Question

if 13 sin A=12

find sec A-tanA​

Answer 1

Refer the picture for the solution

Hope it helps

Answer 2

$\sec A$ - $\tan A$ $=\dfrac{1}{5}$

Step-by-step explanation:

We have,

13 $\sin A$ = 12

To find, $\sec A$ - $\tan A$ = ?

∴ 13 $\sin A$ = 12

⇒ $\sin A$ = $\dfrac{12}{13}$ = $\dfrac{p}{h}$

p = 12 and h = 13

Where, p = perpendicular and h = hypotaneous

By Pythagoras Theorem,

$h^2$ = $p^2$ + $b^2$

$b^2$ = $h^2$ - $p^2$

Base, b = $\sqrt{h^2-p^2}$

= $\sqrt{13^2-12^2}$ = $\sqrt{169-144}$

= $\sqrt{25}$ = 5

∴ $\sec A$ = $\dfrac{h}{b}$ = $\dfrac{13}{5}$ and

$\tan A$ = $\dfrac{p}{b}$ = $\dfrac{12}{5}$

∴ $\sec A$ - $\tan A$

= $\dfrac{13}{5}$ - $\dfrac{12}{5}$

= $\dfrac{13-12}{5}=\dfrac{1}{5}$

Thus, $\sec A$ - $\tan A$ $=\dfrac{1}{5}$