Question
If $2 \sin A=1=\sqrt{2} \cos B$ and $\frac{\pi}{2}
✓
Answer
Given, $2 \sin \mathrm{A}=1$
$\therefore \sin A=1 / 2$
we know that,
$ \cos ^2 \mathrm{~A}=1-\sin ^2 \mathrm{~A}=1-\left(\frac{1}{2}\right)^2=1-\frac{1}{4}=\frac{3}{4}$
$\therefore \cos \mathrm{A}= \pm \frac{\sqrt{3}}{2}$
$\text { Since } \frac{\pi}{2}<\mathrm{A}<\pi $
A lies in the 2 nd quadrant.
$\begin{aligned}
\therefore \quad \cos \mathrm{A} & = \pm \frac{\sqrt{3}}{2} \\
& \text { Since } \frac{\pi}{2}<\mathrm{A}<\pi,
\end{aligned}$
A lies in the $2^{\text {nd }}$ quadrant.
$\begin{aligned}
\therefore \quad \cos A & =-\frac{\sqrt{3}}{2} \\
\tan A & =\frac{\sin A}{\cos A}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}
\end{aligned}$
Also, $\sqrt{2} \cos \mathrm{B}=1$
$\therefore \quad \cos \mathrm{B}=\frac{1}{\sqrt{2}}$
We know that,
$\sin ^2 B=1-\cos ^2 B=1-\left(\frac{1}{\sqrt{2}}\right)^2 \frac{1}{2}=\frac{1}{2}$
$\therefore \sin B= \pm \frac{1}{\sqrt{2}}$
Since $\frac{3 \pi}{2}$B lies in the 4th quadrant,
$ \therefore \quad \sin \mathrm{B}=-\frac{1}{\sqrt{2}}$
$\tan B=\frac{\sin B}{\cos B}=\frac{-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=-1$
$\therefore \quad \frac{\tan A+\tan B}{\cos A-\cos B}=\frac{-\frac{1}{\sqrt{3}}-1}{-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}}=\frac{\frac{-1-\sqrt{3}}{\sqrt{3}}}{\frac{-\sqrt{3}-\sqrt{2}}{2}}$
$=\frac{-(1+\sqrt{3})}{\sqrt{3}} \times \frac{2}{-(\sqrt{3}+\sqrt{2})}$
$=\frac{2(1+\sqrt{3})}{\sqrt{3}(\sqrt{3}+\sqrt{2})}$
$\therefore \sin A=1 / 2$
we know that,
$ \cos ^2 \mathrm{~A}=1-\sin ^2 \mathrm{~A}=1-\left(\frac{1}{2}\right)^2=1-\frac{1}{4}=\frac{3}{4}$
$\therefore \cos \mathrm{A}= \pm \frac{\sqrt{3}}{2}$
$\text { Since } \frac{\pi}{2}<\mathrm{A}<\pi $
A lies in the 2 nd quadrant.
$\begin{aligned}
\therefore \quad \cos \mathrm{A} & = \pm \frac{\sqrt{3}}{2} \\
& \text { Since } \frac{\pi}{2}<\mathrm{A}<\pi,
\end{aligned}$
A lies in the $2^{\text {nd }}$ quadrant.
$\begin{aligned}
\therefore \quad \cos A & =-\frac{\sqrt{3}}{2} \\
\tan A & =\frac{\sin A}{\cos A}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}
\end{aligned}$
Also, $\sqrt{2} \cos \mathrm{B}=1$
$\therefore \quad \cos \mathrm{B}=\frac{1}{\sqrt{2}}$
We know that,
$\sin ^2 B=1-\cos ^2 B=1-\left(\frac{1}{\sqrt{2}}\right)^2 \frac{1}{2}=\frac{1}{2}$
$\therefore \sin B= \pm \frac{1}{\sqrt{2}}$
Since $\frac{3 \pi}{2}$B lies in the 4th quadrant,
$ \therefore \quad \sin \mathrm{B}=-\frac{1}{\sqrt{2}}$
$\tan B=\frac{\sin B}{\cos B}=\frac{-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=-1$
$\therefore \quad \frac{\tan A+\tan B}{\cos A-\cos B}=\frac{-\frac{1}{\sqrt{3}}-1}{-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}}=\frac{\frac{-1-\sqrt{3}}{\sqrt{3}}}{\frac{-\sqrt{3}-\sqrt{2}}{2}}$
$=\frac{-(1+\sqrt{3})}{\sqrt{3}} \times \frac{2}{-(\sqrt{3}+\sqrt{2})}$
$=\frac{2(1+\sqrt{3})}{\sqrt{3}(\sqrt{3}+\sqrt{2})}$
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