Question
If $(\tan\theta+\sin\theta)=\text{m}$ and $(\tan\theta-\sin\theta)=\text{n},$ prove that $\big(\text{m}^2-\text{n}^2\big)^2=16\text{mn}.$
✓
Answer
We have $(\tan\theta+\sin\theta)=\text{m}$ and $(\tan\theta-\sin\theta)=\text{n}$
Now, $\text{LHS}=\big(\text{m}^2-\text{n}^2\big)^2$
$=\Big[(\tan\theta+\sin^2\theta)-(\tan\theta-\sin\theta)^2\Big]^2$
$=\Big[(\tan^2\theta+\sin^2\theta+2\tan\theta\sin\theta\\ \ \ -\big(\tan^2\theta-\sin^2\theta+2\tan\theta\sin\theta\big)\Big]^2$
$=\Big[(\tan^2\theta+\sin^2\theta+2\tan\theta\sin\theta\\ \ \ -\tan^2\theta-\sin^2\theta+2\tan\theta\sin\theta\big)\Big]^2$
$=\big(4\tan\theta\sin\theta\big)^2$
$=16\tan^2\theta\sin^2\theta$
$=16\frac{\sin^2\theta}{\cos^2\theta}\sin^2\theta$
$=16\frac{\big(1-\cos^2\theta\big)\sin^2\theta}{\cos^2\theta}$
$=16\big[\tan^2\theta\big(1-\cos^2\theta\big)\big]$
$=16\big(\tan^2\theta-\sin^2\theta\big)$
$=16(\tan\theta+\sin\theta)(\tan\theta-\sin\theta)$
$=16\text{mn}$ $\big[(\tan\theta+\sin\theta)(\tan\theta-\sin\theta)=\text{mn}\big]$
$\therefore\ \big(\text{m}^2-\text{n}^2\big)\big(\text{m}^2-\text{n}^2\big)=16\text{mn}$
Now, $\text{LHS}=\big(\text{m}^2-\text{n}^2\big)^2$
$=\Big[(\tan\theta+\sin^2\theta)-(\tan\theta-\sin\theta)^2\Big]^2$
$=\Big[(\tan^2\theta+\sin^2\theta+2\tan\theta\sin\theta\\ \ \ -\big(\tan^2\theta-\sin^2\theta+2\tan\theta\sin\theta\big)\Big]^2$
$=\Big[(\tan^2\theta+\sin^2\theta+2\tan\theta\sin\theta\\ \ \ -\tan^2\theta-\sin^2\theta+2\tan\theta\sin\theta\big)\Big]^2$
$=\big(4\tan\theta\sin\theta\big)^2$
$=16\tan^2\theta\sin^2\theta$
$=16\frac{\sin^2\theta}{\cos^2\theta}\sin^2\theta$
$=16\frac{\big(1-\cos^2\theta\big)\sin^2\theta}{\cos^2\theta}$
$=16\big[\tan^2\theta\big(1-\cos^2\theta\big)\big]$
$=16\big(\tan^2\theta-\sin^2\theta\big)$
$=16(\tan\theta+\sin\theta)(\tan\theta-\sin\theta)$
$=16\text{mn}$ $\big[(\tan\theta+\sin\theta)(\tan\theta-\sin\theta)=\text{mn}\big]$
$\therefore\ \big(\text{m}^2-\text{n}^2\big)\big(\text{m}^2-\text{n}^2\big)=16\text{mn}$
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