Question

Prove that: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$

Hint:

Expressing the entire expression in terms of $\tan\theta$ simplifies the algebra considerably compared to working with sines and cosines, which often leads to more complicated fraction handling.

Answer 1

Step 1: Convert everything to sine and cosine from the start, instead of working purely in tan and cot.
Let $s = \sin\theta$, $c = \cos\theta$, so $\tan\theta = \frac{s}{c}$ and $\cot\theta = \frac{c}{s}$. Then $1-\cot\theta = \frac{s-c}{s}$ and $1-\tan\theta = \frac{c-s}{c}$.
Step 2: Rewrite each term of the LHS. \[ \frac{\tan\theta}{1-\cot\theta} = \frac{s/c}{(s-c)/s} = \frac{s^2}{c(s-c)}, \qquad \frac{\cot\theta}{1-\tan\theta} = \frac{c/s}{(c-s)/c} = \frac{c^2}{s(c-s)} = -\frac{c^2}{s(s-c)} \]
Step 3: Combine over a common denominator sc(s-c). \[ \text{LHS} = \frac{s^2}{c(s-c)} - \frac{c^2}{s(s-c)} = \frac{s^3-c^3}{sc(s-c)} \] Using $s^3-c^3=(s-c)(s^2+sc+c^2)$ and cancelling $(s-c)$: \[ \text{LHS} = \frac{s^2+sc+c^2}{sc} \]
Step 4: Split the fraction to match the RHS. \[ \text{LHS} = \frac{s}{c} + 1 + \frac{c}{s} = \tan\theta + 1 + \cot\theta = 1+\tan\theta+\cot\theta = \text{RHS} \]
Since the Left-Hand Side simplifies exactly to the Right-Hand Side, the identity is proved.