Question

In an isosceles triangle $ABC$; $AB=AC=10$ cm and $BC=18$ cm. Find the value of :
${\tan }^{2}C-{\sec }^{2}B+2$

Answer 1

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

$\cos B=\frac{A{B}^2+B{C}^2-A{C}^2}{2AB\times BC}$

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

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

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

Now, $\cos C=\frac{A{C}^2+B{C}^2-A{B}^2}{2AC\times BC}$

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

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

Using Pythagoras Theorem,

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

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

$P=\sqrt{19}$

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

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

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

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