Let’s solve the given problem step-by-step:
1. Given: [tex]\(\sin A = \frac{3}{5}\)[/tex]
2. Find [tex]\(\cos A\)[/tex]:
We know that [tex]\(\sin^2 A + \cos^2 A = 1\)[/tex]. Therefore, we can use the Pythagorean identity to find [tex]\(\cos A\)[/tex]:
[tex]\[
\sin^2 A + \cos^2 A = 1 \\
\left(\frac{3}{5}\right)^2 + \cos^2 A = 1 \\
\frac{9}{25} + \cos^2 A = 1 \\
\cos^2 A = 1 - \frac{9}{25} \\
\cos^2 A = \frac{25}{25} - \frac{9}{25} \\
\cos^2 A = \frac{16}{25} \\
\cos A = \sqrt{\frac{16}{25}} \\
\cos A = \frac{4}{5}
\][/tex]
We chose the positive root since cosine in the first quadrant (where sine is positive) is also positive.
3. Calculate [tex]\(\tan A\)[/tex]:
We know that [tex]\(\tan A = \frac{\sin A}{\cos A}\)[/tex]:
[tex]\[
\tan A = \frac{\frac{3}{5}}{\frac{4}{5}} \\
\tan A = \frac{3}{5} \times \frac{5}{4} \\
\tan A = \frac{3}{4}
\][/tex]
4. Use the double-angle formula for [tex]\(\tan(2A)\)[/tex]:
The double-angle formula for tangent is given by:
[tex]\[
\tan(2A) = \frac{2 \tan A}{1 - \tan^2 A}
\][/tex]
Substitute [tex]\(\tan A = \frac{3}{4}\)[/tex]:
[tex]\[
\tan(2A) = \frac{2 \times \frac{3}{4}}{1 - \left(\frac{3}{4}\right)^2} \\
\tan(2A) = \frac{\frac{6}{4}}{1 - \frac{9}{16}} \\
\tan(2A) = \frac{\frac{6}{4}}{\frac{16}{16} - \frac{9}{16}} \\
\tan(2A) = \frac{\frac{6}{4}}{\frac{7}{16}} \\
\tan(2A) = \frac{6}{4} \times \frac{16}{7} \\
\tan(2A) = \frac{6 \times 16}{4 \times 7} \\
\tan(2A) = \frac{96}{28} \\
\tan(2A) = \frac{24}{7}
\][/tex]
5. Determine [tex]\(-\tan(2A)\)[/tex]:
Since the problem asks for [tex]\(-\tan(2A)\)[/tex]:
[tex]\[
-\tan(2A) = -\frac{24}{7}
\][/tex]
Thus, we have shown that if [tex]\(\sin A = \frac{3}{5}\)[/tex], then [tex]\(-\tan 2A = -\frac{24}{7}\)[/tex].
