Prove that: $\dfrac{\tan \theta}{1 - \cot \theta} + \dfrac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta$.
Step-by-Step Solution
Key Concept: Convert to $\sin, \cos$: $\dfrac{\sin^2\theta}{\cos\theta(\sin\theta - \cos\theta)} - \dfrac{\cos^2\theta}{\sin\theta(\sin\theta - \cos\theta)} = \dfrac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta(\sin\theta - \cos\theta)} = \dfrac{(\sin\theta - \cos\theta)(\sin^2\theta + \sin\theta\cos\theta + \cos^2\theta)}{\sin\theta\cos\theta(\sin\theta - \cos\theta)} = \dfrac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta} = \sec\theta\csc\theta + 1$.
Convert to $\sin\theta, \cos\theta$: LHS $= \dfrac{\sin^2\theta}{\cos\theta(\sin\theta - \cos\theta)} - \dfrac{\cos^2\theta}{\sin\theta(\sin\theta - \cos\theta)}$. [1.0 Mark]
$= \dfrac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta(\sin\theta - \cos\theta)} = \dfrac{(\sin\theta - \cos\theta)(1 + \sin\theta\cos\theta)}{\sin\theta\cos\theta(\sin\theta - \cos\theta)}$. [1.0 Mark]
$= \dfrac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta} = \sec\theta\csc\theta + 1 = $ RHS. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Combining into single fraction over $\sin\theta\cos\theta(\sin\theta-\cos\theta)$: 1.0 Mark
Applying difference of cubes $a^3 - b^3$: 1.0 Mark
Simplifying to $1 + \sec\theta\csc\theta$: 1.0 Mark
Correct Answer: