Prove that: $\dfrac{\sin \theta}{1 - \cos \theta} + \dfrac{\tan \theta}{1 + \cos \theta} = \sec \theta \csc \theta + \cot \theta$.
Step-by-Step Solution
Key Concept: LHS $= \dfrac{\sin\theta(1+\cos\theta) + \tan\theta(1-\cos\theta)}{1 - \cos^2\theta} = \dfrac{\sin\theta + \sin\theta\cos\theta + \tan\theta - \sin\theta}{\sin^2\theta} = \dfrac{\sin\theta\cos\theta + \tan\theta}{\sin^2\theta} = \dfrac{\cos\theta}{\sin\theta} + \dfrac{1}{\sin\theta\cos\theta} = \cot\theta + \sec\theta\csc\theta$.
LHS $= \dfrac{\sin\theta(1+\cos\theta) + \tan\theta(1-\cos\theta)}{\sin^2\theta}$. [1.0 Mark]
$= \dfrac{\sin\theta + \sin\theta\cos\theta + \tan\theta - \sin\theta}{\sin^2\theta} = \dfrac{\sin\theta\cos\theta + \tan\theta}{\sin^2\theta}$. [1.0 Mark]
$= \dfrac{\sin\theta\cos\theta}{\sin^2\theta} + \dfrac{\sin\theta/\cos\theta}{\sin^2\theta} = \cot\theta + \sec\theta\csc\theta = $ RHS. Proved! [1.0 Mark]
---
🎯 Official CBSE Marking Scheme:
Combining fractions over $\sin^2 \theta$: 1.0 Mark
Expanding numerator: 1.0 Mark
Splitting terms to get $\cot \theta + \sec \theta \csc \theta$: 1.0 Mark
Correct Answer: