Question

From the given figure, find ${\sec }^{2}B-{\tan }^{2}B$

• A. $94$
• B. $1$
• C. $54$
• D. $2$

Answer 1

Answer: (B)

$1$

In $△ABC$, $AD\perp BC$ $AB=13$, $BD=5$, $DC=16$
Now, In $△ABD$,
$A{B}^2=A{D}^2+B{D}^2$
${13}^2=A{D}^2+5^2$
$A{D}^2=144$
$AD=12$
Now, in $△ADC$,
$A{C}^2=A{D}^2+C{D}^2$
$A{C}^2={12}^2+{16}^2$
$A{C}^2=144+256$
$AC=20$ cm
Now, ${\sec }^{2}B-{\tan }^{2}B=(\frac{H}{B}{)}^2-(\frac{P}{B}{)}^2=(\frac{AB}{BD}{)}^2-(\frac{AD}{BD}{)}^2$
= $(\frac{13}{5}{)}^2-(\frac{12}{5}{)}^2$
= $\frac{169}{25}-\frac{144}{25}$
= $\frac{25}{25}$
= $1$