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

Given:
\[a\sec\theta + b\tan\theta = m \quad \text{...(1)}\\b\sec\theta + a\tan\theta = n \quad \text{...(2)}\]
We aim to prove: \[a^2 + n^2 = b^2 + m^2\]

Step 1: Square the equations
Square equation (1): \[(a\sec\theta + b\tan\theta)^2 = m^2\\\Rightarrow a^2\sec^2\theta + 2ab\sec\theta\tan\theta + b^2\tan^2\theta = m^2 \quad \text{...(3)}\]
Square equation (2): \[(b\sec\theta + a\tan\theta)^2 = n^2\\\Rightarrow b^2\sec^2\theta + 2ab\sec\theta\tan\theta + a^2\tan^2\theta = n^2 \quad \text{...(4)}\]
Step 2: Sum equations (3) and (4)
Add the left-hand sides (LHS) and right-hand sides (RHS) of (3) and (4): \[a^2\sec^2\theta + b^2\tan^2\theta + b^2\sec^2\theta + a^2\tan^2\theta + 2(2ab\sec\theta\tan\theta) = m^2 + n^2\] Group like terms: \[a^2(\sec^2\theta + \tan^2\theta) + b^2(\sec^2\theta + \tan^2\theta) + 4ab\sec\theta\tan\theta = m^2 + n^2\] Factor: \[(a^2 + b^2)(\sec^2\theta + \tan^2\theta) + 4ab\sec\theta\tan\theta = m^2 + n^2 \quad \text{...(5)}\]
Step 3: Subtract equation (3) from (4)
Equation (4) minus Equation (3): \[[b^2\sec^2\theta + a^2\tan^2\theta + 2ab\sec\theta\tan\theta] - [a^2\sec^2\theta + b^2\tan^2\theta + 2ab\sec\theta\tan\theta] = n^2 - m^2\] Simplify: \[(b^2 - a^2)(\sec^2\theta - \tan^2\theta) = n^2 - m^2 \quad \text{...(6)}\]
Step 4: Apply trigonometric identity
Using the known identity: \[\sec^2\theta - \tan^2\theta = 1\Rightarrow (b^2 - a^2)(1) = n^2 - m^2\Rightarrow b^2 - a^2 = n^2 - m^2\]
Rearrange terms: \[a^2 + n^2 = b^2 + m^2\]

Final Answer:
\[\boxed{a^2 + n^2 = b^2 + m^2}\quad \text{(Proved)}\]