Question

If tan A = 5/12 , find the value of  cosA + sinA/ cosA – sinA​

Answer 1

$\mathrm{\frac{cosA+sinA}{cosA - sinA }} = \frac{17}{7}$

Given:

tan A = $\frac{5}{12}$

So, according to what we are given

Let:

A $\triangle$ABC right angled at B

The sides AB & BC to be of length $5x$ & $12x$.

Now, what we actually need

To find:

• $\huge{\mathrm{\frac{cosA+sinA}{cosA - sinA }}}$ = ?

Solution:

$\because$ ABC is a right angled at B

So, according to Pythagoras theorem,

$\mathrm{AB^2 + BC^2 = AC^2}$

$\implies \mathrm{(5x)^2 + (12x)^2 = AC^2}$

$\implies \mathrm{AC^2 = 25x^2 + 144x^2 }$

$\implies \mathrm{AC^2 = 169x^2 }$

$\implies \mathrm{AC = \sqrt{169x^2}}$

$\implies \mathrm{AC = 13 x}$

This, way we found third side of $\triangle$ABC i.e also called Hypotenuse

So, we are now in position to find other possible trigonometric ratios for this triangle...

$\large{\mathrm{sinA = \frac{5}{13}}}$ &

$\large{\mathrm{cosA = \frac{12}{13}}}$

and finally, we are now in state to solve this expression

$\huge{\mathrm{\frac{cosA+sinA}{cosA - sinA}}}$

$\huge{= \frac{(\frac{12}{13}) + (\frac{5}{13})}{(\frac{12}{13}) - (\frac{5}{13})}}$

$\huge{= \frac{\frac{12+5}{13}}{\frac{12-5}{13}}}$

$\boxed{\green{\huge{= \frac{17}{7}}}}$

and, that's how I evaluated $\mathrm{\frac{cosA+sinA}{cosA - sinA}}$

Check, if my response is enough to clear your doubt,

Happy Learning! :D