Question : If $\tan \theta+\sin \theta=A$ and $\tan \theta-\sin \theta=B$, then what is the value of $A^2-B^2$?
Option 1: $\tan \theta \sin \theta$
Option 2: $4 \cot \theta$
Option 3: $4 \tan \theta \sin \theta$
Option 4: $2 \tan \theta \sin \theta$
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Correct Answer: $4 \tan \theta \sin \theta$
Solution :
$\tan \theta+\sin \theta=A$ and $\tan \theta-\sin \theta=B$
$ A^2-B^2=(A+B)(A-B) $
$⇒ A^2 - B^2 = (2 \tan \theta) × (2 \sin \theta)$
$⇒ A^2 - B^2 = 4 \tan \theta \sin \theta $
Therefore, $A^2 - B^2 = 4 \tan \theta \sin \theta$
Hence, the correct answer is $4 \tan \theta \sin \theta$.
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