Question

In $△ABC$, prove that $\frac{\cos 2A}{a^2}-\frac{\cos 2C}{c^2}=\frac{1}{a^2}-\frac{1}{c^2}$.

Answer 1

$LHS=\frac{cos2A}{a^2}-\frac{cos2C}{c^2}=\frac{2co{s}^{2}A}{a^2}-\frac{1}{a^2}-\frac{2co{s}^{2}c}{c^2}+\frac{1}{c^2}$ $[\because cos2\theta =2co{s}^2\theta -1]$

[using cosine rule $cosA=\frac{b^2+c^2-a^2}{2bc}$ , $cosC=\frac{a^2+b^2-c^2}{ab}]$

$=\frac{(b^2+c^2-a^2{)}^2}{2a^2{b}^2{c}^2}-\frac{(a^2+b^2-c^2{)}^2}{2a^2{b}^2{c}^2}-\frac{1}{a^2}+\frac{1}{c^2}$

$=\frac{b^4+c^4+a^4+2b^2{c}^2-2c^2{a}^2-2a^2{b}^2-(a^4+b^4+c^4+2a^2{b}^2-2b^2{c}^2-2a^2{c}^2)}{2a^2{b}^2{c}^2}-\frac{1}{a^2}+\frac{1}{c^2}$

$=\frac{4b^2{c}^2-4a^2{b}^2}{2a^2{b}^2{c}^2}-\frac{1}{a^2}+\frac{1}{c^2}$

$=\frac{2}{a^2}-\frac{2}{c^2}-\frac{1}{a^2}+\frac{1}{c^2}=\frac{1}{a^2}-\frac{1}{c^2}=RHS$. (Proved)