23-\big(\text{m}\text{n}^2\big)^\frac23=1.$
✓
Answer
We have $(\cot\theta+\tan\theta)=\text{m}$ and $(\sec\theta-\cos\theta)=\text{n}$
Now, $\text{m}^2\text{n}=\Big[\big(\cot\theta+\tan\theta\big)^2\big(\sec\theta-\cos\theta\big)\Big]$
$=\Big[\Big(\frac{1}{\tan\theta}+\tan\theta\Big)^2\Big(\frac{1}{\cos\theta}-\cos\theta\Big)\Big]$
$=\frac{\big(1+\tan^2\theta\big)^2}{\tan^2\theta}\times\frac{\big(1-\cos^2\theta\big)}{\cos\theta}$
$=\frac{\sec^4\theta}{\tan^2\theta}\times\frac{\sin^2\theta}{\cos\theta}$
$=\frac{\sec^4\theta}{\frac{\sin^2\theta}{\cos^2\theta}}\times\frac{\sin^2\theta}{\cos\theta}$
$=\frac{\cos^2\theta\times\sec^4\theta}{\cos\theta}$
$=\cos\theta\sec^4\theta$
$=\frac{1}{\sec\theta}\times\sec^4\theta=\sec^3\theta$
$\therefore\ \big(\text{m}^2\text{n}\big)^\frac23=\big(\sec^3\theta\big)^\frac23=\sec^2\theta$
Again, $\text{mn}^2=\Big[(\cot\theta+\tan\theta)(\sec\theta-\cos\theta)^2\Big]$
$=\bigg[\Big(\frac{1}{\tan\theta}+\tan\theta\Big).\Big(\frac{1}{\cos\theta}-\cos\theta\Big)^2\bigg]$
$=\frac{\big(1+\tan^2\theta\big)}{\tan\theta}\times\frac{\big(1-\cos^2\theta\big)^2}{\cos^2\theta}$
$=\frac{\sec^2\theta}{\tan\theta}\times\frac{\sin^4\theta}{\cos^2\theta}$
$=\frac{\sec^2\theta}{\frac{\sin\theta}{\cos\theta}}\times\frac{\sin^4\theta}{\cos^2\theta}$
$=\frac{\sec^2\theta\times\sin^3\theta}{\cos\theta}$
$=\frac{1}{\cos^2\theta}\times\frac{\sec^3\theta}{\cos\theta}=\tan^3\theta$
$\therefore\ \big(\text{m}\text{n}^2\big)^\frac23=\big(\tan^3\theta\big)^\frac23=\tan^2\theta$
Now, $\big(\text{m}^2\text{n}\big)^\frac23-\big(\text{m}\text{n}^2\big)^\frac23$
$=\sec^2\theta-\tan^2\theta=1$
$=\text{RHS}$
Hence proved.
Now, $\text{m}^2\text{n}=\Big[\big(\cot\theta+\tan\theta\big)^2\big(\sec\theta-\cos\theta\big)\Big]$
$=\Big[\Big(\frac{1}{\tan\theta}+\tan\theta\Big)^2\Big(\frac{1}{\cos\theta}-\cos\theta\Big)\Big]$
$=\frac{\big(1+\tan^2\theta\big)^2}{\tan^2\theta}\times\frac{\big(1-\cos^2\theta\big)}{\cos\theta}$
$=\frac{\sec^4\theta}{\tan^2\theta}\times\frac{\sin^2\theta}{\cos\theta}$
$=\frac{\sec^4\theta}{\frac{\sin^2\theta}{\cos^2\theta}}\times\frac{\sin^2\theta}{\cos\theta}$
$=\frac{\cos^2\theta\times\sec^4\theta}{\cos\theta}$
$=\cos\theta\sec^4\theta$
$=\frac{1}{\sec\theta}\times\sec^4\theta=\sec^3\theta$
$\therefore\ \big(\text{m}^2\text{n}\big)^\frac23=\big(\sec^3\theta\big)^\frac23=\sec^2\theta$
Again, $\text{mn}^2=\Big[(\cot\theta+\tan\theta)(\sec\theta-\cos\theta)^2\Big]$
$=\bigg[\Big(\frac{1}{\tan\theta}+\tan\theta\Big).\Big(\frac{1}{\cos\theta}-\cos\theta\Big)^2\bigg]$
$=\frac{\big(1+\tan^2\theta\big)}{\tan\theta}\times\frac{\big(1-\cos^2\theta\big)^2}{\cos^2\theta}$
$=\frac{\sec^2\theta}{\tan\theta}\times\frac{\sin^4\theta}{\cos^2\theta}$
$=\frac{\sec^2\theta}{\frac{\sin\theta}{\cos\theta}}\times\frac{\sin^4\theta}{\cos^2\theta}$
$=\frac{\sec^2\theta\times\sin^3\theta}{\cos\theta}$
$=\frac{1}{\cos^2\theta}\times\frac{\sec^3\theta}{\cos\theta}=\tan^3\theta$
$\therefore\ \big(\text{m}\text{n}^2\big)^\frac23=\big(\tan^3\theta\big)^\frac23=\tan^2\theta$
Now, $\big(\text{m}^2\text{n}\big)^\frac23-\big(\text{m}\text{n}^2\big)^\frac23$
$=\sec^2\theta-\tan^2\theta=1$
$=\text{RHS}$
Hence proved.
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