Prove that, if \(\tan\theta\tan(\theta+\alpha)=k\), then \[(k+1)\cos(2\theta+\alpha)=(1-k)\cos\alpha.\]

We’ll use the identities \(\tan x= \dfrac{\sin x}{\cos x}, 2\sin A\sin B = \cos(A-B) - \cos(A+B), 2\cos A\cos B = \cos(A-B) + \cos(A+B).\)

We have

\[k = \tan \theta \tan(\theta+\alpha)\]

\[=\dfrac{2\sin (\theta+ \alpha)\sin \theta}{2\cos (\theta+\alpha) \cos \theta}.\]

We want an expression with arguments \((2\theta+\alpha)\) and \(\alpha\), and we can note that these are the sum and difference of the arguments we have here.

Recalling the identities above, we get

\[k= \dfrac{\cos \alpha - \cos (2\theta+\alpha)}{\cos \alpha + \cos (2\theta+\alpha)},\]

which we can then multiply out and rearrange to get \[(k+1)\cos(2\theta+\alpha)=(1-k)\cos\alpha.\]

Solve the equation \[\tan\theta\tan(\theta+\pi/3)=2.\]

Even if we can’t do the first part of this question, we can still attempt this part using what we were told in the first part.

We use the first part of the question, with \(k=2\) and \(\alpha=\pi/3\).

Remember that \(\pi/3\) radians is equal to \(60^\circ\).

So what we now solve is the much simpler problem \[3\cos(2\theta+\pi/3)=-\cos(\pi/3),\] that is, \[\cos(2\theta+\pi/3)=-\dfrac{1}{6}.\] So we have (where \(n\) is an integer) \[2\theta + \dfrac{\pi}{3} = \arccos\left(-\dfrac{1}{6}\right)+2n\pi\]


\[2\theta + \dfrac{\pi}{3} =-\arccos\left(-\dfrac{1}{6}\right) + 2n\pi.\]

So the general solution is

\[\theta=\pm \dfrac{1}{2}\arccos\left(-\dfrac{1}{6}\right)-\dfrac{\pi}{6}+n\pi,\]


\[\theta = 0.3455...+n \pi, -1.393 + n\pi.\]

Discuss the equation \[\tan\theta\tan(\theta+\alpha)+1=0.\]

This time we put \(k=-1\). We find that, from what we proved in the first part, \[0=2\cos\alpha,\] which tells us that there can only exist a solution if \(\alpha\) satisfies \(\cos\alpha=0\), or \(\alpha = \dfrac{\pi}{2} + n\pi\).

Moreover, when \(\cos\alpha = 0\), the equation is satisfied for ANY \(\theta\).

You could plot the curve \(y=\tan x \tan(x+a)\) in Desmos to verify this…