If $\sin \theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$, then $\sin 2\theta$ is a root of $x^2 -44x +36=0$ My own bonafide attempt.

$$\sin(\theta)+\cos(\theta)+\tan(\theta)+\cot(\theta)+\sec(\theta)+\csc(\theta)=7$$ $$\sin(\theta)+\cos(\theta)+\frac{\sin(\theta)}{\cos(\theta)}+\frac{\cos(\theta)}{\sin(\theta)}+\frac{1}{\cos(\theta)}+\frac{1}{\sin(\theta)}=7$$ $$\sin^2\theta\cos\theta+\sin\theta\cos^2\theta+\sin^2\theta+\cos^2\theta+\sin\theta+\cos\theta=7\sin\theta\cos\theta$$ Let $\sin\theta+\cos\theta=u; \sin\theta\cos\theta=v$ $$uv+1+u=7v$$ $$u(1+v)=7v-1$$ $$u=\frac{7v-1}{v+1}$$ $$u^2=\left(\frac{7v-1}{v+1}\right)^2$$ $u^2=(\sin\theta+\cos\theta)^2=1+2\sin\theta\cos\theta=1+2v$ $$1+2v=\left(\frac{7v-1}{v+1}\right)^2$$ where $v=\sin\theta\cos\theta=\frac12 \sin2\theta$

Let $\sin2\theta=x$. Then $$1+x=\left(\frac{\frac72x-1}{\frac12x+1}\right)^2$$ $$1+x=\left(\frac{7x-2}{x+2}\right)^2$$ $$\color{red}{x^2-44x+36=0}$$

With an obvious short notation and setting $x=2cs$, we reduce to the common denominator

$$s+c+\frac sc+\frac cs+\frac 1s+\frac 1c=\frac{sx+cx+2+2c+2s}{x}=\frac{(x+2)(c+s)+2}x=7.$$

As we know that $x\ne0$ and with $(c+s)^2=x+1$, we rewrite

$$(x+2)^2(x+1)=(7x-2)^2,$$

$$x^3-44x^2+36x=0.$$