Prove that, \( \frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=\tan \theta+\cot \theta+1 \)

To prove the identity: \[ \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = \tan \theta + \cot \theta + 1 \] we'll start by expressing everything in terms of sine and cosine. ### Step 1: Express in Terms of Sine and Cosine Recall that: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \cot \theta = \frac{\cos \theta}{\sin \theta} \] Substitute these into the left-hand side (LHS) of the equation: \[ \text{LHS} = \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}} \] ### Step 2: Simplify the Denominators Simplify each denominator: \[ 1 - \frac{\cos \theta}{\sin \theta} = \frac{\sin \theta - \cos \theta}{\sin \theta} \] \[ 1 - \frac{\sin \theta}{\cos \theta} = \frac{\cos \theta - \sin \theta}{\cos \theta} = -\frac{\sin \theta - \cos \theta}{\cos \theta} \] ### Step 3: Rewrite the LHS Substitute the simplified denominators back into the LHS: \[ \text{LHS} = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{-\frac{\sin \theta - \cos \theta}{\cos \theta}} \] Simplify the fractions: \[ \text{LHS} = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)} \] Combine the terms over a common denominator: \[ \text{LHS} = \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)} \] ### Step 4: Factor the Numerator Notice that: \[ \sin^3 \theta - \cos^3 \theta = (\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta) \] Since \(\sin^2 \theta + \cos^2 \theta = 1\), this simplifies to: \[ \sin^3 \theta - \cos^3 \theta = (\sin \theta - \cos \theta)(1 + \sin \theta \cos \theta) \] ### Step 5: Simplify the Expression Substitute back into the LHS: \[ \text{LHS} = \frac{(\sin \theta - \cos \theta)(1 + \sin \theta \cos \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)} \] Cancel out \((\sin \theta - \cos \theta)\): \[ \text{LHS} = \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} + 1 \] ### Step 6: Relate to the Right-Hand Side (RHS) Recognize that: \[ \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} \] Thus: \[ \text{LHS} = \frac{1}{\sin \theta \cos \theta} + 1 = \tan \theta + \cot \theta + 1 = \text{RHS} \] ### Conclusion The identity holds true: \[ \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = \tan \theta + \cot \theta + 1 \] **Answer:** After expanding, both sides reduce to 1/(sinθ cosθ) + 1. Thus, the equation holds: tanθ/(1 – cotθ) + cotθ/(1 – tanθ) = tanθ + cotθ + 1