If $a\sec heta + b an heta = m$ and $b\sec heta + a an heta = n$, prove that $a^2 + n^2 = b^2 + m^2$

Question

If $a\sec	heta + b	an	heta = m$ and $b\sec	heta + a	an	heta = n$, prove that $a^2 + n^2 = b^2 + m^2$

Accepted Answer

## Model Answer **Given:** $a\sec	heta + b	an	heta = m$ … (1) and $b\sec	heta + a	an	heta = n$ … (2) Squaring (1): $a^2\sec^2	heta + b^2	an^2	heta + 2ab\sec	heta	an	heta = m^2$ Squaring (2): $b^2\sec^2	heta + a^2	an^2	heta + 2ab\sec	heta	an	heta = n^2$ Subtracting (2)² from (1)²: $$m^2 - n^2 = (a^2 - b^2)\sec^2	heta - (a^2 - b^2)	an^2	heta$$ $$= (a^2 - b^2)(\sec^2	heta - 	an^2	heta)$$ Using the identity $\sec^2	heta - 	an^2	heta = 1$: $$m^2 - n^2 = a^2 - b^2$$ $$	herefore\quad a^2 + n^2 = b^2 + m^2 \qquad 	extbf{(Proved)}$$ *Source: Chapter 8, Section 8.4 — Trigonometric Identities* --- ## Explanation - The key step is **squaring both equations and subtracting** to eliminate the cross-term $2ab\sec	heta	an	heta$. - The identity $\sec^2	heta - 	an^2	heta = 1$ (from Section 8.4) is what collapses the expression. - Examiners award 1 mark for correct squaring/subtraction and 1 mark for applying the identity and reaching the conclusion. Do not skip steps…

Mathematics — Standard (041) — CBSE Class 10 board question