Question

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

Answer 1

R.E.F image

Given

$c=b=10cm$

$a=18cm$

$CUSB=\frac{c^2+a^2-b^2}{2ca}$

$=\frac{100+364-100}{2\times 10\times 18}$

$CUSB=\frac{18}{20}$

$\therefore CUSB=\frac{9}{10}$

$\therefore CUSC=\frac{9}{10}$

$\sin B=\frac{\sqrt{19}}{10}$

i) ${\sin }^{2}B+co{s}^{2}C=\frac{19}{100}+\frac{81}{100}+\frac{100}{100}=1$

$\therefore {\sin }^{2}B+{\cos }^2=1$

ii) ${\tan }^{2}C-{\sec }^{2}B+2={\left(\frac{\sqrt{19}}{4}\right)}^2-{\left(\frac{10}{9}\right)}^2+2$

$=\frac{19}{81}-\frac{100}{81}+2$

$=-\frac{81}{81}+2$

$=2-1=1$

$\Rightarrow {\tan }^{2}C-{\sec }^{2}B+2=1$