Question

If $tan(A+B)=\sqrt{3}$ and $tan(A-B)=\frac{1}{\sqrt{3}}$, where $0<A+B<{90}^{\circ },A>B$, find A and B. Also calculate $tanA.sin(A+B)+cosA.tan(A-B)$

Answer 1

Given $tan(A+B)=\sqrt{3},tan(A-B)=\frac{1}{\sqrt{3}}$

$\Rightarrow tan(A+B)=\sqrt{3}$

$(A+B)={60}^{\circ }\dots \left(i\right)$

And, $tan(A-B)=\frac{1}{\sqrt{3}}$

$(A-B)={30}^{\circ }\dots \left(ii\right)$

On adding eq. (i) & (ii)

$A+B={60}^{\circ }\underset{–}{A-B={30}^{\circ }}\underset{–}{2A={90}^{\circ }}$

[By adding]

$\Rightarrow A=\frac{{90}^{\circ }}{2}={45}^{\circ }$

From eq. (i), $A+B={60}^{\circ }$

${45}^{\circ }+B={60}^{\circ }$

$B={15}^{\circ }$

$\therefore A={45}^{\circ },B={15}^{\circ }$

Now, $tanA,sin(A+B)+cosA.tan(A-B)$

$=tan{45}^{\circ }.sin\left({60}^{\circ }\right)+cos{45}^{\circ }.tan\left({30}^{\circ }\right)$

$=2\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{3}}$

$=\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{3}}{2}+\frac{\sqrt{6}}{6}$

$=\frac{3\sqrt{3}+\sqrt{6}}{6}$