$tan^{2}\, C\, -\, sec^{2}\, B\, +\, 2 = ?$

Given, $AB = AC = 10$ and $BC = 18$ cm

$\cos B = \dfrac{AB^2 + BC^2 - AC^2}{2 AB \times BC}$

$\cos B = \dfrac{10^2 + 18^2 - 10^2}{2\times10 \times 18}$

$\cos B = \dfrac{9}{10}$

$\cos B = \dfrac{B}{H} = \dfrac{9}{10}$

Now, $\cos C =\dfrac{AC^2 + BC^2 - AB^2}{2 AC \times BC}$

$\cos C = \dfrac{10^2 + 18^2 - 10^2}{2\times10 \times 18}$

$\cos C = \dfrac{9}{10}$

Using Pythagoras Theorem,

$H^2 = P^2 + B^2$

$10^2 = P^2 + 9^2$

$P = \sqrt{19}$

$\tan C = \dfrac{P}{B} = \dfrac{\sqrt{19}}{9}$

Thus, $\tan^2 C - \sec^2 B + 2 = (\dfrac{\sqrt{19}}{9})^2 - (\dfrac{10}{9})^2 + 2$

$\tan^2 C - \sec^2 B= \dfrac{19}{81} - \dfrac{100}{81} + 2$

$\tan^2 C - \sec^2 C + 2= \dfrac{-81}{81} + 2= 1$