Question

In $△ABC,\angle B={90}^{\circ }$ and $sinA=\frac{5}{13}$, then $se{c}^{2}A-ta{n}^{2}A=$ __________________

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

The correct option is C

1

$sinA=\frac{5}{13}$

using $5-12-13$ Pythagoras triplet we get,

$secA=\frac{13}{12}$ and $tanA=\frac{5}{12}$

$se{c}^{2}A-ta{n}^{2}A={\left(\frac{13}{12}\right)}^2-{\left(\frac{5}{12}\right)}^2$

=$\frac{{13}^2-5^2}{{12}^2}$

=$\frac{{12}^2}{{12}^2}$

=1

Hence (C)