Question

In the given figure , $\frac{AB}{AC}$ = $\frac{12}{13}$. Find the value of $se{c}^{2}C$ – $ta{n}^{2}C$

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Answer 1

$se{c}^{2}C$ = $\frac{169}{25}$ $ta{n}^{2}C$ = $\frac{144}{25}$

$se{c}^{2}C$ - $ta{n}^{2}C$ = $\frac{(169-144)}{25}$ = 1