Solve the equations $\tan^2{3θ} = \cot^2{\alpha}$

Question

Solve the equations

$\tan^2{3θ} = \cot^2{\alpha}$

This is in Plane Trigonometry by SL Loney. Please explain how I can solve it according to class 11th student. My working is below:

$\implies \tan^2{3\theta} = \tan^2{(\frac{\pi}2 - \alpha)}\\\implies ({3\theta})^2 = (\pi +\frac{\pi}2 - \alpha)^2\\\implies \theta^2 = (\frac\pi3(n+\frac12) - \frac\alpha3)^2$

I'm unable to understand what should I do next.

The answer given is

$(n+\frac12)\frac\pi3\pm\frac\alpha3$

Answer 1

Your step $\tan^2 A = \tan^2 B \implies A^2=B^2$ is incorrect. Instead, you can do the following: \begin{align*} \tan^2 A & = \tan^2 B\\ (\tan A - \tan B)(\tan A + \tan B) & = 0. \end{align*}

So you have two possibilities $$\tan A = \tan B \qquad \text{ or } \qquad \tan A = \tan(-B).$$ Now proceed from here. For the first one we get $$\tan A = \tan B \implies A=n\pi+B.$$ For the second one we get $$\tan A = \tan (-B) \implies A=n\pi-B.$$

Answer 2

Hint: Your equation can factorized into $$-\frac{4 \sin \left(\frac{\pi }{4}-2 x\right) \sin \left(\frac{\pi }{4}-x\right) \sin \left(x+\frac{\pi }{4}\right) \sin \left(2 x+\frac{\pi }{4}\right) \csc ^2(x) \sec ^2(x)}{(2 \cos (2 x)-1)^2}=0$$

Answer 3

$$[f(x)]^2 \not= f(x^2)$$

So, $$\tan^2(3\theta) = \cot^2\alpha = \tan^2(\pi/2\pm\alpha)$$ $$\tan^2(3\theta)=\tan^2(n\pi+\pi/2\pm\alpha)$$ [As $\cot(\pi/2\pm\alpha) = \mp\tan\alpha\implies \cot^2(\pi/2\pm\alpha) = \tan^2\alpha$ and $\tan$ has periodicity of $n\pi$] $$3\theta = n\pi +\pi/2\pm \alpha$$

$$\theta = \bigg(n+\frac{1}{2}\bigg)\frac{\pi}{3}\pm\frac{\alpha}{3}$$

Answer 4

In general the following four conditions are equivalent (please prove it)

$\tan^2x=\tan^2A$

$\sin^2x=\sin^2A \ \ \ \ (2)$

$\cos^2x=\cos^2A$

$\cos2x=\cos2A \ \ \ \ (4)$

Use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,

$(2)\implies\sin(x+A)\sin(x-A)=0$

which can be arrived at using http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html on $(4)$

Can you finish it from here?