Question

If $\sec A-\tan A=\frac{1}{4}$, then $\sin 2A$ is equal to:

• A. $\frac{230}{289}$
• B. $\frac{240}{289}$
• C. $\frac{250}{289}$
• D. $\frac{260}{289}$

Answer 1

The correct option is B $\frac{240}{289}$

$\sec A-\tan A=\frac{1}{4}.............\left(1\right)$

we know that ${\sec }^{2}A-{\tan }^{2}A=1\Rightarrow \sec A+\tan A=\frac{1}{\sec A-\tan A}\Rightarrow \sec A+\tan A=\mathrm{4...............}\left(2\right)$

Now,

$eq\left(2\right)-eq\left(1\right)\to 2\tan A=4-\frac{1}{4}=\frac{15}{4}\Rightarrow \tan A=\frac{15}{8}$

we know that $\sin 2A=\frac{2\tan A}{1+{\tan }^{2}A}\Rightarrow \sin 2A=\frac{2\times \frac{15}{8}}{1+(\frac{15}{8}{)}^2}=\frac{240}{289}$

Therefore, Answer is $\frac{240}{289}$