If $\csc \theta - \sin \theta = a^3$ and $\sec \theta - \cos \theta = b^3$, prove that $a^2 b^2(a^2 + b^2) = 1$.
Step-by-Step Solution
Key Concept: $a^3 = \dfrac{1-\sin^2\theta}{\sin\theta} = \dfrac{\cos^2\theta}{\sin\theta} \Rightarrow a = \left(\dfrac{\cos^2\theta}{\sin\theta}\right)^{1/3}$.<br>$b^3 = \dfrac{1-\cos^2\theta}{\cos\theta} = \dfrac{\sin^2\theta}{\cos\theta} \Rightarrow b = \left(\dfrac{\sin^2\theta}{\cos\theta}\right)^{1/3}$.<br>$a^2 b^2 = \left(\dfrac{\cos^2\theta}{\sin\theta} \cdot \dfrac{\sin^2\theta}{\cos\theta}\right)^{2/3} = (\sin\theta \cos\theta)^{2/3}$.<br>$a^2 + b^2 = \dfrac{\cos^{4/3}\theta}{\sin^{2/3}\theta} + \dfrac{\sin^{4/3}\theta}{\cos^{2/3}\theta} = \dfrac{\cos^2\theta + \sin^2\theta}{\sin^{2/3}\theta \cos^{2/3}\theta} = \dfrac{1}{(\sin\theta \cos\theta)^{2/3}}$.<br>Product $a^2 b^2(a^2 + b^2) = (\sin\theta \cos\theta)^{2/3} \times \dfrac{1}{(\sin\theta \cos\theta)^{2/3}} = 1$.
$a^3 = \dfrac{\cos^2\theta}{\sin\theta} \Rightarrow a = \dfrac{\cos^{2/3}\theta}{\sin^{1/3}\theta}$ and $b^3 = \dfrac{\sin^2\theta}{\cos\theta} \Rightarrow b = \dfrac{\sin^{2/3}\theta}{\cos^{1/3}\theta}$. [1.5 Marks]
$a^2 b^2 = (\sin\theta \cos\theta)^{2/3}$. [1.0 Mark]
$a^2 + b^2 = \dfrac{\cos^{4/3}\theta}{\sin^{2/3}\theta} + \dfrac{\sin^{4/3}\theta}{\cos^{2/3}\theta} = \dfrac{\cos^2\theta + \sin^2\theta}{(\sin\theta \cos\theta)^{2/3}} = \dfrac{1}{(\sin\theta \cos\theta)^{2/3}}$. [1.5 Marks]
$a^2 b^2 (a^2 + b^2) = (\sin\theta \cos\theta)^{2/3} \times \dfrac{1}{(\sin\theta \cos\theta)^{2/3}} = 1$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Expressing $a^3$ and $b^3$ in terms of $\sin\theta, \cos\theta$: 1.5 Marks
Evaluating $a^2 b^2 = (\sin\theta \cos\theta)^{2/3}$: 1.0 Mark
Evaluating $a^2 + b^2 = 1/(\sin\theta \cos\theta)^{2/3}$: 1.5 Marks
Evaluating product $= 1$: 1.0 Mark
Correct Answer: