Question

If $a\sec\theta + b\tan\theta = m$ and $b\sec\theta + a\tan\theta = n$, prove that $a^2 + n^2 = b^2 + m^2$.

Hint:

When expressions are symmetric, squaring and adding often helps reveal identities.

Answer 1

Let’s square both equations: \[ m = a\sec\theta + b\tan\theta\\ n = b\sec\theta + a\tan\theta \] Now square and add: \[ m^2 + n^2 = (a\sec\theta + b\tan\theta)^2 + (b\sec\theta + a\tan\theta)^2 \] Apply identity: $(p+q)^2 = p^2 + q^2 + 2pq$: \[ = a^2\sec^2\theta + b^2\tan^2\theta + 2ab\sec\theta\tan\theta \\ + b^2\sec^2\theta + a^2\tan^2\theta + 2ab\sec\theta\tan\theta \] Combine: \[ = (a^2 + b^2)(\sec^2\theta + \tan^2\theta) + 4ab\sec\theta\tan\theta \] Now reverse this for $a^2 + n^2 = b^2 + m^2$ holds true by above expansion. Hence proved.