Question

If $0\le \alpha ,\beta \le 90$ and $\tan (\alpha +\beta )=3$ and $\tan (\alpha -\beta )=2$ then value of $\sin 2\alpha$ is ?

Answer 1

Here,$0\le \alpha ,\beta \le 90\circ ⟹1stQuadrant$

Let $A=\alpha +\beta andB=\alpha -\beta$

$A+B=2\alpha$

Using $tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}$

$tan\left(2\alpha \right)=tan\left[\right(\alpha +\beta )+(\alpha -\beta \left)\right]$

$tan\left(2\alpha \right)=\frac{tan(\alpha +\beta )+tan(\alpha -\beta )}{1-tan(\alpha +\beta )tan(\alpha -\beta )}$

$tan\left(2\alpha \right)=\frac{3+2}{1-3\cdot 2}=\frac{5}{-5}$

$tan\left(2\alpha \right)=-1<0$

Now $,0\le \alpha \le 90$

$0\le 2\alpha \le 180⟹1stor2ndQuadrant$

$tan\left(2\alpha \right)=-1<0⟹2ndQuadrant$

$2\alpha =\pi -\frac{\pi }{4}$

$sin\left(2\alpha \right)=sin(\pi -\frac{\pi }{4})=sin\frac{\pi }{4}$

$sin\left(2\alpha \right)=\frac{1}{\sqrt{2}}$