# Thread: Find θ for given condition?(seems to be easy but I can't seem to proceed)?

1. ## Find θ for given condition?(seems to be easy but I can't seem to proceed)?

(sin θ + cos θ - 1)/(2 sin θ cos θ) = √(6) - 2

Find θ ?

2. I got a very messy arctan.

3. ## its 15

try to put it yaar

4. Hi fardeen_gen,

Here is my very blunt approach:

Let's define $\displaystyle \sin(\theta)=s~,~\cos(\theta)=c$

We have $\displaystyle \frac{s+c-1}{2sc}=\sqrt{6}-2$ Call this equation $\displaystyle (*)$ Note that $\displaystyle 2sc=\sin(2\theta)~\forall \theta \in \mathbb{Z}$ and thus $\displaystyle \frac{s+c-1}{\sin(2\theta)}=\sqrt{6}-2$ Now we can square both sides of the equation and do a bit of rearranging;

$\displaystyle \frac{(s+c-1)^2}{\sin^2(2\theta)}=(\sqrt{6}-2)^2$

$\displaystyle \implies \frac{(s^2+c^2)+2sc-2(s+c)+1}{\sin^2(2\theta)}=6-4\sqrt{6}+4$

$\displaystyle \implies \frac{2+\sin(2\theta)-2(s+c)}{\sin^2(2\theta)}=10-4\sqrt{6}$ Call this equation $\displaystyle (**)$

Note that from equation $\displaystyle (*)$ We have;

$\displaystyle s+c=\sin(2\theta)(\sqrt{6}-2)+1$

Now subbing this into equation $\displaystyle (**)$ gives;

$\displaystyle \frac{2+\sin(2\theta)-2[\sin(2\theta)(\sqrt{6}-2)+1]}{\sin^2(2\theta)}=10-4\sqrt{6}$

$\displaystyle \implies \frac{2+\sin(2\theta)-2\sin(2\theta)(\sqrt{6}-2)-2}{\sin^2(2\theta)}=10-4\sqrt{6}$

$\displaystyle \implies \frac{1-2(\sqrt{6}-2)}{\sin(2\theta)}=10-4\sqrt{6}$

$\displaystyle \implies \sin(2\theta)=\frac{5-2\sqrt{6}}{10-4\sqrt{6}}=\frac{5-2\sqrt{6}}{10-4\sqrt{6}}\cdot\frac{10+4\sqrt{6}}{10+4\sqrt{6}}$

$\displaystyle \implies \sin(2\theta)=\frac{1}{2}$

Maybe there is an easier 'technique'