Question

$$
\text{Let } \cos(\alpha+\beta)=\frac45
\text{ and }
\sin(\alpha-\beta)=\frac5{13},
\quad
0\leq\alpha,\beta\leq\frac{\pi}{4}.
$$

$$
\text{Then } \tan 2\alpha =
$$

(A) \( \frac{25}{16} \)

(B) \( \frac{56}{33} \)

(C) \( \frac{19}{12} \)

(D) \( \frac{20}{7} \)

---

Solution

$$
\cos(\alpha+\beta)=\frac45
\Rightarrow
\sin(\alpha+\beta)=\frac35
$$

$$
\sin(\alpha-\beta)=\frac5{13}
\Rightarrow
\cos(\alpha-\beta)=\frac{12}{13}
$$

Using:

$$
\tan 2\alpha =
\frac{
\sin(\alpha+\beta)\cos(\alpha-\beta)
+
\cos(\alpha+\beta)\sin(\alpha-\beta)
}{
\cos(\alpha+\beta)\cos(\alpha-\beta)
–
\sin(\alpha+\beta)\sin(\alpha-\beta)
}
$$

Substituting values:

$$
\tan 2\alpha
=
\frac{
\left(\frac35\right)\left(\frac{12}{13}\right)
+
\left(\frac45\right)\left(\frac5{13}\right)
}{
\left(\frac45\right)\left(\frac{12}{13}\right)
–
\left(\frac35\right)\left(\frac5{13}\right)
}
$$

$$
=
\frac{\frac{36}{65}+\frac{20}{65}}
{\frac{48}{65}-\frac{15}{65}}
=
\frac{56}{33}
$$

Correct Answer: (B) \( \frac{56}{33} \)