Prove the following:
If $ \tan \mathrm{A}=\frac{3}{4} $, then $ \sin \mathrm{A} \cos \mathrm{A}=\frac{12}{25} $

Given:

\( \tan \mathrm{A}=\frac{3}{4} \)

To do:

We have to prove that \( \sin \mathrm{A} \cos \mathrm{A}=\frac{12}{25} \).

Solution:

Let, in a triangle $ABC$ right-angled at $B$, $\tan\ A=\frac{3}{4}$.

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,

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

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

$sec\ \theta=\frac{Hypotenuse}{Adjacent}=\frac{AC}{AB}$

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

Here,

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

$\Rightarrow (AC)^2=(4)^2+(3)^2$

$\Rightarrow AC^2=16+9$

$\Rightarrow AC=\sqrt{25}=5$

Therefore,

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

$=\frac{3}{5}$

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

$=\frac{4}{5}$

Therefore,

$\sin \mathrm{A} \cos \mathrm{A}=\frac{3}{5} \times \frac{4}{5}$

$=\frac{3\times4}{5\times5}$

$=\frac{12}{25}$

Hence proved.

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Updated on: 10-Oct-2022

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