If n. `sin(A+2B)=sinA`, then prove that: `tan(A+B)=(1+n)/(1-n).tanB`

1.	If n. `sin(A+2B)=sinA`, then prove that: `tan(A+B)=(1+n)/(1-n).tanB`
Answer» `nsin(A+2B)=sinA` `rArr n=(sinA)/(sin(A+2B))` `therefore` RHS `=(1+n)/(1-n).tanB=(1+(sinA)/(sin(A+2B)))/(1-(sinA)/(sin(A+2B))).tanB` `(sin(A+2B)+sinA)/(sin(A+2B)-sinA).tanB` `=(2sinA(A+2B+A)/(2)cos(A+2B-A)/(2))/(2cos(A+2B+A)/(2)sin(A+2B-A)/(2)).tanB` `=(sin(A+B)cosBsinB)/(cos(A+B)sinBcosB)` `=tan(A+B)`=LHS Hence Proved.

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