Question

if 13 Sin A =12 find sec A-tan A
​

Answer 1

Step-by-step explanation:

if sin a =12 then p/h= 12/13 now we have given p and h .

so by Pythagoras theorem p

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

169=144+ base Ka square

then b= 5

sec a - tan a =131/65

Answer 2

The value of $\sec A-\tan A$ is equal to $\dfrac{1}{5}$.

Step-by-step explanation:

We have,

$13\sin A =12$

∴ $\sin A =\dfrac{12}{13}$

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

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

Where p = perpendicular and h = hypotaneous

∴ 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

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

∴ $\sec A-\tan A$ = $\dfrac{13}{5}-\dfrac{12}{5}$

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

= $\dfrac{1}{5}$

∴ The value of $\sec A-\tan A$ = $\dfrac{1}{5}$

Thus, the value of $\sec A-\tan A$ is equal to $\dfrac{1}{5}$.