Question

If $4\sin \theta =3\cos \theta$, find $\frac{{\sec }^2\theta }{4(1-{\tan }^2\theta )}$

• A. $\frac{25}{16}$
• B. $\frac{25}{28}$
• C. $\frac{1}{4}$
• D. $\frac{5}{6}$

Answer 1

Answer: (B)

$\frac{25}{28}$

$4\sin \theta =3\cos \theta$

$\Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{3}{4}$

$\Rightarrow \tan \theta =\frac{3}{4}$

$\Rightarrow \frac{{\sec }^2\theta }{4(1-{\tan }^2\theta )}=\frac{1+{\tan }^2\theta }{4(1-{\tan }^2\theta )}=\frac{1+{\left(\frac{3}{4}\right)}^2}{4(1-{\left(\frac{3}{4}\right)}^2)}=\frac{1+\frac{9}{16}}{4(1-\frac{9}{16})}$

$\Rightarrow \frac{\frac{25}{16}}{4\times \frac{7}{16}}=\frac{25}{16}\times \frac{4}{7}=\frac{25}{28}$

$\therefore \frac{{\sec }^2\theta }{4(1-{\tan }^2\theta )}=\frac{25}{28}$

Hence, the answer is $\frac{25}{28}.$