Question

If sin (A + B) = 1 and $\tan \left(A-B\right)=\frac{1}{\sqrt{3}},0°<\left(A+B\right)\le 90°\mathrm{and}A>B$then find the values of A and B.

Answer 1

As we know that,
$$\begin{gathered} \sin 90°=1 \\ \mathrm{Thus}, \\ \mathrm{if}\sin \left(A+B\right)=1 \\ \Rightarrow A+B=90°...\left(1\right) \\ \mathrm{and}\tan 30°=\frac{1}{\sqrt{3}} \\ \mathrm{Thus}, \\ \mathrm{if}\tan \left(A-B\right)=\frac{1}{\sqrt{3}} \\ \Rightarrow A-B=30°...\left(2\right) \\ \mathrm{Solving}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ A=60°\mathrm{and}B=30° \\ \mathrm{Hence},\mathrm{the}\mathrm{values}\mathrm{of}A\mathrm{and}B\mathrm{are}60°\mathrm{and}30°,\mathrm{respectively}. \end{gathered}$$