If `sin(A-B)=1/(sqrt(10 ,))cos(A+B)=2/(sqrt(29)),fin dt h ev a l u eof2Aw h e r eAa n dBl i eb e t w e e n0a n dpi/4dot`

Explanation

`tan 2A=tan[(A+B)+(A-B)]`
`=(tan(A+B)+tan(A-B))/(1-tan(A+B)tan (A-B))`
Given that, `0 ltAltpi//4` and `0ltBltpi//4`. Therefore,
`0ltA+Blt(pi)/(2)`
Also, `-(pi)/(4)ltA-Blt(pi)/(4)`
and `sin(A-B)=(1)/sqrt(10)`
`therefore 0ltA-Blt(pi)/(4)`
Now, `sin(A-B)=(1)/sqrt(10)`
`rArr tan (A-B)=(1)/(3)`
`cos (A+B)=(2)/sqrt(29)`
`rArr tan (A+B)=(5)/(2)`
From Eqs. (i),(ii) and (iii), we get
`tan 2A=((5)/(2)+(1)/(3))/(1-(5)/(2)xx(1)/(3))=(17)/(6)xx(6)/(1)=17`