If 8 tan A = 15, find sin A − cos A.

Solution:

Given:

$8 \tan A=15$

Therefore,

$\tan A=\frac{15}{8}$....(1)

To find:

$\sin A-\cos A$

Now we know $\tan \theta$ is defined as follows

$\tan A=\frac{\text { Perpendicular side opposite to } \angle A}{\text { Base side adjacent to } \angle A}$.....(4)

Now by comparing equation (1) and (2)

We get

Perpendicular side opposite to $\angle A=15$

Base side adjacent to $\angle A=8$

Therefore triangle representing angle A is as shown below

Side AC is unknown and can be found using Pythagoras theorem

Therefore,

$A C^{2}=A B^{2}+B C^{2}$

Now by substituting the value of known sides from figure (a)

We get,

$A C^{2}=15^{2}+8^{2}$

$=225+64$

 

$=289$

Now by taking square root on both sides

We get,

$A C=\sqrt{289}$

$=17$

Therefore Hypotenuse side AC = 17 …… (3)

Now we know, $\sin A$ is defined as follows

$\sin A=\frac{\text { Perpendicular side opposite to } \angle A}{\text { Hypotenuse }}$

Therefore from figure (a) and equation (3)

We get,

$\sin A=\frac{B C}{A C}$

$=\frac{15}{17}$

$\sin A=\frac{15}{17}$...(4)

Now we know, $\cos A$ is defined as follows

$\cos A=\frac{\text { Base side adjacent to } \angle A}{\text { Hypotenuse }}$

Therefore from figure (a) and equation (3)

We get,

$\cos A=\frac{A B}{A C}$

$=\frac{8}{17}$

$\cos A=\frac{8}{17}$....(5)

Now we need to find the value of expression $\sin A-\cos A$

Therefore by substituting the value of $\sin A$ and $\cos A$ from equation (4) and (5) respectively, we get,

$\sin A-\cos A=\frac{15}{17}-\frac{8}{17}$

$=\frac{15-8}{17}$

$=\frac{7}{17}$

Hence $\sin A-\cos A=\frac{7}{17}$