Q.Prove that: $(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1$.

The left side is a difference-of-squares pattern that collapses to exactly the derived identity connecting $\sec\theta$ and $\tan\theta$.

Step 1 — recognise the difference-of-squares pattern.

Step 2 — apply the derived identity $1+\tan^2\theta=\sec^2\theta$, rearranged as $\sec^2\theta-\tan^2\theta=1$.

$\sec^2\theta-\tan^2\theta = 1$ …

Alternatively, rewrite $\sec\theta$ and $\tan\theta$ in terms of sine and cosine: $\sec\theta-\tan\theta = \dfrac{1-\sin\theta}{\cos\theta}$ and $\sec\theta+\tan\theta = \dfrac{1+\sin\theta}{\cos\theta}$; their product is $\dfrac{1-\sin^2\theta …

Expanding the product term by term (getting $\sec^2\theta - \sec\theta\tan\theta+\sec\theta\tan\theta-\tan^2\theta$) and then mishandling the two middle terms is a common slip — recognising the expr …