Find the value of "k"

Question

Let

$$\frac{3 \pi}{4}<\theta<\pi$$ and $$\sqrt{2 \cot \theta+\frac{1}{\sin ^{2} \theta}}=k-\cot \theta,$$ then $k=?$

I have been able to reduce this to $$2\cos{\theta}=\sin{\theta}*(k-1)$$

How to proceed further?

Answer 1

Let $c:=\cot\theta$ so your surd is $\sqrt{1+2c+c^2}$. Hence $k=-1$.

Answer 2

Choose $x=\dfrac{5\pi}6$. Then

$$\sqrt{-2\sqrt3+4}=k+\sqrt3.$$

By unnesting,

$$\sqrt{-2\sqrt3+4}=\sqrt3-1$$ and $k=-1$.

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Note that this also works with $x=\dfrac{3\pi}4$, giving

$$0=k+1.$$

Answer 3

Using the identity $\frac{1}{\sin^2\theta} = 1+\cot^2\theta$ you have:

$$k = \cot\theta +\sqrt{2\cot\theta + \frac{1}{\sin^2\theta}} = $$ $$=\cot\theta +\sqrt{2\cot\theta + 1+\cot^2\theta} = \cot\theta + |1+\cot\theta| = \cot\theta -(1+\cot\theta) = -1$$

because $1+\cot\theta < 0$ in $(3/4\pi,\pi)$. Your mistake was something along $|1+\cot\theta| = 1+\cot\theta$.