Question

If sin (A-B)= 1/2 ,tan (A+B)=√3 then prove that A=45°

Answer 1

Step-by-step explanation:

LHS

put A=45 then we get

sin (45 -B) =1/2

and

tan (45 - B )=√3

Answer 2

If sin (A-B)= 1/2 ,tan (A+B)=√3 then it is proved  that A=45°

Step-by-step explanation:

Given $\textrm{sin(A-B)}=\frac{1}{2}$

        $\Rightarrow \textrm {sin(A-B)}=\textrm {sin}30^{0}$

           $\Rightarrow \textrm{A-B}=30^{0}$

and it is given that $\textrm{tan(A+B)}=\sqrt{3}$

         $\Rightarrow \textrm {tan(A+B)}=\textrm {tan}60^{0}$

             $\Rightarrow \textrm{A+B}=60^{0}$

Now we have   $\textrm{A-B}=30^{0}$        ..............Equ(1)

         and $\textrm{A+B}=60^{0}$             ...................Equ(2)

Adding  both equation we get

          $\textrm{(A-B)}+\textrm{(A+B)}=30^{0}+60^{0}$

    $\Rightarrow \textrm{2A}=90^{0}$

    $\Rightarrow \textrm{A}=\frac{90^{0}}{2}$

    $\Rightarrow \textrm{A}=45^{0}$            

Hence proved