Question

In triangle $ABC$, $AB=AC=15$ cm and $BC=18$ cm. Find:

${\tan }^{2}B-{\sec }^{2}B+2$

Answer 1

$AB=AC=15$, $BC=18$
Using cosine rule,
$A{C}^2=B{C}^2+A{B}^2-2\left(AB\right)\left(BC\right)\cos B$
${15}^2={18}^2+{15}^2-2\left(15\right)\left(18\right)\cos B$
$\cos B=\frac{9}{15}$
$\cos B=\frac{3}{5}$
$\cos B=\frac{B}{H}=\frac{3}{5}$
Now, using Pythagoras theorem.
$H^2=P^2+B^2$
$5^2=P^2+3^2$
$P=4$
${\tan }^{2}B-{\sec }^{2}B+2$
= $(\frac{P}{B}{)}^2-(\frac{H}{B}{)}^2+2$
= $(\frac{4}{3}{)}^2-(\frac{5}{3}{)}^2+2$
= $\frac{16-25+18}{9}$
= $1$