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 are to prove: \[ a^2 + n^2 = b^2 + m^2 \]

Step 1: Square both 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: Add equations (3) and (4)
Add LHS and 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: Now subtract equation (3) from (4)
Equation (4) - 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: Use identity
We know: \[ \sec^2\theta - \tan^2\theta = 1 \Rightarrow (b^2 - a^2)(1) = n^2 - m^2 \Rightarrow b^2 - a^2 = n^2 - m^2 \]
Bring terms to one side: \[ a^2 + n^2 = b^2 + m^2 \]

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