# Math Help - Solving of Trigonometry equation

1. ## Solving of Trigonometry equation

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 !

2. have you played with the fact that $\sin (x+y) = \sin x \cos y + \sin y \cos x$ ?

3. What if you write $\tan{x}-9\tan{y}=0$

Does this help?

Also, think about what the range of tan(x)?

4. Hello, liukawa!

$\text{If }\sin(x+y) = 1\,\text{ and }\,\tan x \,=\, 9\tan y,\,\text{ find the value of }\tan x.$

I tried all kinds of gymnastics and got nowhere.
Then my two brain cells got together . . .

Since $\sin(x+y) \,=\,1$, then: . $x + y \,=\,\frac{\pi}{2}$
. . That is: . $y \:=\:\frac{\pi}{2} - x$

Then we have:
. . $\tan x \:=\:9\tan y \quad\Rightarrow\quad \tan x \:=\:9\tan(\frac{\pi}{2} - x) \quad\Rightarrow\quad \tan x \:=\:9\cot x$
. . . .
I hope you followed that.

Multiply both sides by $\tan x\!:\;\;\tan^2\!x \:=\:9$

Therefore: . $\tan x \:=\:\pm3$

5. Originally Posted by liukawa
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 !
Alternatively

$sin(x+y)=1\Rightarrow\ sin^2(x+y)+cos^2(x+y)=1\Rightarrow\ cos(x+y)=0$

$\displaystyle\ tan(x+y)=\frac{sin(x+y)}{cos(x+y)}$

$\displaystyle\ tan(x+y)=\frac{tanx+tany}{1-tanxtany}\Rightarrow\ tanxtany=1\Rightarrow\ tanx=\frac{1}{tany}$

$\displaystyle\ 9tany=\frac{1}{tany}\Rightarrow\ 9tan^2y=1\Rightarrow\ tan^2y=\frac{1}{9}\Rightarrow\ tany=\pm\frac{1}{3}$

$\Rightarrow\ tanx=9tany=\pm3$