Question : If $\sin \theta+\tan \theta=\mathrm{p}$ and $\tan \theta-\sin \theta=\mathrm{q}$, then which of the relation satisfies for given values of $\mathrm{p}$ and $\mathrm{q}$?
Option 1: $\mathrm{p}^2-\mathrm{q}^2=4 \sqrt{\mathrm{pq}}$
Option 2: $\mathrm{p}^2-\mathrm{q}^2=8 \sqrt{\mathrm{pq}}$
Option 3: $\mathrm{p}^2+\mathrm{q}^2=4 \sqrt{\mathrm{pq}}$
Option 4: $\mathrm{p}^2+\mathrm{q}^2=4 \mathrm{pq}$
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Correct Answer: $\mathrm{p}^2-\mathrm{q}^2=4 \sqrt{\mathrm{pq}}$
Solution :
Given, $\sin \theta+\tan \theta=\mathrm{p}$ and $\tan \theta-\sin \theta=\mathrm{q}$
Now, $\mathrm{pq}=(\tan \theta-\sin \theta)(\tan \theta+\sin \theta)$
Using $(a-b)(a+b)=a^2-b^2$
⇒ $\mathrm{pq}=\tan^2\theta-\sin^2\theta$
⇒ $\mathrm{pq}=\frac{\sin^2\theta}{\cos^2\theta}-\sin^2\theta$
⇒ $\mathrm{pq}=\sin^2\theta(\frac{1}{\cos^2\theta}-1)$
⇒ $\mathrm{pq}=\sin^2\theta(\frac{1-\cos^2\theta}{\cos^2\theta})$
We know, $\sin^2\theta+\cos^2\theta = 1$
⇒ $\mathrm{pq}=\sin^2\theta(\frac{\sin^2\theta}{\cos^2\theta})$
⇒ $\mathrm{pq}=\sin^2\theta\tan^2\theta$
And, $\mathrm{p^2}=(\sin \theta+\tan \theta)^2$
$=\sin^2 \theta+\tan^2 \theta+2\sin\theta\tan\theta$
Also, $\mathrm{q^2}=(\tan \theta-\sin \theta)^2$
$=\tan^2\theta+\sin^2\theta-2\tan\theta\sin\theta$
$\therefore$ $\mathrm{p^2-q^2}=(\sin^2 \theta+\tan^2 \theta+2\sin\theta\tan\theta)-(\tan^2\theta+\sin^2\theta-2\tan\theta\sin\theta)=4\sin\theta\tan\theta$
$\therefore$ we can write $\mathrm{p^2-q^2}=4\sqrt{\mathrm{pq}}$
Hence, the correct answer is $\mathrm{p}^2-\mathrm{q}^2=4 \sqrt{\mathrm{pq}}$.
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