If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n$, show that $m^2 - n^2 = 4 \sqrt{m n}$.
Step-by-Step Solution
Key Concept: LHS $= (m+n)(m-n) = (2\tan\theta)(2\sin\theta) = 4 \tan\theta \sin\theta$.<br>RHS $= 4\sqrt{(\tan\theta+\sin\theta)(\tan\theta-\sin\theta)} = 4\sqrt{\tan^2\theta - \sin^2\theta} = 4\sqrt{\dfrac{\sin^2\theta}{\cos^2\theta} - \sin^2\theta} = 4\sqrt{\sin^2\theta(\sec^2\theta - 1)} = 4 \sin\theta \tan\theta$. LHS $=$ RHS.
LHS $= (m+n)(m-n) = (2\tan\theta)(2\sin\theta) = 4 \tan\theta \sin\theta$. (1) [1.0 Mark]
RHS $= 4\sqrt{\tan^2\theta - \sin^2\theta} = 4\sqrt{\sin^2\theta\left(\dfrac{1 - \cos^2\theta}{\cos^2\theta}\right)} = 4\sqrt{\sin^2\theta \tan^2\theta} = 4 \sin\theta \tan\theta$. (2) [1.5 Marks]
From (1) & (2), LHS $=$ RHS. Proved! [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Evaluating LHS $= 4\tan\theta \sin\theta$: 1.0 Mark
Evaluating RHS $= 4\tan\theta \sin\theta$: 1.5 Marks
Concluding $m^2 - n^2 = 4\sqrt{mn}$: 0.5 Mark
Correct Answer: