Question

Prove that : $(1 + \cot \theta - \csc \theta)(1 + \tan \theta + \sec \theta) = 2$

Hint:

Using the difference of squares identity $(a-b)(a+b) = a^2 - b^2$ is an extremely elegant and fast shortcut for simplifying products of trigonometric sums!

Answer 1

Step 1: Convert every ratio to sine and cosine, and multiply out fully.
$(1+\cot\theta-\csc\theta)(1+\tan\theta+\sec\theta)$, when every term is multiplied out, gives nine products: $1, \tan\theta, \sec\theta, \cot\theta, \cot\theta\tan\theta, \cot\theta\sec\theta, -\csc\theta, -\csc\theta\tan\theta, -\csc\theta\sec\theta$.
Step 2: Simplify the mixed products individually.
$\cot\theta \tan\theta = 1$. Also $\cot\theta \sec\theta = \frac{\cos\theta}{\sin\theta}\cdot\frac{1}{\cos\theta} = \csc\theta$, and $\csc\theta\tan\theta = \frac{1}{\sin\theta}\cdot\frac{\sin\theta}{\cos\theta}=\sec\theta$.
Step 3: Add everything together and watch terms cancel.
Adding all nine terms: $1+\tan\theta+\sec\theta+\cot\theta+1+\csc\theta-\csc\theta-\sec\theta-\csc\theta\sec\theta$. The $\sec\theta$ terms cancel and the $\csc\theta$ terms cancel, leaving $2+\tan\theta+\cot\theta-\csc\theta\sec\theta$.
Step 4: Simplify the remaining trig terms.
$\tan\theta+\cot\theta = \frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}$, which is exactly $\csc\theta\sec\theta$. So $\tan\theta+\cot\theta-\csc\theta\sec\theta = 0$, leaving just $2$.
This confirms $(1+\cot\theta-\csc\theta)(1+\tan\theta+\sec\theta) = 2$, as required.
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