Question

If $sinA=\frac{1}{2},cosB=\frac{\sqrt{3}}{2},$ where $\frac{\pi }{2}$ (i) tan(A+B)
(ii) tan(A-B)

Answer 1

We have,
$sinA=\frac{1}{2},cosB=\frac{\sqrt{3}}{2}$
$\therefore cosA=-\sqrt{1-si{n}^{2}A}$
$sinB=\sqrt{1-co{s}^{2}B}$
[$\because$ cosine is negetive in second quadrant]
$\Rightarrow cosA=-\sqrt{1-{\left(\frac{1}{2}\right)}^2}$
$sinB=\sqrt{1-{\left(\frac{\sqrt{3}}{2}\right)}^2}$
$\Rightarrow cosA=-\sqrt{1-\frac{1}{4}}$ and
$sinB=\sqrt{1-\frac{3}{4}}$
$\Rightarrow cosA=-\sqrt{\frac{3}{4}}$ and
$sinB=\sqrt{\frac{1}{4}}$
$\Rightarrow cosA=-\frac{\sqrt{3}}{2}\text{and}sinB=\frac{1}{2}$
$\therefore tanA=\frac{sinA}{cosA}=\frac{\frac{1}{2}}{\frac{-\sqrt{3}}{2}}=\frac{-1}{\sqrt{3}}$
Now,
$tan(A+B)=\frac{tanA+tanB}{1-tanA.tanB}$
$=\frac{\frac{-1}{\sqrt{3}}+\frac{1}{\sqrt{3}}}{1-\left(\frac{-1}{\sqrt{3}}\right)\times \left(\frac{1}{\sqrt{3}}\right)}$
=0
$\therefore$ tan(A+B)=0
(ii) We have,
$sinA=\frac{1}{2}$ and $cosB=\frac{\sqrt{3}}{2}$
$\therefore cosA=-sqrt1-si{n}^{2}A$ and
$sinB=\sqrt{1-co{s}^{2}B}$
[$\because$ cosine is negetive in second quadrant]
$\Rightarrow cosA=-\sqrt{1-{\left(\frac{1}{2}\right)}^2}$
$sinB=\sqrt{1-{\left(\frac{\sqrt{3}}{2}\right)}^2}$
$\Rightarrow cosA=-\sqrt{1-\frac{1}{4}}$ and
$sinB=\sqrt{1-\frac{3}{4}}$
$\Rightarrow cosA=-\sqrt{\frac{3}{4}}$ and
$sinB=\sqrt{\frac{1}{4}}$
$\Rightarrow cosA=-\frac{\sqrt{3}}{2}$ and
$sinB=\frac{1}{2}$
$\therefore tanA=\frac{sinA}{cosA}=\frac{\frac{1}{2}}{\frac{-\sqrt{3}}{2}}=\frac{-1}{\sqrt{3}}$
and, $tanB=\frac{sinB}{cosB}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}$
Now, $tan(A-B)=\frac{tanA-tanB}{1+tanA.tanB}$
$=\frac{\frac{-1}{\sqrt{3}}-\left(\frac{1}{\sqrt{3}}\right)}{1+\left(\frac{-1}{\sqrt{3}}\right)\times \frac{1}{\sqrt{3}}}$
$=\frac{\frac{-2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{\frac{-2}{\sqrt{3}}}{\frac{3-1}{3}}$
$=\frac{\frac{-2}{\sqrt{3}}}{\frac{2}{3}}$ = $\frac{-3}{\sqrt{3}}$
$=\frac{-\sqrt{3}\times \sqrt{3}}{\sqrt{3}}=-\sqrt{3}$
$\therefore tan(A-B)=-\sqrt{3}$