Q4(iii):

Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (iii) $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{cosec } \theta$ [Hint : Write the expression in terms of $\sin \theta$ and $\cos \theta$]

Given: The trigonometric identity $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}$, where $\theta$ is an acute angle.

To Prove: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{cosec } \theta$

Step 1: Expressing the identity in terms of $\sin \theta$ and $\cos \theta$

We know the fundamental trigonometric identities: $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$. Substituting these into the Left Hand Side (LHS):

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: Simplifying the denominators

Find a common denominator for the terms in the denominators of the fractions:

LHS = $\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}$

Step 3: Performing division of fractions

Using the rule $\frac{a/b}{c/d} = \frac{a}{b} \times \frac{d}{c}$:

LHS = $\left( \frac{\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\sin \theta - \cos \theta} \right) + \left( \frac{\cos \theta}{\sin \theta} \times \frac{\cos \theta}{\cos \theta - \sin \theta} \right)$

LHS = $\frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}$

Step 4: Aligning the denominators

To combine the fractions, we need a common denominator. Note that $(\cos \theta - \sin \theta) = -(\sin \theta - \cos \theta)$.

LHS = $\frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}$

Step 5: Combining the fractions

The common denominator is $\sin \theta \cos \theta (\sin \theta - \cos \theta)$:

LHS = $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$

Step 6: Applying the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Here, $a = \sin \theta$ and $b = \cos \theta$:

LHS = $\frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$

Canceling the common term $(\sin \theta - \cos \theta)$:

LHS = $\frac{\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta}{\sin \theta \cos \theta}$

Step 7: Final simplification

Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$:

LHS = $\frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}$

LHS = $\frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta}$

LHS = $\left( \frac{1}{\cos \theta} \right) \left( \frac{1}{\sin \theta} \right) + 1$

LHS = $\sec \theta \text{cosec } \theta + 1$

Conclusion: Since LHS = RHS, the identity is proven.

Final Answer: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{cosec } \theta$