Question

If $\sin (A-B)=\frac{1}{\sqrt{10}},\cos (A+B)=\frac{2}{\sqrt{29}}$, fid the value of $\tan 2A$ where A and B lies between 0 and $\pi /4$

Answer 1

$\tan 2A=\tan [(A+B)+(A-B)]=\frac{\tan (A+B)+\tan (A-B)}{1-\tan (A+B)\tan (A-B)}$ (i)
Given that, $0<A<\pi /4$ and $0<B<\pi /4$. Therefor,
$0<A+B<\frac{\pi }{2}$
Also, $-\frac{\pi }{4}<A-B<\frac{\pi }{4}$ and $\sin (A-B)=\frac{1}{\sqrt{10}}=$(+)ve
$\therefore$ $0<A-B<\frac{\pi }{4}$
Now, $\sin (A-B)=\frac{1}{\sqrt{10}}$
$\Rightarrow$ $\tan (A-B)=\frac{1}{3}$ (ii)
$\cos (A+B)=\frac{2}{\sqrt{29}}$
$\Rightarrow$ $\tan (A-B)=\frac{5}{2}$ (iii)
From Eqs. (i), (ii) and (iii), we get
$\tan 2A=\frac{\frac{5}{2}+\frac{1}{3}}{1-\frac{5}{2}\times \frac{1}{3}}=\frac{17}{6}\times \frac{6}{1}=17$
Ans: 17