Question

if tan (a+b)=√3 and tan (a-b)=1/√3
Find A and B.

Please fast !

Answer 1

$Given tan(A+B)=3⟹A+B=60∘⋯(1) </p><p></p><p> tan(A−B)=31⟹A−B=3(2) </p><p></p><p>(1)+(2)⟹2A=90∘⟹A=45∘</p><p></p><p>Put A value in (1)</p><p></p><p>⟹B=60∘−45∘=15∘</p><p></p><p>$

Answer 2

Given :-

• tan (a+b)=√3 and tan (a-b)=1/√3

Need to find :-

• Value of A and B ?

Answer :-

• A = 45° and B = 15°

Step - By - Step - Explanation :-

• Refer in the attachment !

More Information :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}$