In given figure we have right angled triangle,
with
Hypotenuse=5
perpendicular=3
Base=4
Apply the trignometric expression for tanΘ
[tex]\tan \emptyset=\frac{Perpendicular}{\text{base}}[/tex]
Substitute the value of Perendicular and base,
[tex]\tan \emptyset=\frac{3}{4}[/tex]
Multiply bothe side by arc(tan)
[tex]\begin{gathered} \arctan (\tan \emptyset)=\arctan \frac{3}{4} \\ \emptyset=\arctan \frac{3}{4} \end{gathered}[/tex]
The trignometric expression for tan2Θ is:
[tex]\tan 2\emptyset=\frac{2\tan \emptyset}{1-\tan ^2\emptyset}[/tex]
Put the value of Θ in the tan2Θ
[tex]\begin{gathered} \tan 2\emptyset=\frac{2\tan (\arctan \frac{3}{4)}}{1-\tan ^2(\arctan \frac{3}{4})} \\ \tan 2\emptyset=\frac{2\frac{3}{4}}{1-(\frac{3}{4})^2} \\ \tan 2\emptyset=\frac{\frac{6}{4}}{1-\frac{9}{16}} \\ \tan 2\emptyset=\frac{\frac{6}{4}}{\frac{16-9}{16}} \\ \tan 2\emptyset=\frac{\frac{6}{4}}{\frac{7}{16}} \\ \tan 2\emptyset=\frac{6}{4}\times\frac{16}{7} \\ \tan 2\emptyset=\frac{6\times4}{7} \\ \tan 2\emptyset=\frac{24}{7} \end{gathered}[/tex]
ANSWER : (D). 24/7
