Question

if 13 sin A= 12

find sec A - tan A​

Answer 1

$\textbf{Given:}$

$sinA=\displaystyle\frac{12}{13}$

$\text{Using,}$

$\boxed{\bf\,cos^2\theta=1-sin^2\theta}$

$\implies\,cos^2A=1-sin^2A$

$\implies\,cos^2A=1-\displaystyle\frac{144}{169}$

$\implies\,cos^2A=\displaystyle\frac{25}{169}$

$\implies\,cosA=\displaystyle\frac{5}{13}$

$\text{Now,}$

$sec\,A-tan\,A$

$=\displaystyle\frac{1}{cos\,A}-\frac{sin\,A}{cos\,A}$

$=\displaystyle\frac{1-sin\,A}{cos\,A}$

$=\displaystyle\frac{1-(\frac{12}{13})}{\frac{5}{13}}$

$=\displaystyle\frac{\frac{1}{13}}{\frac{5}{13}}$

$=\displaystyle\frac{1}{5}$

$\therefore\boxed{\bf\,sec\,A-tan\,A=\frac{1}{5}}$

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Answer 2

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

Step-by-step explanation:

We have,

$13\sin A=12\\\\\sin A=\dfrac{12}{13}$

It is required to find the value of $\sec A-\tan A$. Firstly, we will simplify it. As we know :

$\sec A=\dfrac{1}{\cos A}\\\\\tan A=\dfrac{\sin A}{\cos A}$

So,

$=\sec A-\tan A\\\\=\dfrac{1}{\cos A}-\dfrac{\sin A}{\cos A}\\\\=\dfrac{1-\sin A}{\cos A}$............(1)

We know that :

$\sin^2 A+\cos^2 A=1\\\\\cos^2 A=1-\sin^2 A\\\\\cos A=\sqrt{1-\sin^2 A} \\\\\cos A=\sqrt{1-(\dfrac{12}{13})^2}\\\cos A=\dfrac{5}{13}$

Equation (1) becomes :

$\sec A-\tan A=\dfrac{1-\sin A}{\cos A}\\\\=\dfrac{1-(12/13)}{(5/13)}\\\\=\dfrac{1}{5}$

So, the value of $\sec A-\tan A=\dfrac{1}{5}$.

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