Question

If $\sin A=\frac{3}{5}$ and A is not in the first quadrant then find $\frac{\cos A+\sin 2A}{\tan A+\sec A}.$

• A. $\frac{16}{25}$
• B. $\frac{17}{25}$
• C. $\frac{22}{25}$
• D. $\frac{19}{25}$

Answer 1

Answer: (C)

$\frac{22}{25}$

$\cos A=\sqrt{1-{\sin }^{2}A}=\sqrt{1-{\left(\frac{3}{5}\right)}^2}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=-\frac{4}{5}$ (A is in the second Quadrant)

Now, $\frac{\cos A+\sin 2A}{\tan A+\sec A}=\frac{\cos A+2\sin A\cos A}{\frac{\sin A}{\cos A}+\frac{1}{\cos A}}=\frac{{\cos }^{2}A(1+2\sin A)}{1+\sin A}=\frac{(-\frac{4}{5}{)}^2[1+2\ast \frac{3}{5}]}{1+\frac{3}{5}}=\frac{16}{25}\ast \frac{\frac{11}{5}}{\frac{8}{5}}=\frac{22}{25}$

Therefore, Answer is $\frac{22}{25}$