Introduction to Trigonometry and Its Applications
NCERT Exemplar Ch 08
CBSE_NCERT_EXEMPLAR_CH08
Grade 10
Question:
If $\tan A + \sin A = m$ and $\tan A - \sin A = n$, show that $m^2 - n^2 = 4 \sqrt{mn}$.
Step-by-Step Solution
Key Concept: Compute $m^2 - n^2 = 4 \tan A \sin A$ and $4\sqrt{mn} = 4\sqrt{\tan^2 A - \sin^2 A} = 4 \tan A \sin A$.
Stepwise Solution:
\text{LHS} = m^2 - n^2 = (\tan A + \sin A)^2 - (\tan A - \sin A)^2 = 4 \tan A \sin A$. [1.0 Mark]
\text{RHS} = 4 \sqrt{mn} = 4 \sqrt{(\tan A + \sin A)(\tan A - \sin A)} = 4 \sqrt{\tan^2 A - \sin^2 A}$. [1.0 Mark]
$4 \sqrt{\dfrac{\sin^2 A}{\cos^2 A} - \sin^2 A} = 4 \sqrt{\sin^2 A \left(\dfrac{1}{\cos^2 A} - 1\right)} = 4 \sqrt{\sin^2 A \tan^2 A} = 4 \sin A \tan A = \text{LHS}$. Proved! [1.0 Mark]
Marking Scheme:
• Evaluating $m^2 - n^2 = 4 \tan A \sin A$: 1.0 Mark
• Evaluating $mn = \tan^2 A - \sin^2 A$: 1.0 Mark
• Proving $\sqrt{\tan^2 A - \sin^2 A} = \tan A \sin A$: 1.0 Mark
Correct Answer:
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