If $ 8 \tan A=15 $, find $ \sin A-\cos A $

Given:

\( 8 \tan A=15 \).

To do:

We have to find \( \sin A-\cos A \).

Solution:  

Let, in a triangle $ABC$ right-angled at $B$, $8\ tan\ A = 15$.

This implies,

$tan\ A=\frac{15}{8}$

We know that,

In a right-angled triangle $ABC$ with right angle at $B$,

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

By trigonometric ratios definitions,

$sin\ A=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$

$cos\ A=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$

$tan\ A=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$

Here,

$AC^2=AB^2+BC^2$

$\Rightarrow AC^2=(8)^2+(15)^2$

$\Rightarrow AC^2=64+225$

$\Rightarrow AC=\sqrt{289}=17$

Therefore,

$sin\ A=\frac{BC}{AC}=\frac{15}{17}$

$cos\ A=\frac{AB}{AC}=\frac{8}{17}$

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

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

$=\frac{7}{17}$
 The value of $\sin A-\cos A$ is $\frac{7}{17}$.

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Updated on: 10-Oct-2022

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