If sin

Solution:

Given :

$\sin A=\frac{4}{5}$ and $\cos B=\frac{5}{13}$

We know that

$\cos A=\sqrt{1-\sin ^{2} A} \quad$ and $\quad \sin B=\sqrt{1-\cos ^{2} B} \quad, \quad$ where $0

$\Rightarrow \cos A=\sqrt{1-\left(\frac{4}{5}\right)^{2}} \quad$ and $\quad \sin B=\sqrt{1-\left(\frac{5}{13}\right)^{2}}$

$\Rightarrow \cos A=\sqrt{1-\frac{16}{25}} \quad$ and $\quad \sin B=\sqrt{1-\frac{25}{169}}$

$\Rightarrow \cos A=\sqrt{\frac{9}{25}} \quad$ and $\sin B=\sqrt{\frac{144}{169}}$

$\Rightarrow \quad \cos A=\frac{3}{5} \quad$ and $\quad \sin B=\frac{12}{13}$

Now,

(i) $\sin (A+B)=\sin A \cos B+\cos A \sin B$

$=\frac{4}{5} \times \frac{5}{13}+\frac{3}{5} \times \frac{12}{13}$

$=\frac{20}{65}+\frac{36}{65}$

$=\frac{56}{65}$

(ii) $\cos (A+B)=\cos A \cos B-\sin A \sin B$

$=\frac{3}{5} \times \frac{5}{13}-\frac{4}{5} \times \frac{12}{13}$

$=\frac{15}{65}-\frac{48}{55}$

$=\frac{-33}{65}$

(iii) $\sin (A-B)=\sin A \cos B-\cos A \sin B$

$=\frac{4}{5} \times \frac{5}{13}-\frac{3}{5} \times \frac{12}{13}$

$=\frac{20}{65}-\frac{36}{65}$

$=\frac{-16}{65}$

(iv) $\cos (A-B)=\cos A \cos B+\sin A \sin B$

$=\frac{3}{5} \times \frac{5}{13}+\frac{4}{5} \times \frac{12}{13}$

$=\frac{15}{65}+\frac{48}{65}$

$=\frac{63}{65}$