Hi Guys !
Am Currently stuck with this question :
If sin(x+y) = 1 and if tanx = 9tany , find the value of tanx.
Need help guys thanks !
Hello, liukawa!
$\displaystyle \text{If }\sin(x+y) = 1\,\text{ and }\,\tan x \,=\, 9\tan y,\,\text{ find the value of }\tan x.$
I tried all kinds of gymnastics and got nowhere.
Then my two brain cells got together . . .
Since $\displaystyle \sin(x+y) \,=\,1$, then: .$\displaystyle x + y \,=\,\frac{\pi}{2}$
. . That is: .$\displaystyle y \:=\:\frac{\pi}{2} - x $
Then we have:
. . $\displaystyle \tan x \:=\:9\tan y \quad\Rightarrow\quad \tan x \:=\:9\tan(\frac{\pi}{2} - x) \quad\Rightarrow\quad \tan x \:=\:9\cot x $
. . . . I hope you followed that.
Multiply both sides by $\displaystyle \tan x\!:\;\;\tan^2\!x \:=\:9$
Therefore: .$\displaystyle \tan x \:=\:\pm3$
Alternatively
$\displaystyle sin(x+y)=1\Rightarrow\ sin^2(x+y)+cos^2(x+y)=1\Rightarrow\ cos(x+y)=0$
$\displaystyle \displaystyle\ tan(x+y)=\frac{sin(x+y)}{cos(x+y)}$
$\displaystyle \displaystyle\ tan(x+y)=\frac{tanx+tany}{1-tanxtany}\Rightarrow\ tanxtany=1\Rightarrow\ tanx=\frac{1}{tany}$
$\displaystyle \displaystyle\ 9tany=\frac{1}{tany}\Rightarrow\ 9tan^2y=1\Rightarrow\ tan^2y=\frac{1}{9}\Rightarrow\ tany=\pm\frac{1}{3}$
$\displaystyle \Rightarrow\ tanx=9tany=\pm3$