(a) If $sin mid A= rac{12}{13}$ and $sin B= rac{4}{5}$, where $ rac{pi}{2}<A<pi$ and $0<B< rac{pi}{2}$, find the following:

Solution:

(a) Given :

$\sin A=\frac{12}{13}$ and $\sin B=\frac{4}{5}$

When, $\frac{\pi}{2}

$\cos A=-\sqrt{1-\sin ^{2} A}$ and $\cos B=\sqrt{1-\sin ^{2} B}$

( As cosine function is negative in second qudrant and positive in first quadrant)

$\Rightarrow \cos A=-\sqrt{1-\left(\frac{12}{13}\right)^{2}}$ and $\cos B=\sqrt{1-\left(\frac{4}{5}\right)^{2}}$

$\Rightarrow \cos A=-\sqrt{1-\frac{144}{169}}$ and $\cos B=\sqrt{1-\frac{16}{25}}$

$\Rightarrow \cos A=-\sqrt{\frac{25}{169}}$ and $\cos B=\sqrt{\frac{9}{25}}$

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

Now,

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

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

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

$=\frac{16}{65}$

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

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

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

$=\frac{-63}{65}$

(b) Given:

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

and that $A$ and $B$ both lie in second qudrant.

We know that in second quadrant $\sin e$ function is positive and cosine function is negative.

Therefore,

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

$\Rightarrow \cos A=-\sqrt{1-\left(\frac{3}{5}\right)^{2}} \quad$ and $\quad \sin B=\sqrt{1-\left(\frac{-12}{13}\right)^{2}}$

$\Rightarrow \cos A=-\sqrt{1-\frac{9}{25}} \quad$ and $\quad \sin B=\sqrt{1-\frac{144}{169}}$

$\Rightarrow \cos A=-\sqrt{\frac{16}{25}} \quad$ and $\quad \sin B=\sqrt{\frac{25}{69}}$

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

Now,

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

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

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

$=\frac{-56}{65}$